A 12.1.
Model: spinning skater, whose arms are outstretched, is a rigid rotating body.
Visualize:
Solve: The speed v = r , where r = 140 cm/2 = 0.70 m. Also, 180 rpm = (180)2 /60 rad/s = 6 rad/s. Thus, v = (0.70 m)(6 rad/s) = 13.2 m/s. Assess: A speed of 13.2 m/s 26 mph for the hands is a little high, but reasonable.
12.2.
Solve:
Model:
Assume constant angular acceleration.
2 rad min (a) The final angular velocity is f = ( 2000 rpm ) = 209.4 rad/s. The definition of rev 60 s angular acceleration gives us
=
f i 209.4 rad/s 0 rad/s = = = 419 rad/s t t 0.50 s
The angular acceleration of the drill is 4.2 102 rad/s. (b)
f = i + i t + ( t ) = 0 rad + 0 rad +
2
1 2
1 2 ( 419 rad/s )( 0.50 s ) = 52.4 rad 2
rev The drill makes (52.4 rad) = 8.3 revolutions. 2 rad
12.3.
Model: Visualize:
Assume constant angular acceleration.
Solve:
(a) Since at = r , find first. With 90 rpm = 9.43 rad/s and 60 rpm = 6.28 rad/s,
=
9.43 rad 6.28 rad/s = = 0.314 rad/s 2 t 10 s
The angular acceleration of the sprocket and pedal are the same. So
at = r = ( 0.18 m ) ( 0.314 rad/s 2 ) = 0.057 m/s2
(b) The length of chain that passes over the sprocket during this time is L = r . Find :
f = i + i t + ( t )
1 2
2
f i = = ( 6.28 rad/s )(10 s ) +
1 ( 0.314 rad/s2 ) (10 s )2 = 78.5 rad 2
The length of chain which has passed over the top of the sprocket is
L = (0.10 m)(78.5 rad) = 7.9 m
12.4.
Model: Visualize:
Assume constant angular acceleration.
2 rad min The initial angular velocity is i = ( 60 rpm ) = 2 rad/s. rev 60 s The angular acceleration is
Solve:
=
f i
t
=
0 rad/s 2 rad/s = 0.251 rad/s 2 25 s
The angular velocity of the fan blade after 10 s is
f = i + ( t t0 ) = 2 rad/s+ ( 0.251 rad/s 2 ) (10 s 0 s ) = 3.77 rad/s
The tangential speed of the tip of the fan blade is
vt = r = ( 0.40 m ) ( 3.77 rad/s ) = 1.51 m/s
(b)
1 2 2 The fan turns 78.6 radians = 12.5 revolutions while coming to a stop.
f = i + i t + ( t ) = 0 rad + ( 2 rad/s )( 25 s ) +
1 ( 0.251 rad/s 2 ) ( 25 s )2 = 78.6 rad 2
12.5.
Model: Visualize:
The earth and moon are particles.
Choosing xE = 0 m sets the coordinate origin at the center of the earth so that the center of mass location is the distance from the center of the earth. Solve: 24 22 8 m x + mM xM ( 5.98 10 kg ) ( 0 m ) + ( 7.36 10 kg ) ( 3.84 10 m ) xcm = E E = mE + mM 5.98 1024 kg + 7.36 10 22 kg
= 4.67 106 m Assess: The center of mass of the earth-moon system is called the barycenter, and is located beneath the surface of the earth. Even though xE = 0 m the earth influences the center of mass location because mE is in the denominator of the expression for xcm .
12.6. Visualize: Please refer to Figure EX12.6. The coordinates of the three masses mA , mB , and mC are
(0 cm, 0 cm), (0 cm, 10 cm), and (10 cm, 0 cm), respectively. Solve: The coordinates of the center of mass are
xcm = ycm =
mA xA + mB xB + mC xC (100 g)(0 cm) + (200 g)(0 cm) + (300 g)(10 cm) = = 5.0 cm mA + mB + mC (100 g + 200 g + 300 g) mA yA + mB yB + mC yC (100 g)(0 cm) + (200 g)(10 cm) + (300 g)(0 cm) = = 3.3 cm mA + mB + mC (100 g + 200 g + 300 g)
12.7. Visualize: Please refer to Figure EX12.7. The coordinates of the three masses mA , mB , and mC are
(0 cm, 10 cm), (10 cm, 10 cm), and (10 cm, 0 cm), respectively. Solve: The coordinates of the center of mass are
xcm = ycm =
mA xA + mB xB + mC xC (200 g)(0 cm) + (300 g)(10 cm) + (100 g)(10 cm) = = 6.7 cm mA + mB + mC (200 g + 300 g + 100 g) mA yA + mB yB + mC yC (200 g)(10 cm) + (300 g)(10 cm) + (100 g)(0 cm) = = 8.3 cm mA + mB + mC (200 g + 300 g + 100 g)
12.8.
Model: Visualize:
The balls are particles located at the balls respective centers.
Solve:
The center of mass of the two balls measured from the left hand ball is
xcm =
(100 g ) ( 0 cm ) + ( 200 g ) ( 30 cm ) = 20 cm
100 g + 200 g
The linear speed of the 100 g ball is
2 rad min v1 = r = xcm = ( 0.20 m )(120 rev/min ) = 2.5 m/s rev 60 s
12.9.
Solve:
Model: The earth is a rigid, spherical rotating body.
The rotational kinetic energy of the earth is K rot = 1 I 2 . The moment of inertia of a sphere about its 2
2 rad = 7.27 10 5 rad/s 24 3600 s
2 diameter (see Table 12.2) is I = 5 M earth R 2 and the angular velocity of the earth is
=
Thus, the rotational kinetic energy is
1 2 K rot = M earth R 2 2 2 5 1 = (5.98 1024 kg)(6.37 106 m) 2 (7.27 105 rad/s) 2 = 2.57 1029 J 5
12.10. Model: The triangle is a rigid body rotating about an axis through the center. Visualize: Please refer to Figure EX12.10. Each 200 g mass is a distance r away from the axis of rotation, where r is given by
0.20 m 0.20 m = cos30 r = = 0.2309 m r cos30
Solve: The moment of inertia of the triangle is I = 3 mr 2 = 3(0.200 kg)(0.2309 m) 2 = 0.0320 kg m 2 . The frequency of rotation is given as 5.0 revolutions per s or 10 rad/s. The rotational kinetic energy is
K rot. =
121 I = (0.0320 kg m 2 )(10.0 rad/s) 2 = 15.8 J 2 2
12.11.
Model: The disk is a rigid body rotating about an axis through its center.
Visualize:
Solve:
The speed of the point on the rim is given by vrim = R . The angular velocity of the disk can be
determined from its rotational kinetic energy which is K = 1 I 2 = 0.15 J. The moment of inertia I of the disk 2 about its center and perpendicular to the plane of the disk is given by
1 1 I = MR 2 = (0.10 kg)(0.040 m) 2 = 8.0 105 kg m 2 2 2 2(0.15 J) 0.30 J 2 = = = 61.237 rad/s I 8.0 105 kg m 2 Now, we can go back to the first equation to find vrim . We get vrim = R = (0.040 m)(61.237 rad/s) = 2.4 m/s.
12.12.
The baton is a thin rod rotating about a perpendicular axis through its center of mass. 1 Solve: The moment of inertia of a thin rod rotating about its center is I = ML2 . For the baton, 12 1 2 I = ( 0.400 kg )( 0.96 m ) = 0.031 kg m 2 12 The rotational kinetic energy of the baton is
K rot = 121 2 rad min I = ( 0.031 kg m 2 ) (100 rev/min ) = 1.68 J 2 2 rev 60 s
2
Model:
12.13.
Model: The structure is a rigid body rotating about its center of mass.
Visualize:
We placed the origin of the coordinate system on the 300 g ball. Solve: First, we calculate the center of mass:
xcm =
(300 g)(0 cm) + (600 g)(40 cm) = 26.67 cm 300 g + 600 g
Next, we will calculate the moment of inertia about the structures center of mass:
I = (300 g)( xcm ) 2 + (600 g)(40 cm xcm ) 2 = (0.300 kg)(0.2667 m) 2 + (0.600 kg)(0.1333 m) 2 = 0.032 kg m 2 Finally, we calculate the rotational kinetic energy: 2 1 1 100 2 K rot = I 2 = (0.032 kg m 2 ) rad/s = 1.75 J 2 2 60
12.14. Model: The moment of inertia of any object depends on the axis of rotation. In the present case, the rotation axis passes through mass A and is perpendicular to the page. Visualize: Please refer to Figure EX12.14. mi xi = mA xA + mB xB + mC xC + mD xD Solve: (a) xcm = mA + mB + mC + mD mi
= ycm (100 g)(0 m) + (200 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 m) = 0.057 m 100 g + 200 g + 200 g + 200 g m y + mB yB + mC yC + mD yD = AA mA + mB + mC + mD =
(100 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 cm) + (200 g )(0 m) = 0.057 m 700 g (b) The distance from the axis to mass C is 14.14 cm. The moment of inertia through A and perpendicular to the page is
2 2 2 I A = mi ri 2 = mA rA + mB rB + mC rC2 + mD rD i
= (0.100 kg)(0 m) 2 + (0.200 kg)(0.10 m) 2 + (0.200 kg)(0.1414 m) 2 + (0.200 kg)(0.10 m) 2 = 0.0080 kg m 2
12.15.
Model: The moment of inertia of any object depends on the axis of rotation.
Visualize:
Solve:
(a) xcm =
m
mi xi
i
=
mA xA + mB xB + mC xC + mD xD mA + mB + mC + mD
ycm
(100 g)(0 m) + (200 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 m) = = 0.057 m 100 g + 200 g + 200 g + 200 g m y + mB yB + mC yC + mD yD = AA mA + mB + mC + mD =
(100 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 cm) + (200 g )(0 m) = 0.057 m 700 g (b) The moment of inertia about a diagonal that passes through B and D is
2 I BD = mA rA + mC rC2
where rA = rC = (0.10 m)cos 45 = 7.07 cm and are the distances from the diagonal. Thus,
2 I BD = (0.100 kg) rA + (0.200 kg)rC2 = 0.0015 kg m 2
Assess:
Note that the masses B and D, being on the axis of rotation, do not contribute to the moment of inertia.
12.16.
Solve:
Model: The three masses connected by massless rigid rods is a rigid body.
