08 homework BAUTISTA, ALDO Due: Oct 23 2006, 4:00 am
1
Mechanics - Basic Physical Concepts
Math: Circle: 2 r, r 2 ; Sphere: 4 r2 , (4/ r3 3) b2 Quadratic Eq.: a x2 + b x + c = 0, x = b 2 a4 a c
Cartesian and polar coordinates: y x = r cos , y = r sin , r 2 = x2 + y 2 , tan = x Trigonometry: cos cos + sin sin = cos( ) sin + sin = 2 sin + cos 2 2 cos + cos = 2 cos + cos 2 2 sin 2 = 2 sin cos , cos 2 = cos2 sin2 1 cos = 2 sin2 2 , 1 + cos = 2 cos2 2 Vector algebra: A = (Ax , Ay ) = Ax + Ay Resultant: R = A + B = (Ax + Bx , Ay + By ) Dot: A B = A B cos = Ax Bx + Ay By + Az Bz Cross product: = k , k = , k = Ax Bx Ay By k Az Bz
B F s = F s cos = F s A F ds
(in Joules)
Eects due to work done: Fext = m a Fc fnc Wext |AB = KB KA + UB UA + Wdiss |AB B Kinetic energy: KB KA = A m a ds, K = 1 m v 2 2
B K (conservative F ): UB UA = A F ds 1 k x2 Uspring = 2 Ugravity = m g y , From U to F : Fx = U , Fy = U , Fz = U x y z
Fspring = U = k x x U = 0, 2 U > 0 stable, < 0 unstable Equilibrium: x x2 W Power: P = ddt = F v = F v cos = F v (Watts)
Fgravity = U = m g , y
Collision
C =AB =
d d 1 Calculus: dx xn = n xn1 , dx ln x = x , d sin = cos , d cos = sin , d d d dx const = 0
C = A B sin = A B = A B , use right hand rule
Measurements
Impulse: I = p = pf pi tif F dt Momentum: p = m v x Two-body: xcm = m1m1 +m2 x2 1 +m2 pcm M vcm = p1 + p2 = m1 v1 + m2 v2 Fcm F1 + F2 = m1 a1 + m2 a2 = M acm K1 + K2 = K1 + K2 + Kcm Two-body collision: pi = pf = (m1 + m2 ) vcm vi = vi vcm , vi = vi + vcm Elastic: v1 v2 = (v1 v2 ), vi = vi , vi = 2 vcm vi Many body center of mass: rcm = Force on cm: Fext = dp = M acm , dt
mi r i = mi r dm mi
t
Dimensional analysis: e.g., 2 F = m a [M ][L][T ]2 , or F = m v [M ][L][T ]2 r N (a x + b) = a N x + b N Summation: i i=1 i=1 i
p=
pi
Rotation of Rigid-Body
Motion
One dimensional motion: v = d s , a = d v dt dt s s v v Average values: v = tf tii , a = tf tii f f One dimensional motion (constant acceleration): v (t) : v = v0 + a t + s(t) : s = v t = v0 t + 1 a t2 , v = v02 v 2 2 = v2 + 2 a s v (s) : v 0 Nonuniform acceleration: x = x0 + v0 t + 1 a t2 + 2 1 j t3 + 1 s t4 + 1 k t5 + 1 p t6 + . . ., (jerk, snap,. . .) 6 24 120 720 Projectile motion: trise = tf all =
ttrip v0y 2=g 1 g t2 , R = v t h = 2 f all ox trip 2 1 Circular: ac = v , v = 2 T r , f = T (Hertz=s1 ) r 2 + a2 Curvilinear motion: a = at r
Isphere = 2 M R2 , Ispherical shell = 2 M R2 5 3 I = M (Radius of gyration)2 , I = Icm + M D2 Kinetic energies: Krot = 1 I 2 , K = Krot + Kcm 2 Angular momentum: L = r m v = r m r = I Torque: = d L = m d v r = F r = I d = I dt dt dt Wext = K +U + Wf , K = Krot + 1 m v 2 , 2 P =