Please refer to Figure EX12.16. (a)
Visualize:
xcm =
m x m
i
ii
=
(0.100 kg)(0 m) + (0.200 kg)(0.06 m) + (0.100 kg)(0.12 m) = 0.060 m 0.100 kg + 0.200 kg + 0.100 kg
ycm =
m y m
i
ii
=
(0.100 kg)(0 m) + (0.200 kg)
(
(0.10 m) 2 (0.06 m) 2 + (0.100 kg)(0 m)
)
0.100 kg + 0.200 kg + 0.100 kg
2
= 0.040 m
(b) The moment of inertia about an axis through A and perpendicular to the page is
I A = mi ri 2 = mB (0.10 m) + mC (0.10 m) 2 = (0.100 kg)[(0.10 m) 2 + (0.10 m) 2 ] = 0.0020 kg m 2
(c) The moment of inertia about an axis that passes through B and C is
I BC = mA
Assess:
(
(0.10 m) 2 (0.06 m) 2
)
2
= 0.00128 kg m 2
Note that mass mA does not contribute to I A , and the masses mB and mC do not contribute to I BC .
12.17.
Solve:
Model: The door is a slab of uniform density. (a) The hinges are at the edge of the door, so from Table 12.2, 1 2 I = ( 25 kg )( 0.91 m ) = 6.9 kg m 2 3 (b) The distance from the axis through the center of mass along the height of the door is 0.91 m d = 0.15 m = 0.305 m. Using the parallelaxis theorem, 2 1 2 2 I = I cm + Md 2 = ( 25 kg )( 0.91 m ) + ( 25 kg )( 0.305 cm ) = 4.1 kg m 2 12 Assess: The moment of inertia is less for a parallel axis through a point closer to the center of mass.
12.18.
Solve:
Model: The CD is a disk of uniform density. (a) The center of the CD is its center of mass. Using Table 12.2, 1 1 2 I cm = MR 2 = ( 0.021 kg )( 0.060 m ) = 3.8 105 kg m 2 2 2 (b) Using the parallelaxis theorem with d = 0.060 m,
I = I cm + Md 2 = 3.8 105 kg m 2 + ( 0.021 kg )( 0.060 m ) = 1.14 104 kg m 2
2
12.19. Visualize:
Solve:
Torque by a force is defined as = Fr sin where is measured counterclockwise from the r vector to
the F vector. The net torque on the pulley about the axle is the torque due to the 30 N force plus the torque due to the 20 N force:
(30 N)r1 sin1 + (20 N)r2 sin2 = (30 N)(0.02 m) sin ( 90) + (20 N)(0.02 m) sin (90) = ( 0.60 N m) + (0.40 N m) = 0.20 N m
Assess:
A negative torque causes a clockwise acceleration of the pulley.
12.20. Visualize:
The two equal but opposite 50 N forces, one acting at point P and the other at point Q, make a couple that causes a net torque. Solve: The distance between the lines of action is l = d cos30. The net torque is given by
= lF = (d cos30)F = (0.10 m)(0.866)(50 N) = 4.3 N m
12.21. Visualize:
Solve:
The net torque on the spark plug is = Fr sin = 38 N m = F (0.25 m)sin(120) F = 176 N
That is, you must pull with a force of 176 N to tighten the spark plug. Assess: The force applied on the wrench leads to its clockwise motion. That is why we have used a negative sign for the net torque.
12.22.
Model: The disk is a rotating rigid body.
Visualize:
The radius of the disk is 10 cm and the disk rotates on an axle through its center. Solve: The net torque on the axle is
= FA rA sin A + FBrB sin B + FC rC sin C + FD rD sin D
= (30 N)(0.10 m)sin(90) + (20 N)(0.050 m)sin 90 + (30 N)(0.050 m)sin135 + (20 N)(0.10 m)sin 0 = 3 N m + 1 N m + 1.0607 N m = 0.94 N m Assess: A negative torque means a clockwise rotation of the disk.
12.23.
Model: The beam is a solid rigid body.
Visualize:
The steel beam experiences a torque due to the gravitational force on the construction worker FG gravitational force on the beam FG
()
()
C
and the
B
. The normal force exerts no torque since the net torque is calculated about the
point where the beam is bolted into place. Solve: The net torque on the steel beam about point O is the sum of the torque due to FG
()
(F )
G
C
and the torque due to
B
. The gravitational force on the beam acts at the center of mass.
= (( FG )C )(4.0 m)sin( 90) + (( FG ) B )(2.0 m)sin(90)
= (70 kg)(9.80 m/s 2 )(4.0 m) (500 kg)(9.80 m/s 2 )(2.0 m) = 12.5 kN m The negative torque means these forces would cause the beam to rotate clockwise. The magnitude of the torque is 12.5 kN m.
12.24.
Model: Model the arm as a uniform rigid rod. Its mass acts at the center of mass.
Visualize:
Solve:
(a) The torque is due both to the gravitational force on the ball and the gravitational force on the arm:
= ball + arm = ( mb g )rb sin 90 + (ma g )ra sin 90
= (3.0 kg)(9.8 m/s 2)(0.70 m)+(4.0 kg)(9.8 m/s 2)(0.35 m) = 34 N m
(b) The torque is reduced because the moment arms are reduced. Both forces act at = 45 from the radial line, so
= ball + arm = (mb g )rb sin 45 + ( ma g ) ra sin 45
= (3.0 kg)(9.8 m/s 2 )(0.70 m)(0.707) + (4.0 kg)(9.8 m/s 2)(0.35 m)(0.707) = 24 N m
12.25. Solve: = I is the rotational analog 2 = (2.0 kg m )(4.0 rad/s 2 ) = 8.0 kg m 2 /s 2 = 8.0 N m.
of
Newtons
second
law
F = ma.
We
have
12.26.
Visualize:
Since
= /I , a graph of the angular acceleration looks just like the torque graph with
the numerical values divided by I = 4.0 kg m 2 .
Solve: From the discussion about Figure 4.47
f = i + area under the angular acceleration curve between ti and tf
The area under the curve between t = 0 s and t = 3 s is 0.75 rad/s. With 1 = 0 rad/s, we have
f = 0 rad/s + 0.75 rad/s = 0.75 rad/s
12.27. Model: Two balls connected by a rigid, massless rod are a rigid body rotating about an axis through the center of mass. Assume that the size of the balls is small compared to 1 m. Visualize:
We placed the origin of the coordinate system on the 1.0 kg ball. Solve: The center of mass and the moment of inertia are
xcm =
(1.0 kg)(0 m) + (2.0 kg)(1.0 m) = 0.667 m and ycm = 0 m (1.0 kg + 2.0 kg)
I about cm = mi ri 2 = (1.0 kg)(0.667 m) 2 + (2.0 kg)(0.333 m) 2 = 0.667 kg m 2
2 We have f = 0 rad/s, tf ti = 5.0 s, and i = 20 rpm = 20(2 rad/60 s) = 3 rad/s, so f = i + (tf ti ) becomes
2 2 0 rad/s = rad/s + (5.0 s) = rad/s 2 15 3 Having found I and , we can now find the torque that will bring the balls to a halt in 5.0 s:
= I about cm = kg m 2
2 3
2 4 rad/s 2 = N m = 0.28 N m 15 45
The magnitude of the torque is 0.28 N m, applied in the counterclockwise direction.
12.28. Model: A circular plastic disk rotating on an axle through its center is a rigid body. Assume axis is perpendicular to the disk. Solve: To determine the torque () needed to take the plastic disk from i = 0 rad/s
and the disks moment of inertia (I ) about the axle in its center. The radius of the disk is R = 10.0 cm. We have 1 1 I = MR 2 = (0.200 kg)(0.10 m) 2 = 1.0 103 kg m 2 2 2 i 60 rad/s 0 rad/s f = i + (tf ti ) = f = = 15 rad/s 2 t f ti 4.0 s Thus, = I = (1.0 103 kg m 2 )(15 rad/s 2 ) = 0.047 N m.
to
f = 1800 rpm = (1800)(2 ) / 60 rad/s = 60 rad/s in tf ti = 4.0 s, we need to determine the angular acceleration ( )
12.29.
Model: The compact disk is a rigid body rotating about its center.
Visualize:
Solve:
(a) The rotational kinematic equation 1 = 0 + (t1 t0 ) gives
200 2 (2000 rpm) rad/s 2 rad/s = 0 rad + (3.0 s 0 s) = 9 60 The torque needed to obtain this operating angular velocity is 200 rad/s 2 = 1.75 103 N m 9
= I = (2.5 105 kg m 2 )
(b) From the rotational kinematic equation,
1 = 0 + 0 (t1 t0 ) + (t1 t0 ) 2 = 0 rad + 0 rad + 2 2 9 100 revolutions = 50 rev = 100 rad = 2
Assess:
1
1 200
2 rad/s 2 ( 3.0 s 0 s )
Fifty revolutions in 3 seconds is a reasonable value.
12.30. Model: The rocket attached to the end of a rigid rod is a rotating rigid body. Assume the rocket is small compared to 60 cm. Visualize: Please refer to Figure EX12.30. Solve: We can determine the rockets angular acceleration from the relationship = I . The torque can be found from the thrust (F) using = Fr sin . The moment of inertia (I) can be calculated from equations given in
Table 12.2. Specifically, I = I rod about one end + I rocket becomes
1 1 M rod L2 + ML2 = (0.100 kg)(0.60 m) 2 + (0.200 kg)(0.60 m) 2 3 3 = 0.012 kg m 2 + 0.072 kg m 2 = 0.0840 kg m 2
=
Assess:
I
=
Fr sin (4.0 N)(0.60 m)sin(45) = = 20 rads/s 2 I 0.0840 kg m 2
The rocket will accelerate counterclockwise since is positive.
12.31.
Model: The rod is in rotational equilibrium, which means that net = 0.
Visualize:
As the gravitational force on the rod and the hanging mass pull down (the rotation of the rod is exaggerated in the figure), the rod touches the pin at two points. The piece of the pin at the very end pushes down on the rod; the right end of the pin pushes up on the rod. To understand this, hold a pen or pencil between your thumb and forefinger, with your thumb on top (pushing down) and your forefinger underneath (pushing up). Solve: Calculate the torque about the left end of the rod. The downward force exerted by the pin acts through this point, so it exerts no torque. To prevent rotation, the pins normal force npin exerts a positive torque (ccw about the left end) to balance the negative torques (cw) of the gravitational force on the mass and rod. The gravitational force on the rod acts at the center of mass, so
net = 0 N m = pin (0.40 m)(2.0 kg)(9.8 m/s 2) (0.80 m)(0.50 kg)(9.8 m/s 2)
pin = 11.8 N m
12.32.