s Kinematics: = r , = v , = at r r 2 Moment of inertia: I = mi ri = r2 dm 2 2 Idisk = 1 M R2 , Iring = 1 M (R1 + R2 ) 2 2 1 M 2, I 1 M (a2 + b2 ) Irod = 12 rectangle = 12
Rolling, angular momentum and torque
I Rolling: K = 1 Ic + M R2 2 = 1 Rc + M v 2 2 2 2 Angular momentum: L = r p, L = r p = I
Relative velocity: v = v + u
Law of Motion and applications
Force: F = m a, Fg = m g, F12 = F21
2 Circular motion: ac = v , v = 2 T r = 2 r f r Friction: Fstatic s N Fkinetic = k N
Torque: = d L = r d p = r F , = r F = I dt dt 1 Gyroscope: p = d = L d L = L = m g h I dt dt
Static equilibrium
Equilibrium (concurrent forces):
i Fi = 0
Energy
Work (for all F): W = WAB = WB WA
i = 0 Fi = 0, about any point Subdivisions: rcm = mA rAcm +mB rBcm mA +mB Elastic modulus = stress/strain stress: F/A strain: L/L, x/h, V /V
homework 08 BAUTISTA, ALDO Due: Oct 23 2006, 4:00 am Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 part 1 of 1 10 points Initially a wheel rotating about a xed axis at a constant angular deceleration of 0.7 rad/s2 has an angular velocity of 2.91 rad/s and an angular position of 8.4 rad. What is the angular position of the wheel after 4 s? Correct answer: 14.44 rad (tolerance 1 %). 1 Apply 0 = 0 t + t2 . 2 problem it reads 1 2 1 = 1 t2 + t 2 22 and by taking 1 to the right side, we nd the answer 2 . Explanation: For our
2
Solution: First nd the speed attained before the wheel begins to slow down. With 0 = 0 rad/s , we have 1 = 1 t1 = (7.06 rad/s2 ) (4.46 s) = 31.4876 rad/s . Next, consider the wheel as it comes to rest. Now 1 = 31.4876 rad/s, 2 = 0 rad/s , so
2 2 2 = 0 = 1 + 2 2 1 2 1 2 = 2 1 (31.4876 rad/s)2 = 2 (81.6815 rad) = 6.06912 rad/s2 2 = 6.06912 rad/s2 .
(1)
Question 3 part 2 of 2 10 points Determine the time needed to bring the wheel to rest. Correct answer: 5.18817 s (tolerance 1 %).
Question 2 part 1 of 2 10 points A grinding wheel, initially at rest, is rotated with constant angular acceleration of 7.06 rad/s2 for 4.46 s. The wheel is then brought to rest, with uniform deceleration, in 13 rev. Find the magnitude of the angular deceleration required to bring the wheel to rest. Correct answer: 6.06912 rad/s2 (tolerance 1 %). Explanation: Let : 1 = 7.06 rad/s2 , t = 4.46 s , and 1 = 81.6815 rad . Basic Concepts = 0 + t 2 2 = 0 + 2 = t.
Explanation: Since 1 = 31.4876 rad/s , 2 = 0 rad/s , and 2 from Eq. 1, we have 2 = 1 + 2 t t= 1 2 2 1 = 1 2 (81.6815 rad) = (31.4876 rad/s) = 5.18817 s .
Question 4 part 1 of 3 10 points A beetle takes a joy ride on a pendulum. The string supporting the mass of the pendulum is 184 cm long. If the beetle rides through a swing of 28 , how far has he traveled along the path of the pendulum?
homework 08 BAUTISTA, ALDO Due: Oct 23 2006, 4:00 am Correct answer: 89.9194 cm (tolerance 1 %). Explanation: Arc length is dened as s = r = (184 cm)(0.488692 rad) = 89.9194 cm , where theta is in radians. Dimensional analysis for radians cm = cm 180
3
Linear and angular velocity are related by v = r = (184 cm)(2.9 rad/s) = 533.6 cm/s , where is in radians per unit time. Question 7 part 1 of 2 10 points Four particles with masses 3 kg, 3 kg, 3 kg, and 2 kg are connected by rigid rods of negligible mass as shown. y 3 kg 2 kg
Question 5 part 2 of 3 10 points If the he swings through same angle 28 , what is the displacement experienced by the beetle? Correct answer: 89.0273 cm (tolerance 1 %). Explanation: Using the law of cosines, we have R= =r r 2 + r 2 2 r r cos 2 (1 cos )
5m
O
x
3 kg
7m
3 kg
= (184 cm) 2 [(1 cos(28 )] = 89.0273 cm . Alternative Solution: Divide the isosceles triangle in half. Then R = 2.0 r sin 2 = 2.0 (184 cm) sin (0.244346 rad) = 89.0273 cm . Question 6 part 3 of 3 10 points If the pendulum at some instant is swinging at 2.9 rad/s, how fast is the beetle traveling? Correct answer: 533.6 cm/s (tolerance 1 %). Explanation:
The origin is at the center of the rectangle, which is 7 m wide and 5 m long. If the system rotates in the xy plane about the z axis (origin, O ) with an angular speed of 8 rad/s, calculate the moment of inertia of the system about the z axis. Correct answer: 203.5 kg m2 (tolerance 1 %). Explanation:
Let : m1 m2 m3 m4 w
= 3 kg , = 3 kg , = 3 kg , = 2 kg , = 7 m, = 5 m,
top left bottom left bottom right top right and
From I=
j 2 mj r j ,
homework 08 BAUTISTA, ALDO Due: Oct 23 2006, 4:00 am where in this case all distances are equal to r= w 2
2 2
4
3. max = 4. max = 5. max = 6. max = 7. max = 8. max = 9. max = 10. max =
+
2
5g 6L 12 g 5L 6g 5L g 4L 4g 5L 8g correct 3L 3g 2L 2g L
2
2
(7 m) (5 m) = + 2 2 = 4.30116 m , we obtain
Iz = [m1 + m2 + m3 + m4 ] r 2 = [(3 kg) + (3 kg) + (3 kg) + (2 kg)] (4.30116 m)2 = 203.5 kg m2 . Question 8 part 2 of 2 10 points Find the rotational energy of the system. Correct answer: 6512 J (tolerance 1 %).
Explanation: The rotational energy of the system is K= 1 I 2 2 1 = (203.5 kg m2 ) (8 rad/s)2 2 = 6512 J . Question 9 part 1 of 1 10 points
Explanation: The mechanical energy of the system is conserved. Measuring heights from the point at the bottom of the rod when it is vertical, the initial potential energy of the system is Ui = (3 m) g L. The potential energy at the bottom of the swing, Uf = 2 m g (L/2) = m g L . To calculate the nal kinetic energy, we need the moment of inertia of the system, I = 2 m (L/2)2 + m L2 = 3 m L2 /2 . Therefore, 3mgL = mg + max = 13 m L2 2 , 22 8g . 3L
A massless rod of length L has a mass 2 m fastened at its center and another mass m fastened at one end. On the opposite end from the mass m, the rod is hinged with a frictionless hinge. The rod is released from rest from an initial horizontal position; then it swings down. What is the angular velocity max as the rod swings through its lowest (vertical) position? 1. max = 2. max = g 2L g L
Question 10 part 1 of 1 10 points An 873 kg car is held in place by an unrealiable winch. The gearbox of the winch breaks and at that moment the car-drum system free-falls from rest, shown in the gure. During the cars fall, there is no slipping between the (massless) rope, the pulley, and the winch drum. The moment of inertia of the winch drum is 257 kg m2 and that of the pulley is 14 kg m2 .
homework 08 BAUTISTA, ALDO Due: Oct 23 2006, 4:00 am The radius of the drum is 86 cm and the radius the of pulley is 9 cm . The acceleration due to gravity is 9.81 m/s2 . 14 kg m2 Question 11 part 1 of 4 10 points
5
86 cm
257 kg m2 873 kg
Consider a rod of length L and mass m which is pivoted at one end. An object with mass m is attached to the free end of the rod. The acceleration of gravity g = 9.8 m/s2 . Note: Contray to the diagram shown below, consider the mass at the end of the rod to be a point particle.
9m
m
Find the speed of the car as it hits the ground. Correct answer: 7.23018 m/s (tolerance 1 %). Explanation: Let : Id = 257 kg m2 , Ip = 14 kg m2 , rd = 86 cm = 0.86 m , rp = 9 cm = 0.09 m , mb = 873 kg , and h = 9 m . Applying conservation of mechanical energy (Ki = Uf = 0) and v = r , 1 1 1 2 2 m v 2 + Id d + Ip p m g h = 0 2 2 2 1 1 v2 1 v2 m v 2 + Id 2 + Ip 2 m g h = 0 2 2 2 rp rd v= 2 m g h Id Ip m+ 2 + 2 rd rp 2 (873 kg) (9.81 m/s2 ) (9 m) (257 kg m2 ) (14 kg m2 ) (873 kg) + + (0.86 m)2 (0.09 m)2
C L
28
m
Determine the moment of inertia, I , of the system with respect to the pivot point. 1. None of these. 13 m L2 12 4 3. I = m L2 correct 3 2. I = 4. I = L2 3 m L2 2 5 6. I = m L2 4 5 7. I = m L2 3 5. I = Explanation: Basic Concepts: The moment of inertia, the center of mass and the energy conservation. The rotational kinetic energy is Krot = 1 I 2 . 2
=
= 7.23018 m/s .