Model: The massless rod is a rigid body.
Visualize:
Solve:
equilibrium ( net = 0 Nm). We have ( Fnet ) y = (40 N) (100 N) + (60 N) = 0 N, so the object is in translational
equilibrium. Measuring net about the left end,
To be in equilibrium, the object must be in both translational equilibrium ( Fnet = 0 N) and rotational
net = (60 N)(3.0 m)sin( +90) + (100 N)(2.0 m)sin(90) = 20 N m
The object is not in equilibrium.
12.33.
Model: The object balanced on the pivot is a rigid body.
Visualize:
Since the object is balanced on the pivot, it is in both translational equilibrium and rotational equilibrium. Solve: There are three forces acting on the object: the gravitational force FG acting through the center of
()
mass of the long rod, the gravitational force FG
()
1
2
acting through the center of mass of the short rod, and the
normal force P on the object applied by the pivot. The translational equilibrium equation ( Fnet ) y = 0 N is
( FG )1 ( FG )2 + P = 0 N P = ( FG )1 + ( FG )2 = (1.0 kg)(9.8 m/s 2 ) + (4.0 kg)(9.8 m/s 2 ) = 49 N
Measuring torques about the left end, the equation for rotational equilibrium net = 0 Nm is
Pd w1 (1.0 m) w2 (1.5 m) = 0 Nm (49 N)d (1.0 kg)(9.8 m/s 2 )(1.0 m) (4.0 kg)(9.8 m/s2 )(1.5 m) = 0 N d = 1.40 m Thus, the pivot is 1.40 m from the left end.
12.34.
Model: Visualize:
The see-saw is a rigid body. The cats and bowl are particles.
Solve:
The see-saw is in rotational equilibrium. Calculate the net torque about the pivot point.
m2 gd = m1 g ( 2.0 m ) mB g ( 2.0 m ) d= m2
net = 0 = ( FG )1 ( 2.0 m ) ( FG )2 ( d ) ( FG ) B ( 2.0 m )
( m1 mB )( 2.0 m ) = ( 5.0 kg 2.0 kg )( 2.0 m ) = 1.5 m
4.0 kg
Assess: The smaller cat is close but not all the way to the end by the bowl, which makes sense since the combined mass of the smaller cat and bowl of tuna is greater than the mass of the larger cat.
12.35.
Solve:
(a) According to Equation 12.35, the speed of the center of mass of the tire is
vcm = R = 20 m/s =
vcm 20 m/s 60 2 = = 66.67 rad/s = ( 66.7 ) rpm = 6.4 10 rpm R 0.30 m 2
(b) The speed at the top edge of the tire relative to the ground is vtop = 2vcm = 2(20 m/s) = 40 m/s. (c) The speed at the bottom edge of the tire relative to ground is vbottom = 0 m/s.
12.36. Model: The can is a rigid body rolling across the floor. Assume that the can has uniform mass distribution. Solve: The rolling motion of the can is a translation of its center of mass plus a rotation about the center of mass. The moment of inertia of the can about the center of mass is 1 MR 2, where R is the radius of the can. Also 2
vcm = R , where is the angular velocity of the can. The total kinetic energy of the can is
K = K cm + K rot = 1 1 1 1 1 v 2 2 Mvcm + I cm 2 = Mvcm + MR 2 cm 2 2 2 2 2 R
2
3 3 2 = Mvcm = (0.50 kg)(1.0 m/s) 2 = 0.38 J 4 4
12.37.
Model: The sphere is a rigid body rolling down the incline without slipping.
Visualize:
The initial gravitational potential energy of the sphere is transformed into kinetic energy as it rolls down. Solve: (a) If we choose the bottom of the incline as the zero of potential energy, the energy conservation equation will be K f = U i . The kinetic energy consists of both translational and rotational energy. This means
Kf =
1 1 1 2 1 2 I cm 2 + Mvcm = Mgh MR 2 2 + M ( R ) 2 = Mgh 2 2 2 5 2 7 MR 2 2 = Mg (2.1 m)sin 25 10
10 7
=
(b) From part (a)
g (2.1 m)(sin 25) = R2
10 7
g (2.1 m)(sin 25) = 88 rad/s (0.04 m) 2
K total =
1 1 7 1 1 2 1 2 I cm 2 + Mvcm = MR 2 2 and K rot = I cm 2 = MR 2 2 = MR 2 2 2 2 10 2 2 5 5 22 1 MR K 1 10 2 5 rot = 7 = = K total 10 MR 2 2 5 7 7
12.38. Visualize: Please refer to Figure EX12.38. To determine angle , put the tails of the vectors together. Solve: (a) The magnitude of A B is AB sin = (6)(4)sin 45 = 17. The direction of A B, using the right
hand rule, is out of the page. Thus, A B = (17, out of the page).
(b) The magnitude of C D is CD sin = (6)(4)sin180 = 0. Thus C D = 0.
12.39.
Solve:
Visualize: Please refer to Figure EX12.39. (a) The magnitude of A B is AB sin = (6)(4)sin 60 = 20.78. The direction of A B is given by the
right hand rule. To curl our fingers from A to B, we have to point our thumb out of the page. Thus, A B = (21, out of the page). (b) C D = ((6)(4)sin 90, into the page) = (24, into the page).
12.40.
j j (a) (i ) i = k i = j j j (b) i ( i ) = i ( k ) = i k = ( ) =
Solve:
j (a) i (i ) = i k = j j (b) (i ) k = k k = 0
12.41.
Solve:
12.42.
Solve:
j j (a) A B = (3i + ) (3i 2 + 2k ) j = 9i i 6i + 6i k + 3 i 2 + 2 k j jj j = 0 6k + 6( ) + 3(k ) 0 + 2i = 2i 6 9k j j
(b)
12.43. Solve: (a) C D = 0 implies that D must also be in the same or opposite direction as the C vector or
zero, because i i = 0. Thus D = ni , where n could be any real number. implies that E must be along the vector, because i = k . Thus E = 2 . j j j (b) C E = 6k implies that F must be along the k vector, because i k = . Thus F = 1k . (c) C F = 3 j j
12.44.
Solve:
= r F = (5i + 5 ) (10 ) N m j j
j = [50(i ) 50( )] N m = [50(+ k ) 0] N m = 50k N m jj
12.45.
Solve:
= r F = (5 ) (10i + 10 ) N m = (5 i + 50 ) N m j j j jj
= [50( k ) + 0] N m = 50k N m
12.46.
Solve:
Visualize: Please refer to Figure EX12.46. L = r mv = (1.0i + 2.0 ) m (0.200 kg)(3.0 m/s) cos45i sin 45 j j
(
)
j j jj = (0.42i i 0.42i + 0.85 i 0.85 ) kg m 2 /s = (1.27 k ) kg m 2 /s or (1.27 kg m 2 /s, into page)
12.47.
Solve:
Visualize: Please refer to Figure EX12.47. L = r mv = (3.0i + 2.0 ) m (0.1 kg)(4.0 ) m/s j j 2 2 j = 1.20(i ) kg m /s + 0.8( ) kg m /s = 1.20k kg m 2 /s + 0 kg m 2 /s jj = 1.20k kg m 2 /s or (1.20 kg m 2 /s, out of page)
12.48. Model: The bar is a rotating rigid body. Assume that the bar is thin. Visualize: Please refer to Figure EX12.48. Solve: The angular velocity = 120 rpm = (120)(2 ) / 60 rad/s = 4 rad/s. From Table 12.2, the moment of
1 inertial of a rod about its center is I = 12 ML2 . The angular momentum is
1 L = I = (0.50 kg)(2.0 m) 2 (4 rad/s) = 2.1 kg m 2 /s 12
If we wrap our fingers in the direction of the rods rotation, our thumb will point in the z direction or out of the page. Consequently, L = (2.1 kg m 2 /s, out of the page)
12.49. Model: The disk is a rotating rigid body. Visualize: Please refer to Figure EX12.49. Solve: From Table 12.2, the moment of inertial of the disk about its center is
1 1 I = MR 2 = (2.0 kg)(0.020 m) 2 = 4.0 104 kg m 2 2 2 is 600 rpm = 600 2 /60 rad/s = 20 rad/s. velocity
2 2
The
angular
4
Thus,
L = I = (4.0 10 kg m )(20 rad/s) = 0.025 kg m /s. If we wrap our right fingers in the direction of the disks rotation, our thumb will point in the x direction. Consequently, L = 0.025 i kg m 2 /s = (0.025 kg m 2 /s, into page)
12.50.
Model: The beach ball is a spherical shell. Solve: From Table 12.2, the moment of inertia about a diameter of a spherical shell is 2 2 1 2 I = MR 2 = ( 0.100 kg )( 0.50 m ) = kg m 2 3 3 60 Require 1 kg m 2 L = 0.10 kg m 2 /s = I = 60 = ( 60 rad/s )
rev 60 s In rpm, this is ( 60 rad/s ) = 57 rpm. 2 rad min
12.51.
Model: The wheel is a rigid rolling body.
Visualize:
Solve:
The front of the disk is moving forward at velocity vcm . Also, because of rotation the point is moving
2 2 v = vcm + vcm = 2vcm = 2(20 m/s) = 28 m/s
downward at velocity vrel = R = vcm . So, this point has a speed
Assess:
The speed v is independent of the radius of the wheel.
12.52
Model: The triangle is a rigid body rotating about its center of mass perpendicular to the plane of the triangle. The center of mass of any symmetrical object of uniform density is at the physical center of the object. Visualize:
The distance to one tip of the triangle from the center of mass is given by r cos30 = (5.0/2) cm, which yields r = (2.5 cm)/ cos30 = 2.9 cm. Solve: The speed of the tip is v = r = (2.9 cm)(120 rpm) = (2.9 cm)(4 rad/s) = 36 cm/s
12.53. Visualize:
Solve: We will consider a vertical strip of width dx and of mass dm at a position x from the origin. The formula for the x component of the center of mass is 1 xcm = x dm M
The area of the steel plate is A = 1 (0.2 m)(0.3 m) = 0.030 m 2 . Mass dm in the strip is the same fraction of M as 2 dA is of A. Thus dm dA M 0.800 kg = dm = dA = dA = (26.67 kg/m 2 )l dx 2 M A A 0.030 m
The relationship between l and x is l x 2 = l = x 0.20 m 0.30 m 3 Therefore,
xcm 1 (17.78 kg/m 2 ) x3 2 (26.67 kg/m 2 ) x 2 dx = = M M 3 3
0.3 m
=
0m
(17.78 kg/m 2 ) (0.3 m)2 = 20 cm 0.8 kg 3
Due to symmetry ycm = 0 cm.
12.54.
Visualize:
Solve: Build the plate from the three shapes 1, 2, and 3. The center of mass of the complete shape is the center of mass of the center of masses of the three smaller shapes. m1 ( xcm )1 + m2 ( xcm )2 + m3 ( xcm )3 xcm = m1 + m2 + m3
m1 + m2 + m3 The masses of each of the smaller shapes are a fraction of the larger shapes mass by the ratio of areas. The following table will be useful in the solution. Shape A mass xcm ycm 2 All M xcm ycm 32 cm
1
24 cm 2 2.0 cm 6.0 cm
2 2 24 32 2.0 32 6.0 32
ycm =
m1 ( ycm )1 + m2 ( ycm )2 + m3 ( ycm )3
M M M
1.0 cm 2.0 cm 2.0 cm
0 cm 2.5 cm 1.5 cm
2 3 Thus,
24 2.0 6.0 M ( 1.0 cm ) + M ( 2.0 cm ) + M ( 2.0 cm ) 32 32 32 = 0.25 cm xcm = M Similarly, ycm = 0.125 cm. Assess: The plates center of mass of (0.25 cm, 0.125 cm) is reasonable. The cutout means more of the mass is left of center and above the center.
12.55.
Model:
The disk is a rigid rotating body. The axis is perpendicular to the plane of the disk.
Visualize:
Solve:
(a) From Table 12.2, the moment of inertia of a disk about its center is
1 1 I = MR 2 = (2.0 kg)(0.10 m) 2 = 0.010 kg m 2 2 2 (b) To find the moment of inertia of the disk through the edge, we can make use of the parallel axis theorem:
I = I center + Mh 2 = (0.010 kg m 2 ) + (2.0 kg)(0.10 m) 2 = 0.030 kg m 2
Assess: The larger moment of inertia about the edge means there is more inertia to rotational motion about the edge than about the center.
12.56.
Model: The object is a rigid rotating body. Assume the masses m1 and m2 are small and the rod is thin. Visualize: Please refer to P12.56. Solve: The moment of inertia of the object is the sum of the moment of inertia of the rod, mass m1 , and mass
m2 . Using Table 12.2 for the moment of inertia of the rod, we get
I rod = I rod about center + I m1 + I m2 = =
Assess:
1 L L ML2 + m1 + m2 12 2 4
2
2
1 1 1 L2 M m ML2 + m1L2 + m2 L2 = + m1 + 2 12 4 16 4 3 4
1 With m1 = m2 = 0 kg, I rod 12 ML2 , as expected.
12.57. Visualize:
We chose the origin of the coordinate system to be on the axis of rotation, that is, at a distance d from one end of the rod. Solve: The moment of inertia can be calculated as follows:
I = x 2 dm
x1
x2
and
dm dx M = dm = dx M L L M 3 3 3 3 [( L d ) ( d ) ] = [( L d ) + d ] 3L
I =
2
M L
Ld
d
3 M x x 2 dx = L3
Ld
d
1 M = 3 L
For d = 0 m, I = 1 ML , and for d = 1 L, 3 2
I=
3 3 M L L 1 2 + = ML 3L 2 2 12
Assess: The special cases d = 0 m and d = L/2 of the general formula give the same results that are found in Table 12.2.
12.58. Visualize:
Solve: We solve this problem by dividing the disk between radii r1 and r2 into narrow rings of mass dm. Let dA = 2 rdr be the area of a ring of radius r. The mass dm in this ring is the same fraction of the total mass M as dA is of the total area A. (a) The moment of inertia can be calculated as follows:
I disk = r 2 dm and dm =
r
M M dA = ( 2 r ) dr A ( r22 r12 )
r r2
I disk =
2 M 2M 2 2M r 4 r 2 (2 r ) dr = 2 2 r 3dr = 2 2 2 2 ( r2 r1 ) r1 ( r2 r1 ) r1 ( r2 r1 ) 4
r1
M 2M r 4 r14 ) = ( r22 + r12 ) = 2 2 (2 2 4 ( r2 r1 ) Replacing r1 with r and r2 with R, the moment of inertia of the disk through its center is I disk = 1 M ( R 2 + r 2 ). 2
(b) For r = 0 m, I disk = 1 MR 2 . This is the moment of inertia for a solid disk or cylinder about the center. 2
Additionally, for r R, we have I = MR 2 . This is the expression for the moment of inertia of a cylindrical hoop or ring about the center. (c) The initial gravitational potential energy of the disk is transformed into kinetic energy as it rolls down. If we choose the bottom of the incline as the zero of potential energy, and use vcm = R, the energy conservation equation K f = U i is 121 1 M v2 1 2 2 I + Mvcm = Mgh ( R 2 + r 2 ) cm + Mvcm = Mgyi = Mg (0.50 m)sin 20 2 2 2 2 R2 2 2 2 12 1 r2 2 R +r 2 1 22 vcm + vcm = vcm + + 2 = 1.6759 m /s 2 4R 2 2 4 4R (0.015 m) 2 2 3 22 vcm + = 1.6759 m /s vcm = 1.37 m/s 4 4(0.020 m) 2 For a sliding particle on a frictionless surface K f = U i , so
12 v mvf = mgyi vf = 2 gyi = 2 g (0.50 m)sin 20 = 1.83 m/s cm = 0.75 2 vf That is, vcm is 75% of the speed of a particle sliding down a frictionless ramp.
12.59.
Model: Visualize:
The plate has uniform density.
Solve:
The moment of inertia is
I = r 2 dm. Let the mass of the plate be M. Its area is L2 . A region of area dA located at (x,y) has mass M M dm = dA = 2 dx dy. The distance from the axis of rotation to the point (x, y) is r = x 2 + y 2 . With A L L L L L x and y , 2 2 2 2
I=
L L 2 2 L 2
(x
L 2
L 2
2
M + y2 ) 2 L
M dx dy = 2 L
L 2
3 + y x
2 L 2
L 2
x3
L 2
dy
L 2
M =2 L =
L3 y 2 L L3 y 2 L M L 24 + 2 24 2 dy = L2
2
L3 M L4 y3 2 + Ly dy = 2 + L L 12 L 12 3 2
L 2
L 2
L3 L3 1 2 M L4 2 12 + L 24 + 24 = 6 ML L
12.60.
Solve:
From Equation 12.16,
I = ( x 2 + y 2 )dm
dm =
A small region of area dA has mass dm, and
M M dA = dx dy A A The area of the plate is 1 (0.20 m)(0.30 m) = 0.030 m 2 . So 2
M 0.800 kg 2 = = 26.67 kg m A 0.030 m 2 1 1 1 The limits for x are 0 x 30 cm. For a particular value of x, x y x. Note that is the slope of the 3 3 3 top and bottom edges of the triangle. Therefore,
30 cm
I=
(x
0 1 x 3
1 x 3
2
+ y 2 ) ( 26.67 kg/m 2 ) dx dy
30 cm
= ( 26.67 kg/m
2
)
0
2 y3 3 x y + dx 3 1 x
3
1
x
= ( 26.67 kg/m 2 )
30 cm
0
2 3 2 x3 x+ dx 81 3
30 cm
56 1 = ( 26.67 kg/m 2 ) x 4 81 4 0
= 0.037 kg m 2
12.61.
Model: The ladder is a rigid rod of length L. To not slip, it must be in both translational equilibrium ( Fnet = 0 N) and rotational equilibrium ( net = 0 N m). We also apply the model of static friction.
Visualize:
Since the wall is frictionless, the only force from the wall on the ladder is the normal force n2 . On the other hand, the floor exerts both the normal force n1 and the static frictional force f s . The gravitational force FG on the ladder acts through the center of mass of the ladder. Solve: The x- and y-components of Fnet = 0 N are
F
x
= n2 fs = 0 N fs = n2
F
y
= n1 FG = 0 N n1 = FG
The minimum angle occurs when the static friction is at its maximum value f s max = s n1. Thus we have n2 = f s = s n1 = s mg . We choose the bottom corner of the ladder as a pivot point to obtain net , because two forces pass through this point and have no torque about it. The net torque about the bottom corner is
net = d1mg d 2 n2 = (0.5L cos min )mg ( L sin min ) s mg = 0 N m
0.5cos min = s sin min tan min = 0.5
s
=
0.5 = 1.25 min = 51 0.4
12.62.
Model: The beam is a rigid body of length 3.0 m and the student is a particle.
Visualize:
Solve:
To stay in place, the beam must be in both translational equilibrium ( Fnet = 0 N) and rotational
equilibrium ( net = 0 Nm). The first condition is
F
y
= ( FG )beam ( FG )student + F1 + F2 = 0 N
F1 + F2 = ( FG ) beam + ( FG )student = (100 kg + 80 kg)(9.80 m/s 2 ) = 1764 N
Taking the torques about the left end of the beam, the second condition is ( FG ) beam (1.5 m) ( FG )student (2.0 m) + F2 (3.0 m) = 0 N m (100 kg)(9.8 m/s 2 )(1.5 m) (80 kg)(9.8 m/s 2 )(2.0 m) + F2 (3.0 m) = 0 N m F2 = 1013 N From F1 + F2 = 1764 N, we get F1 = 1764 N 1013 N = 0.75 kN. Assess: To establish rotational equilibrium, the choice for the pivot is arbitrary. We can take torques about any point on the body of interest.
12.63.
Model: The structure is a rigid body.
Visualize:
Solve:
We pick the left end of the beam as our pivot point. We dont need to know the forces Fh and Fv
because the pivot point passes through the line of application of Fh and Fv and therefore these forces do not exert a torque. For the beam to stay in equilibrium, the net torque about this point is zero. We can write
about left end = ( FG ) B (3.0 m) ( FG ) W (4.0 m) + (T sin150)(6.0 m) = 0 N m
Using ( FG ) B = (1450 kg)(9.8 m/s 2 ) and ( FG ) W = (80 kg)(9.8 m/s 2 ), the torque equation can be solved to yield T = 15,300 N. The tension in the cable is slightly more than the cable rating. The worker should be worried.
12.64. Model: Model the beam as a rigid body. For the beam not to fall over, it must be both in translational equilibrium ( Fnet = 0 N) and rotational equilibrium ( net = 0 N m).