homework 08 BAUTISTA, ALDO Due: Oct 23 2006, 4:00 am Solution: The moment of inertia of the rod, Irod , with respect to the pivot point is Irod = 1 m L2 , 3 Question 13 part 3 of 4 10 points
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and the moment of inertia Im of the mass m with respect to the pivot point is Imass = m L2 . Then, the moment of inertia of the system I is I = Irod + Im 1 = m L2 + m L 2 3 4 = m L2 . 3 Question 12 part 2 of 4 10 points Note: The length C in the gure represents the location of the center-of-mass of the rod plus mass system (but is not drawn to scale). Determine the position of the center of mass from the pivot point; i.e., nd C . 1. C = 5 L 8
The unit is released from rest in the horizontal position. What is the kinetic energy of the unit when the rod momentarily has a vertical orientation? 1. K = 5 mgL 2
2. None of these. 3. K = 3 m g L correct 2
4. K = 2 m g L 5. K = m g L 1 mgL 2 Explanation: The potential energy, U , released can be computed in two ways. Method 1: 6. K = U = U
cm of rod
2. C = L 3. None of these. 3 L correct 4 1 5. C = L 2 7 6. C = L 8 Explanation: 4. C =
+ U
mass
L = mg + mgL 2 3 = mgL. 2 Method 2: U |rod+m = U |cm of = (2 m) g = 3 mgL. 2
rod+m
3 L 4
1 The center of mass of the rod is L. Then, 2 the center of mass of the rod plus mass system is 1 L+L m 2 C= m+m 1 1 = L+ L 4 2 3 = L. 4
Question 14 part 4 of 4 10 points Given: The kinetic energy is KV at the vertical position and the moment of inertia of the system is I . Find the angular velocity of the system in terms of KV and I .
homework 08 BAUTISTA, ALDO Due: Oct 23 2006, 4:00 am KV I KV I = 4 m, x1 = 0.9 m , xd = 3.1 m . and
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1. = 2. =
Consider rotational equilibrium about B : FB xd A FA x2 = x1 B mg Mg x2 x3
3. None of these. 4. = 5. = KV 2I
2 KV correct I 2 KV 6. = I Explanation: The rotational kinetic energy of the system is given by 1 KV = I 2 , 2 where is the angular velocity of the system. Then, 2 KV . = I Question 15 part 1 of 2 10 points The diving board shown in gure has a mass of 35 kg. The acceleration of gravity is 9.81 m/s2 .
x1 = 1.1 m 2 Applying rotational equilibrium at point B , = F A x1 m g x 2 M g x d = 0 FA = x2 m + x d M g x1 (1.1 m) (35 kg) + (3.1 m) (67 kg) = 0.9 m 1 kN (9.81 m/s2 ) 1000 N = 2.68358 kN .
The direction of FA is downward, so it is tension. Question 16 part 2 of 2 10 points
0.9 m A B
3.1 m
Find the force on the support B at that same instant. Correct answer: 3.6842 kN (tolerance 1 %). Explanation: Applying translational equilibrium, Fy = F B F A m g M g = 0 FB = FA + (m + M ) g = 2.68358 kN + (35 kg + 67 kg) (9.81 m/s2 ) 1 kN 1000 N = 3.6842 kN .