Visualize:
The boy walks along the beam a distance x, measured from the left end of the beam. There are four forces acting on the beam. F1 and F2 are from the two supports, FG is the gravitational force on the beam, and FG is
()
b
()
B
the gravitational force on the boy. Solve: We pick our pivot point on the left end through the first support. The equation for rotational equilibrium is ( FG ) b (2.5 m) + F2 (3.0 m) ( FG ) B x = 0 N m (40 kg)(9.80 m/s 2 )(2.5 m) + F2 (3.0 m) (20 kg)(9.80 m/s 2 ) x = 0 N m The equation for translation equilibrium is
F1 + F2 = ( FG )b + ( FG )B = (40 kg + 20 kg)(9.8 m/s 2 ) = 588 N Just when the boy is at the point where the beam tips, F1 = 0 N. Thus F2 = 588 N. With this value of F2 , we can simplify the torque equation to: (40 kg)(9.80 m/s 2 )(2.5 m) + (588 m) N)(3.0 (20 kg)(9.80 m/s 2 ) x = 0 N m x = 4.0 m Thus, the distance from the right end is 5.0 m 4.0 m = 1.0 m.
F
y
= 0 N = F1 + F2 ( FG )b ( FG )B
12.65.
Visualize:
Please refer to Figure P12.65.
Solve: The bricks are stable when the net gravitational torque on each individual brick or combination of bricks is zero. This is true as long as the center of gravity of each individual brick and any combination is over a base of support. To determine the relative positions of the bricks, work from the top down. The top brick can extend past the second brick by L 2 . For maximum extension, their combined center of gravity will be at the edge of the third brick, and the combined center of gravity of the three upper bricks will be at the edge of the fourth brick. The combined center of gravity of all four bricks will be over the edge of the table. Measuring from the left edge of the brick 2, the center of gravity of the top two bricks is L m + mL m1 x1 + m2 x2 3 2 ( x12 )com = = = L. 2m 4 m1 + m2
Thus the top two bricks can extend L 4 past the edge of the third brick. The top three bricks have a center of mass L 3L 5L m + m + m m1 x1 + m2 x2 + m3 x3 2 4 4 = 5 L. = ( x123 )com = 3m 6 m1 + m2 + m3 Thus the top three bricks can extend past the edge of the fourth brick by L 6. Finally, the four bricks have a combined center of mass at L 4L 11L 17 L m + m + m + m 2 6 12 12 = 7 L. ( x1234 )com = 4m 8 The center of gravity of all four bricks combined is 7 L 8 from the left edge of the bottom brick, so brick 4 can extend L 8 past the table edge. Thus the maximum distance to the right edge of the top brick from the table edge is L L L L 25 d max = + + + = L. 8 6 4 2 24 Thus, yes, it is possible that no part of the top brick is directly over the table because d max > L. Assess: As crazy as this seems, the center of gravity of all four bricks is stably supported, so the net gravitational torque is zero, and the bricks do not fall over.
12.66.
Model: Visualize:
The pole is a uniform rod. The sign is also uniform.
Solve:
The geometry of the rod and cable give the angle that the cable makes with the rod. 250 = tan 1 = 51.3 200 The rod is in rotational equilibrium about its left-hand end. 1 1 net = 0 = (100 cm )( FG )P ( 80 cm ) ( FG )S ( 200 cm ) ( FG )S + ( 200 cm ) T sin 51.3 2 2
= (100 cm )( 5.0 kg ) ( 9.8 m/s 2 ) mS ( 9.8 m/s 2 ) (140 cm ) + (156 cm ) T
With T = 300 N , mS = 30.6 kg. Assess: A mass of 30.6 kg is reasonable for a sign.
12.67.
Model: The hollow cylinder is a rigid rotating body.
Visualize:
We placed the origin of the coordinate system on the ground. Solve: (a) Newtons second law for the block is ( FG ) B + T = mB a y , where T is the tension in the string, ( FG ) B = mB g is the gravitational force on the block, and a y is the acceleration of block. The string tension exerts a negative (cw) torque on the cylinder, so the rotational form of Newtons second law for the hollow cylinder is
= TR = I = I
ay R
T =
Ia y R2
where we used the acceleration constraint a y = R. With this expression for T, Newtons second law for the block becomes
mB g
mB g Ia = mB a y a y = 2 R (mB + I / R 2 )
The moment of inertia of a hollow cylinder is I = mC R 2 , so the equation for a y simplifies to
ay = mB g (3.0 kg)(9.8 m/s 2 ) = = 5.88 m/s 2 mB + mC 3.0 kg + 2.0 kg
The speed of the block just before it hits the ground can now be found using kinematics:
2 v12 = v0 + 2a y ( y1 y0 ) = 0 m 2 /s 2 + 2(5.88 m/s 2 )(0 m 1.0 m) v1 = 3.4 m/s
(b) The conservation of energy equation K1 + U g1 = K 0 + U g0 for the system (block + cylinder + earth) is
1 1 1 12 2 mBv12 + I12 + mB gy1 = mBv0 + I 0 + mB gy0 2 2 2 2 v12 m 1 1 m mBv12 + ( mC R 2 ) 2 + 0 J = 0 J + 0 J + mB gy0 v12 B + C = mB gy0 R 2 2 2 2 v12 = 2mB gy0 2(3.0 kg)(9.8 m/s 2 )(1.0 m) = = 11.76 m 2 /s 2 v1 = 3.4 m/s mB + mC (3.0 kg) + (2.0 kg)
Assess: Newtons second law and the conservation of energy method give the same result for the blocks final velocity.
12.68.
Model: The bar is a solid body rotating through its center.
Visualize:
Solve:
(a) The two forces form a couple. The net torque on the bar about its center is
I L where F is the force produced by one of the air jets. We can find I and as follows:
net = LF = I F =
1 1 ML2 = (0.50 kg)(0.60 m) 2 = 0.015 kg m 2 12 12 1 = 0 + (t1 t0 ) 150 rpm = 5.0 rad/s = 0 rad + (10 s 0 s) = 0.50 rad/s 2 I= F= (0.015 kg m 2 )(0.5 rad/s 2 ) = 0.0393 N (0.60 m)
The force F = 39 mN. (b) The torque of a couple is the same about any point. It is still net = LF . However, the moment of inertia has changed.
net = LF = I =
LF 1 1 where I = ML2 = (0.500 kg)(0.6 m) 2 = 0.060 kg m 2 I 3 3 (0.0393 N) (0.60 m) = = 0.393 rad/s 2 0.060 kg m 2
Finally,
1 = 0 + (t1 t0 ) = 0 rad/s + (0.393 rad/s 2 )(10 s 0 s)
= 3.93 rad/s = (3.93)(60) rpm = 37.5 rpm 2
The angular speed is 38 rpm. Assess: Note that and 1/ I . Thus, 1/ I . I about the center of the rod is 4 times smaller than I about one end of the rod. Consequently, is 4 times larger.
12.69.
Model: The flywheel is a rigid body rotating about its central axis.
Visualize:
Solve: (a) The radius of the flywheel is R = 0.75 m and its mass is M = 250 kg. The moment of inertia about the axis of rotation is that of a disk: 1 1 I = MR 2 = (250 kg)(0.75 m) 2 = 70.31 kg m 2 2 2 The angular acceleration is calculated as follows:
net = I = net / I = (50 N m)/(70.31 kg m 2 ) = 0.711 rad/s 2
Using the kinematic equation for angular velocity gives
1 = 0 + (t1 t0 ) = 1200 rpm = 40 rad/s = 0 rad/s + 0.711 rad/s 2 (t1 0 s)
t1 = 177 s
(b) The energy stored in the flywheel is rotational kinetic energy: 1 1 K rot = I12 = (70.31 kg m 2 )(40 rad/s) 2 = 5.55 105 J 2 2 5 The energy stored is 5.6 10 J. energy delivered (5.55 105 J)/2 = = 1.39 105 W = 139 kW (c) Average power delivered = time interval 2.0 s (d) Because = I , = I
half energy = I full energy . full energy =1 (from part (a)) = 40 rad/s. half energy t t
5.55 105 J K rot = = 88.85 rad/s 70.31 kg m 2 I
can be obtained as:
12 1 I half energy = K rot half energy = 2 2 Thus
= (70.31 kg m 2 )
40 rad/s 88.85 rad/s = 1.30 kN m 2.0 s
12.70. Model: The pulley is a rigid rotating body. We also assume that the pulley has the mass distribution of a disk and that the string does not slip. Visualize:
Because the pulley is not massless and frictionless, tension in the rope on both sides of the pulley is not the same. Solve: Applying Newtons second law to m1 , m2 , and the pulley yields the three equations:
T1 ( FG )1 = m1a1
( FG )2 + T2 = m2 a2
T2 R T1R 0.50 N m = I
Noting that a2 = a1 = a, I = 1 mp R 2 , and = a/R, the above equations simplify to 2
T1 m1 g = m1a m2 g T2 = m2 a 0.50 N m 1 a 1 0.50 N m 1 T2 T1 = mp R 2 + = mp a + R 2 0.060 m 2 R R
Adding these three equations,
1 (m2 m1 ) g = a m1 + m2 + mp + 8.333 N 2 (m m1 ) g 8.333 N (4.0 kg 2.0 kg)(9.8 m/s 2 ) 8.333 N a= 2 = = 1.610 m/s 2 2.0 kg + 4.0 kg + (2.0 kg/2) m1 + m2 + 1 mp 2 We can now use kinematics to find the time taken by the 4.0 kg block to reach the floor: 1 1 y1 = y0 + v0 (t1 t0 ) + a2 (t1 t0 ) 2 0 = 1.0 m + 0 + (1.610 m/s 2 )(t1 0 s) 2 2 2 2(1.0 m) t1 = = 1.11 s (1.610 m/s 2 )
12.71.
Model: Assume the string does not slip on the pulley.
Visualize:
The free-body diagrams for the two blocks and the pulley are shown. The tension in the string exerts an upward force on the block m2 , but a downward force on the outer edge of the pulley. Similarly the string exerts a force on block m1 to the right, but a leftward force on the outer edge of the pulley. Solve: (a) Newtons second law for m1 and m2 is T = m1a1 and T m2 g = m2 a2 . Using the constraint a2 = + a1 = a, we have T = m1a and T + m2 g = m2 a. Adding these equations, we get m2 g = (m1 + m2 )a, or
a=
m2 g mm g T = m1a = 1 2 m1 + m2 m1 + m2
(b) When the pulley has mass m, the tensions (T1 and T2 ) in the upper and lower portions of the string are different. Newtons second law for m1 and the pulley are:
T1 = m1a
and
T1R T2 R = I
I a aI = R R R2
We are using the minus sign with because the pulley accelerates clockwise. Also, a = R . Thus, T1 = m1a and
T2 T1 =
Adding these two equations gives
I T2 = a m1 + 2 R
Newtons second law for m2 is T2 m2 g = m2 a2 = m2 a. Using the above expression for T2 ,
I m2 g a m1 + 2 + m2 a = m2 g a = R m1 + m2 + I / R 2 Since I = 1 mp R 2 for a disk about its center, 2 a= With this value for a we can now find T1 and T2 : m2 g m1 + m2 + 1 mp 2
1 m2 g 1 m2 ( m1 + 2 mp ) g m1 + mp = ( m1 + m2 + 12 mp ) 2 m1 + m2 + 12 mp
T1 = m1a =
Assess:
m1m2 g m1 + m2 + 1 mp 2
T2 = a (m1 + I / R 2 ) =
For m = 0 kg, the equations for a, T1 , and T2 of part (b) simplify to
a= m2 g m1 + m2 and T1 = m1m2 g m1 + m2 and T2 = m1m2 g m1 + m2
These agree with the results of part (a).
12.72. Model: The disk is a rigid spinning body. Visualize: Please refer to Figure P12.72. The initial angular velocity is 300 rpm or (300)(2 )/60 = 10 rad/s. After 3.0 s the disk stops. Solve: Using the kinematic equation for angular velocity,
1 = 0 + (t1 t0 ) = 1 0
t1 t0 = (0 rad/s 10 rad/s) 10 = rad/s 2 (3.0 s 0 s) 3
Thus, the torque due to the force of friction that brings the disk to rest is
= I = fR f =
( 12 mR 2 ) = 1 (mR) = 1 (2.0 kg)(0.15 m) 10 rad/s2 = 1.57 N I = R R 2 2 3
The minus sign with = fR indicates that the torque due to friction acts clockwise.
12.73.
Model: The entire structure is a rigid rotating body. The two thrust forces are a couple that exerts a torque on the structure about its center of mass. We will assume the thrust forces are perpendicular to the connecting tunnel. Visualize: Please refer to Figure 12.30. We chose a coordinate system in which m1 and m2 are on the x-axis
and m1 = 100,000 kg is at the origin. m3 is the mass of the tunnel, whose center is at x3 = 45 m. Solve: (a) Assuming the center of mass of the tunnel is at the center of the tunnel, we get
xcm =
m1 x1 + m2 x2 + m3 x3 m1 + m2 + m3
xcm =
(1.0 105 kg)(0 m) + (2.0 105 kg)(90 m) + (5.0 104 kg)(45 m) = 57.9 m 1.0 105 kg + 2.0 105 kg + 5.0 104 kg
The center of mass of the entire structure is 58 m from the 100,000 kg rocket. (b) Initially, the angular velocity is zero. The structures angular velocity after 30 s is
1 = 0 + (t1 t0 ) = 0 rad/s + (30 s 0 s) = (30 s)
The angular acceleration can be found from = I , where is the net torque on the structure and I is its moment of inertia. The two thrusts form a couple with torque
= lF = (90 m)(50,000 N) = 4.50 106 N m
2 2 1 2 I = I m1 + I m2 + I tunnel = m1 x12 + m2 x2 + m3 ( 90 m ) + m3 ( 58 m 45 m ) 12 1 2 5 2 5 2 4 = (1.0 10 kg)(57.9 m) + (2.0 10 kg)(32.1 m) + (5 10 kg)(90 m) 2 + ( 5 104 kg ) (13 m ) 12 = 5.83 108 kg m 2
=
I
=
4.5 106 N m = 7.71 103 rad/s 2 5.83 108 kg m 2
1 = (30 s)(7.71 103 rad/s 2 ) = 0.23 rad/s
Assess: Note that the parallel axis theorem was used in finding the moment of inertia of the tunnel.
12.74. Model: Assume that the hollow sphere is a rigid rolling body and that the sphere rolls up the incline without slipping. We also assume that the coefficient of rolling friction is zero. Visualize:
The initial kinetic energy, which is a combination of rotational and translational energy, is transformed in gravitational potential energy. We chose the bottom of the incline as the zero of the gravitational potential energy. Solve: The conservation of energy equation K f + U gf = K i + U gi is
1 1 1 1 2 2 M (v1 )cm + I cm (1 ) 2 + Mgy1 = M (v0 )cm + I cm ( 0 ) 2 + Mgy0 2 2 2 2 1 1 2 1 1 (v ) 2 2 2 2 0 J + 0 J + Mgy1 = M (v0 )cm + MR 2 ( 0 )cm + 0 J Mgy1 = M (v0 )cm + MR 2 0 2cm R 2 2 3 2 3
5 (v ) 2 5 5 (5.0 m/s) 2 2 gy1 = (v0 )cm y1 = 6 0 cm = = 2.126 m g 6 6 9.8 m/s 2
The distance traveled along the incline is
2.126 m y1 = = 4.3 m sin 30 0.5 Assess: This is a reasonable stopping distance for an object rolling up an incline when its speed at the bottom of the incline is approximately 10 mph. s=
12.75.
Model: Visualize:
The masses are particles.
Solve:
(a) The moment of inertia of the barbell is 2 I = I M + I m = Mx 2 + m ( L x )
K rot = 121 2 I = Mx 2 + m ( L x ) 2 2 2
The rotational kinetic energy is therefore
(
)
To find x such that K rot is a minimum, set
1 dK rot = 0 = ( 2 Mx 2m ( L x ) ) 2 dx 2 (b) This is the center of mass location measured from the mass M.
x=
m L m+M
12.76. Model: The disk is a rigid body rotating on an axle passing through one edge. The gravitational potential energy is transformed into rotational kinetic energy as the disk is released. Visualize:
We placed the origin of the coordinate system at a distance R just below the axle. In the initial position, the center of mass of the disk is at the same level as the axle. The center of mass of the disk in the final position is coincident with the origin of the coordinate system. Solve: (a) The torque is due to the gravitational force on the disk acting at the center of mass. Thus
= (mg ) R = (5.0 kg)(9.8 m/s 2 )(0.30 m) = 14.7 N m
The moment of inertia about the disks edge is obtained using the parallel-axis theorem:
1 3 3 I = I cm + mR 2 = mR 2 + mR 2 = mR 2 = (5.0 kg)(0.30 m)2 = 0.675 kg m 2 2 2 2 14.7 N m = = = 22 rad/s 2 I 0.675 kg m 2
(b) The energy conservation equation K f + U gf = K i = U gi is
12 12 1 I1 + mgy1 = I 0 + mgy0 I12 + 0 J = 0 J + mgR 2 2 2
1 =
2mgR 2(5.0 kg)(9.8 m/s 2 )(0.30 m) = = 6.6 rad/s I 0.675 kg m 2
Assess: An angular velocity of 6.6 rad/s (or 1.05 revolutions/s) as the center of mass of the disk reaches below the axle is reasonable.
12.77. Model: The hoop is a rigid body rotating about an axle at the edge of the hoop. The gravitational torque on the hoop causes it to rotate, transforming the gravitational potential energy of the hoops center of mass into rotational kinetic energy. Visualize:
We placed the origin of the coordinate system at the hoops edge on the axle. In the initial position, the center of mass is a distance R above the origin, but it is a distance R below the origin in the final position. Solve: (a) Applying the parallel-axis theorem, I edge = I cm + mR 2 = mR 2 + mR 2 = 2mR 2 . Using this expression in
the energy conservation equation K f + U gf = K i + U gi yields:
1 1 2 I edge12 + mgy1 = I edge 0 + mgy0 2 2
(b) The speed of the lowest point on the hoop is
1 2g (2mR 2 )12 mgR = 0 J + mgR 1 = 2 R
Assess:
2g (2 R ) = 8 gR R Note that the speed of the lowest point on the loop involves a distance of 2R instead of R. v = (1 )(2 R) =
12.78. Model: The long, thin rod is a rigid body rotating about a frictionless pivot on the end of the rod. The gravitational torque on the rod causes it to rotate, transforming the gravitational potential energy of the rods center of mass into rotational kinetic energy. Visualize:
We placed the origin of the coordinate system at the pivot point. In the initial position, the center of mass is a distance 1 L above the origin. In the final position, the center of mass is at y = 0 m and thus has zero gravitational potential 2 energy. Solve: (a) The energy conservation equation for the rod K f + U gf = K i + U gi is
12 12 I1 + mgy1 = I 0 + mgy0 2 2
11 2 2 mL 1 + 0 J = 0 J + mg ( L /2) 1 = 3 g / L 23
(b) The speed at the tip of the rod is vtip = (1 ) L = 3 gL .
12.79. Model: The sphere attached to a thin rod is a rigid body rotating about the rod. Assume the rod is vertical and the sphere solid. Visualize: Please refer to Figure P12.79. The sphere rotates because the string wrapped around the rod exerts a torque . Solve: The torque exerted by the string on the rod is = Tr. From the parallel-axis theorem, the moment of inertia of the sphere about the rods axis is
MR 2 13 R 2 I off center = I cm + M = MR 2 + = MR 2 4 20 2 5 From Newtons second law,
2
=
I
=
Tr 20Tr = (13MR 2 /20) 13MR 2
12.80.
Model: The pulley is a rigid rotating body.
Visualize:
We placed the origin of the coordinate system on the floor. The pulley rotates about its center. Solve: Using kinematics for the physics book (mass = m1 ), 1 1 y1 = y0 + v0 (t1 t0 ) + a (t1 t0 ) 2 0 m = 1.0 m + 0 m + a(0.71 s 0 s) 2 a = 3.967 m/s 2 2 2 Since the torque acting on the pulley is = TR = I , we have
I =
TR
=
TR TR 2 = a /R a
We can compare the measured value of I for the pulley with the theoretical value. We first must find the tension T. From the free-body diagram, Newtons second law of motion is T m1 g = m1a. This means T = m1 ( g + a ) = (1.0 kg)(9.80 m/s 2 3.967 m/s 2 ) = 5.833 N With these values of T and a, we can now find I as:
I = TR 2 (5.833 N)(0.06 m) 2 = = 5.3 103 kg m 2 a (3.967 m/s 2 )
Let us now calculate the theoretical values of I:
hoop about center: I = MR 2 = (2.0 kg)(0.06 m) 2 = 7.2 10 3 kg m 2 1 1 disk about center: I disk = MR 2 = (2.0 kg)(0.06 m) 2 = 3.6 10 3 kg m 2 2 2 < I < I hoop , the mass of the disk is not uniformly distributed. The mass is concentrated near the rim.
Since I disk
12.81.
Model: The angular momentum of the satellite in the elliptical orbit is a constant.
Visualize:
Solve: (a) Because the gravitational force is always along the same direction as the direction of the moment arm vector, the torque = r Fg is zero at all points on the orbit. (b) The angular momentum of the satellite at any point on the elliptical trajectory is conserved. The velocity is perpendicular to r at points a and b, so = 90 and L = mvr. Thus
r Lb = La mvb rb = mva ra vb = a va rb 30,000 km 30,000 km 9000 km = 6000 km and rb = + 9000 km = 24,000 km ra = 2 2 6000 km vb = (8000 m/s) = 2000 m/s 24,000 km
(c) Using the conservation of angular momentum Lc = La , we get
r mvc rc sin c = mva ra vc = a va /sin c rc
rc = (9000 km)2 + (12,000 km) 2 = 1.5 107 m
From the figure, we see that sin c = 12,000 15,000 = 0.80. Thus
6000 km (8000 m/s) vc = = 4000 m/s 0.80 15,000 km
12.82.
Model: The clay balls are particles and undergo a totally inelastic collision. Linear momentum is conserved during the collision. Visualize:
Solve:
measured counterclockwise 1 1 = 180 + tan 1 (1) = 225, and 2 = tan 1 = 26.6. So 2 L = L1 + L2 + m1r1v1 sin 1 + m2 r2v2 sin 2
(a)
The
angle
is
from
r
to
v.
From
geometry,
= ( 0.015 kg )
(
2 m ( 2 m/s ) sin 225 + ( 0.025 kg )
)
(
5 m ( 2.0 m/s ) sin 26.6
)
= 0.020 kg m /s.
2
Note that the signs of L1 and L2 agree with those determined by the right-hand rule. (b) At the instant before the clay balls collide they are located at (1.5 m, 1.0 m). Here, 2 2 = tan 1 = 33.7 1 = 180 + 2 = 146.3 3 So
L = ( 0.015 kg )
( (1.0 m) + (1.5 m) )( 2.0 m/s) sin 213.7 + ( 0.025 kg ) ( (1.0 m ) + (1.5 m ) ) ( 2.0 m/s ) sin 33.7
2 2 2 2
= 0.020 kg m 2 /s (c) The clay balls have a final speed v after the collision. Linear momentum is conserved.
p1i + p2i = p(1+ 2) f
( 0.015 kg )( 2.0 m/s ) + ( 0.025 kg )( 2.0 m/s ) = ( 0.015 kg + 0.025 kg ) v
v = 0.50 m/s. The balls are moving to the left. The angle = 2 from part (b). The angular momentum after the collision is
L = ( 0.040 kg ) Assess:
(
(1.0 m )
2
+ (1.5 m )
2
)(0.50 m/s )sin 33.7 = 0.020 kg m /s
2
Angular momentum is also conserved since net = 0.
12.83.
Model: For the (bullet + block + rod) system, angular momentum is conserved. After the bullet is stuck in the block, the mechanical energy of the system is conserved. Assume the block is small. Visualize:
The origin of the coordinate system was placed at the center-of-mass of the block as it freely hangs from the bottom of the rod. Solve: The initial angular momentum of the system about the pivot is due only to the bullet:
Li = mb vb L = (0.010 kg)vb (1.0 m) = (0.010 kg m)vb The angular momentum immediately after the bullet hits and sticks in the block is equal to I . The moment of inertia is 1 1 I = I rod + I block + I bullet = mR L2 + mB L2 + mb L2 = mR + mB + mb L2 3 3 1 = (1.0 kg) + 2.0 kg + 0.010 kg (1.0 m) 2 = 2.343 kg m 2 3 Angular momentum is conserved in the collision:
Lf = (2.343 kg m 2 ) = Li = (0.010 kg m)vb We need to determine before we can find vb . To find we use the conservation of mechanical energy equation K f + U gf = K i + U gi as the pendulum swings out, which is
12 I + 0 J 2 1 L (0.010 kg + 2.0 kg)(9.8 m/s 2 ) L(1 cos30) + (1.0 kg)(9.8 m/s 2 ) (1 cos30) = (2.343 kg m 2 ) 2 2 2 0 J + (mb + mB ) gyB + mR gyR = The energy equation can be further simplified to
2.6390 kg m 2 /s 2 + 0.6565 kg m 2 /s 2 = (1.1715 kg m 2 ) 2 = 1.677 rad/s
Finally, we can use the conservation of angular momentum equation to obtain the speed of the bullet:
vb = (2.343 kg m 2 )(1.677 rad/s) = 3.9 102 m/s 0.010 kg m
Assess: A speed of 390 m/s for a bullet is reasonable.
12.84.
Model: For the (bullet + door) system, the angular momentum is conserved in the collision.
Visualize:
Solve: As the bullet hits the door, its velocity v is perpendicular to r . Thus the initial angular momentum about the rotation axis, with r = L, is
Li = mBvB L = (0.010 kg)(400 m/s)(1.0 m) = 4.0 kg m 2 /s After the collision, with the bullet in the door, the moment of inertia about the hinges is
1 1 I = I door + I bullet = mD L2 + mB L2 = (10.0 kg)(1.0 m) 2 + (0.010 kg)(1.0 m) 2 = 3.343 kg m 2 3 3
Therefore,
Lf = I = (3.343 kg m 2 ) .
2 2
Using
the
angular
momentum
conservation
equation
Lf = Li (3.343 kg m ) = 4.0 kg m /s and thus = 1.20 rad/s.
12.85.
Model: The mechanical energy of both the hoop (h) and the sphere (s) is conserved. The initial gravitational potential energy is transformed into kinetic energy as the objects roll down the slope. The kinetic energy is a combination of translational and rotational kinetic energy. We also assume no slipping of the hoop or of the sphere. Visualize:
The zero of gravitational potential energy is chosen at the bottom of the slope. Solve: (a) The energy conservation equation for the sphere or hoop K f + U gf = K i + U gi is
1 1 1 1 I (1 ) 2 + m(v1 ) 2 + mgy1 = I ( 0 ) 2 + m(v0 ) 2 + mgy0 2 2 2 2
For the sphere, this becomes
2 1 2 1 2 (v1 )s 2 mR 2 + m(v1 )s + 0 J = 0 J + 0 J + mghs 2 5 2 R
7 2 (v1 )s = gh (v1 )s = 10 gh /7 = 10(9.8 m/s 2 )(0.30 m)/7 = 2.05 m/s 10
1 (v ) 2 1 (mR 2 ) 1 2 h + m(v1 ) 2 + 0 J = 0 J + 0 J + mghhoop h 2 R 2 (v ) 2 hhoop = 1 h g
For the hoop, this becomes
For the hoop to have the same velocity as that of the sphere,
hhoop =
2 (v1 )s (2.05 m/s) 2 = = 42.9 cm g 9.8 m/s 2
The hoop should be released from a height of 43 cm. (b) As we see in part (a), the speed of a hoop at the bottom depends only on the starting height and not on the mass or radius. So the answer is No.
12.86.
Model: Model the turntable as a rigid disk rotating on frictionless bearings. As the blocks fall from above and stick on the turntable, the turntable slows down due to increased rotational inertia of the ( turntable + blocks) system. Any torques between the turntable and the blocks are internal to the system, so angular momentum of the system is conserved. Visualize: The initial moment of inertia is I1 and the final moment of inertia is I 2 .
Solve: The initial moment of inertia is I1 = I disk = 1 mR 2 = 1 (2.0 kg)(0.10 m) 2 = 0.010 kg m 2 and the final 2 2 moment of inertia is
I 2 = I1 + 2mR 2 = 0.010 kg m 2 + 2(0.500 kg) (0.10 m) 2 = 0.010 kg m2 + 0.010 kg m 2 = 0.020 kg m2
Let 1 and 2 be the initial and final angular velocities. Then
Lf = Li 2 I 2 = 1I1 2 = I11 (0.010 kg m 2 )(100 rpm) = = 50 rpm I2 0.020 kg m 2
Model: Model the turntable as a rigid disk rotating on frictionless bearings. For the ( turntable + block) system, no external torques act as the block moves outward towards the outer edge. Angular momentum is thus conserved. Visualize: The initial moment of inertia of the turntable is I1 and the final moment of inertia is I 2 .
Solve: The initial moment of inertia is I1 = I disk = 1 mR 2 = 1 (0.2 kg)(0.2 m) 2 = 0.0040 kg m 2 . As the block 2 2 reaches the outer edge, the final moment of inertia is
12.87.
I 2 = I1 + mB R 2 = 0.0040 kg m 2 + (0.020 kg)(0.20 m) 2 = 0.0040 kg m 2 + 0.0008 kg m 2 = 0.0048 kg m 2 Let 1 and 2 be the initial and final angular velocities, then the conservation of angular momentum equation is
Lf = Li 2 I 2 = 1I1 2 = I11 (0.0040 kg m 2 )(60 rpm) = = 50 rpm I2 (0.0048 kg m 2 )
Assess: A change of angular velocity from 60 rpm to 50 rpm with an increase in the value of the moment of inertia is reasonable.
12.88. Model: Model the merry-go-round as a rigid disk rotating on frictionless bearings about an axle in the center and John as a particle. For the (merry-go-round + John) system, no external torques act as John jumps on the merry-go-round. Angular momentum is thus conserved. Visualize: The initial angular momentum is the sum of the angular momentum of the merry-go-round and the angular momentum of John. The final angular momentum as John jumps on the merry-go-round is equal to I final final . Solve: Johns initial angular momentum is that of a particle: LJ = mJ vJ R sin = mJ vJ R. The angle = 90 since John runs tangent to the disk. The conservation of angular momentum equation Lf = Li is
1 I final final = Ldisk + LJ = MR 2 i + mJ vJ R 2 2 rad 1 2 = (250 kg)(1.5 m) 2 (20 rpm) + (30 kg)(5.0 m/s)(1.5 m) = 814 kg m /s 60 rpm 2 final = 814 kg m 2 /s I final
1 1 I final = I disk + I J = MR 2 + mJ R 2 = (250 kg)(1.5 m) 2 + (30 kg)(1.5 m)2 = 349 kg m 2 2 2 814 kg m 2 /s final = = 2.33 rad/s = 22 rpm 349 kg m 2
12.89.
Model: The toy car is a particle located at the rim of the track. The track is a cylindrical hoop rotating about its center, which is an axis of symmetry. No net torques are present on the track, so the angular momentum of the car and track is conserved. Visualize:
Solve:
The toy cars steady speed of 0.75 m/s relative to the track means that vc vt = 0.75 m/s v c = vt + 0.75 m/s,
where vt is the velocity of a point on the track at the same radius as the car. Conservation of angular momentum implies that Li = Lf 0 = I c c + I t t = ( mr 2 ) c + ( Mr 2 ) t = m c + M t The initial and final states refer to before and after the toy car was turned on. Table 12.2 was used for the track. v v Since c = c , t = t , we have r r 0 = mvc + Mvt
m ( vt + 0.75 m/s ) + Mvt = 0 vt = M ( 0.200 kg ) ( 0.75 m/s ) = ( 0.75 m/s ) = 0.125 m/s m+M ( 0.200 kg + 1.0 kg )
The minus sign indicates that the track is moving in the opposite direction of the car. The angular velocity of the track is v ( 0.125 m/s ) t = t = = 0.417 rad/s clockwise. r 0.30 m In rpm, rev 60 s t = ( 0.417 rad/s ) 2 rad min = 4.0 rpm Assess: The speed of the track is less than that of the car because it is more massive.
12.90.
Model: Model the skater as a cylindrical torso with two rod-like arms that are perpendicular to the axis of the torso in the initial position and collapse into the torso in the final position. Visualize:
Solve: For the initial position, the moment of inertia is I1 = I Torso + 2 I Arm . The moment of inertia of each arm is that of a 66 cm long rod rotating about a point 10 cm from its end, and can be found using the parallel-axis theorem. In the final position, the moment of inertia is I 2 = 1 MR 2 . The equation for the conservation of angular 2
momentum Lf = Li can be written I 2 2 = I11 2 = ( I1 / I 2 )1. Calculating I1 and I 2 , 1 1 I1 = M T R 2 + 2 M A L2 + M A d 2 A 2 12 1 2 1 2 = (40 kg)(0.10 m) + 2 (2.5 kg)(0.66 m) 2 + (2.5 kg) ( 0.33 m + 0.10 m ) = 1.306 kg m 2 2 12 1 1 (1.306 kg m 2 ) I 2 = MR 2 = (45 kg)(0.10 m) 2 = 0.225 kg m 2 2 = (1.0 rev/s) = 5.8 rev /s 2 2 (0.225 kg m 2 )
12.91. Model: Assume that the marble does not slip as it rolls down the track and around a loop-the-loop. The mechanical energy of the marble is conserved. Visualize:
Solve: The balls center of mass moves in a circle of radius R r. The free-body diagram on the marble at its highest position shows that Newtons second law for the marble is
mv12 Rr The minimum height (h) that the track must have for the marble to make it around the loop-the-loop occurs when the normal force of the track on the marble tends to zero. Then the weight will provide the centripetal acceleration needed for the circular motion. For n 0 N, mg + n =
mg = mv 2 v12 = g ( R r ) (R r)
g (R r) r2
Since rolling motion requires v12 = r 212 , we have
12 r 2 = g ( R r ) 12 =
The conservation of energy equation is
1 1 ( K f + U gf ) top of loop = ( K i + U gi )initial mv12 + I12 + mgy1 = mgy0 = mgh 2 2
2 Using the above expressions and I = 5 mr 2 the energy equation simplifies to
1 1 2 g(R r) mg ( R r ) + mr 2 + mg 2( R r ) = mgh h = 2.7 ( R r ) 2 2 2 5 r
12.92.
Model: The Swiss cheese wedge is of uniform densityor at least uniform enough that its center of mass is at the same location as that of a solid piece. To find the angle at which the cheese starts sliding, the cheese will be treated as a particle, and the model of static friction will be used. Visualize:
Solve: The angle at which the cheese starts sliding, S , will be compared to the critical angle c for stability. Use Newtons second law with the free body diagram. ( Fnet ) x = 0 = fs FG sin s
( Fnet ) y = 0 = n FG cos s
With FG = mg , the y-direction equation gives n = mg cos . The cheese starts sliding when s is at its maximum
value. Combining that with the x-direction equation and fs = s n, 0 = s ( mg cos s ) mg sin s s = tan 1 ( s ) = tan 1 ( 0.90 ) = 42 The cheese will start sliding at an angle of 42. The center of mass of the cheese wedge can be found using the result of problem 12.53. There, the center of mass of a triangle with the same proportions as the cheese wedge was found. So xcm is at the center of the cheese ycm ( 30 cm 20 cm ) y = 4.0 cm = cm 12 cm 30 cm Note that here we have measured ycm from the base of the wedge. Stability considerations require that the center of mass be no further than the left corner of the wedge. At the critical angle geometry shown in the figure above, the right triangle formed by the wedges center of mass, lower left corner, and center point of the base is a 45-45-90 triangle. So c = 45.
Assess: The cheese will slide first as the incline reaches 42. It would not topple until the angle reaches 45. So Emily is correct.
wedge (by symmetry). The ycm can be found by proportional reasoning.
12.93.
Model: Define the system as the rod and cube. Energy and angular momentum are conserved in a perfectly elastic collision in the absence of a net external torque. The rod is uniform. Visualize: Please refer to Figure CP12.93. Solve: Let the final speed of the cube be vf , and the final angular velocity of the rod be . Energy is conserved, and angular momentum around the rods pivot point is conserved. 121 1 Ei = Ef mv0 = mvf2 + I rod 2 2 2 2 d d Li = Lf mv0 = mvf + I rod 2 2 This is two equations in the two unknowns vf and . From Table 12.2,
I rod = From the angular momentum equation,
1 1 1 Md 2 = ( 2m ) d 2 = md 2 12 12 6 = 3 ( v0 vf ) d
d v0 = vf + 3
Substituting into the energy equation, 1 1 11 2 9 mv0 2 = mvf 2 + md 2 2 ( v0 vf ) 2 2 2 6 d 3 2 v0 2 = vf 2 + ( v0 vf ) 2 6 1 0 = vf 2 v0vf + v0 2 5 5 This is a quadratic equation in vf . The roots are
v 2 6 6 v0 v0 4 0 5 5 2 5 3 vf = = v0 v0 2 5 5 1 v = 5 0 v0 1 The answer vf = v0 means the ice cube missed the rod. So vf = v0 to the right. 5
2
12.94.
Model: The clay ball is a particle. The rod is a uniform thin rod rotating about its center. Angular momentum is conserved in the collision. Visualize:
Solve: This is a two-part problem. Angular momentum is conserved in the collision, and energy is conserved as the ball rises like a pendulum. The angular momentum conservation equation about the rods pivot point is Li = Lf mv0 r = ( I ball+rod )
L 1 = 15 cm. The rod and ball are a composite object. From Table 12.2, I rod = ML2 , so 2 12 1 L2 1 L2 M I ball+rod = I ball + I rod = mr 2 + ML2 = m + ML2 = m + 12 4 12 4 3 v 2v If vf is the final velocity of the clay ball, = f = f since the ball sticks to the rod. Thus r L 2 mv0 L L M 2v = m + f 2 4 3 L
Note r =
mv0 ( 0.010 kg )( 2.5 m/s ) = 0.714 m/s = M ( 0.075 kg ) m+ ( 0.010 kg ) + 3 3 Energy is conserved as the clay ball rises. Compare the energy of the ball-rod system just after the collision to when the ball reaches the maximum height. Note that the center of mass of the rod does not change position. 1 Ei = Ef ( I rod + ball ) 2 = mgh 2 Thus vf =
1 L2 M 2vf M 2 m + = mgh vf m + = mgL (1 cos ) 2 4 3 L 3 2 v M 1 f m + = cos mgL 3 Using the various values, cos = 0.393 = 67. L Assess: The clay ball rises h = (1 cos ) = 9.1 cm. This is about 2/3 of the height of the pivot point, and is 2 reasonable.
2
12.95. Model: Because no external torque acts on the star during gravitational collapse, its angular momentum is conserved. Model the star as a solid rotating sphere. Solve: (a) The equation for the conservation of angular momentum is
2 2 Li = Lf I ii = I f f mRi2 i = mRf2 f 5 5 Rf = Ri
i f
The angular velocity is inversely proportional to the period T. We can write
Rf = Ri
0.10 s Tf = ( 7.0 108 m ) = 1.3749 105 m = 137 km 2.592 106 s Ti
(b) A point on the equator rotates with r = Rf . Its speed is
v= 2 Rf 2 (137,490 m) = = 8.6 106 m/s 0.10 s T
Model: For the ( turntable + bicycle wheel + professor) system the angular momentum is conserved because the turntable is frictionless and no external torques act on the system. Solve: (a) Nothing happens. The bicycle wheel already has an angular momentum and nothing changes for the wheel when it is handed to the professor. So, nothing happens to the professor. (b) The initial angular momentum is 12.96.
180 2 rad 2 ( Lwheel )i = I = (mR 2 ) = (4.0 kg)(0.32 m) 2 = 7.72 kg m /s s 60 When the wheel is turned upside down, the angular momentum of the wheel becomes
( Lwheel )f = 7.72 kg m 2 /s
Since the initial and final total angular momentum should be equal, the professor must acquire some angular momentum. From the angular momentum conservation equation,
Lprof + ( Lwheel ) f = ( Lwheel )i Lprof = ( Lwheel )i ( Lwheel )f = 7.72 kg m 2 /s (7.72 kg m 2 /s) = 15.44 kg m 2 /s
Let us now calculate the angular velocity of the turntable by using Lprof = I total , where I total = I turntable + I body + I arms +
I wheel in hand . We assume that the axis of the professor and turntable also become the axis of the arms and wheel (i.e., no leaning).
1 1 2 I turntable = mT RT = (5.0 kg)(0.25 m) 2 = 0.156 kg m 2 2 2 1 2 I body = mbody R = (70 kg)(0.125 m) 2 = 0.547 kg m 2 2 1 2 I arms = 2 marms R 2 = (2.5 kg)(0.45 m) 2 = 0.338 kg m 2 3 3 I wheel in hand = mwheel r 2 = (4.0 kg)(0.45 m) 2 = 0.810 kg m 2 Thus, I total = (0.156 + 0.547 + 0.338 + 0.810) kg m 2 = 1.851 kg m 2 . Using this value for I total , we can now find to be
=
Lprof I total
=
15.44 kg m 2 /s = 8.34 rad/s = 79.7 rpm 1.851 kg m 2
12-1