Find the force on the support A when a 67 kg diver stands at the end of the diving board. Correct answer: 2.68358 kN (tolerance 1 %). Explanation: Let : m = 35 kg , M = 67 kg ,
homework 08 BAUTISTA, ALDO Due: Oct 23 2006, 4:00 am The direction of FB is upward, so it is compression. T1 Question 17 part 1 of 3 10 points Two masses, 4.7 kg and 1.4 kg, are suspended by a pulley with a radius of 14 cm and a mass of 3 kg as shown in the gure. The cord has negligible weight and causes the pulley to rotate without slipping. Treat the pulley as a uniform disk. The acceleration of gravity is 9.8 m/s2 . T2
8
4.7 kg m1 g
1.4 kg m2 g
a
Solution: The moment of inertia of the pulley (treated as a uniform disk) about its axis of rotation is I= 1 M r 2 = 0.0294 kg m2 , 2
14 cm 3 kg 4.7 kg 4.54 m 1.4 kg What is the angular acceleration of the pulley? Correct answer: 30.3947 rad/s2 (tolerance 1 %). Explanation:
where M is the mass of pulley and r is its radius. The two masses are attached to one another via a cord. We assume the cord does not stretch, so the motion of the two are coupled such that the displacements are equal in magnitude and opposite. The accelerations are also equal and opposite. Dening up as positive, and since m1 > m2 , the acceleration a will be downwards for m1 and upwards for m2 . The forces are as follows m1 : T1 m1 g = m1 a T2 m2 g = +m2 a
T1 = m1 g m 1 a m2 : T2 = m2 g + m 2 a
Let :
M = 3 kg , R = 0.14 m , m1 = 4.7 kg , m2 = 1.4 kg , h = 4.54 m , v = R, 1 I = M R2 , and 2 1 1 Kdisk = I 2 = M v 2 . 2 4
The net force on the pulley is T1 T2 . Hence the net torque is = I = (T1 T2 ) r , so I = [(m1 g m1 a) (m2 g + m2 a)] r But the cord causes the pulley to rotate witha out slipping, so = r I = [(m1 m2 ) g (m1 + m2 ) r ] r
Consider the free body diagrams
[I + (m1 + m2 ) r 2 ] = (m1 m2 ) r g
a
homework 08 BAUTISTA, ALDO Due: Oct 23 2006, 4:00 am = [m1 m2 ] r g I + [m1 + m2 ] r 2 [m1 m2 ] g = M r + m1 + m2 2 [(4.7 kg) (1.4 kg)] = (3 kg) + (4.7 kg) + (1.4 kg) 2 9.8 m/s2 0.14 m = 30.3947 rad/s2 . Question 18 part 2 of 3 10 points What is the magnitude of the tension in the cord on the side of the m2 = 1.4 kg mass? Correct answer: 19.6774 N (tolerance 1 %). Explanation: From the above we see that the tension on the 1.4 kg mass side is T2 = m 2 g + m 2 a = m2 (g + r ) = 19.6774 N . Question 19 part 3 of 3 10 points Hint: Check your geometry. The masses start from rest h = 4.54 m vertically apart. What is the speed of the m2 = 1.4 kg mass when the two masses pass each other? Correct answer: 4.39533 m/s (tolerance 1 %). Explanation: The displacement when the two masses pass one another is one half of their initial separation 1 1 h h = a t2 t = , 2 2 a since the masses start from rest. Finally v = at h =a a = ah = rh = (30.3947 rad/s2 )(0.14 m)(4.54 m) = 4.39533 m/s . Question 20 part 1 of 1 10 points
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A ywheel of radius 0.61 m and moment of inertia of 18 kg m2 rotates initially at a rate of 6.6 revolutions/sec. If a force of 4.9 N is applied tangentially to the ywheel to slow it down, how much work will be done by this force in bringing the ywheel to a stop? 1. 392.04 J 2. 784.08 J 3. 15477.1 J correct 4. 746.442 J 5. 30954.2 J 6. 118.8 J 7. 2.989 J Explanation: Use work-energy principle, we have W ork = K = So, W= 1 (18 kg m2 ) 2 4 2 (6.6 revolutions/sec)2 = 15477.1 J . Question 21 part 1 of 2 10 points Consider the problem of the solid sphere rolling down an incline without slipping. The 1 1 I 2 = I (2 )2 f 2 . 2 2
homework 08 BAUTISTA, ALDO Due: Oct 23 2006, 4:00 am incline has an angle . The spheres length up the incline is , and its height is h. At the beginning, the sphere rests on the very top of the incline. The sphere has mass M and radius R. The acceleration of gravity is 9.8 m/s2 . Hint: The moment of inertia of a sphere with respect to an axis through its center is 2 M R2 . 5 Choose the instantaneous axis through the contact point P as the axis of origin for the torque equation. R
10
N
f m g sin m g cos
With the origin at the point of contact, : M g R sin = I , (1)
M
h
The acceleration of the center of mass is 1. a = 2. a = 3. a = 4. a = 5. a = 6. a = 7. a = 8. a = 3 7 2 7 5 7 3 5 3 5 5 7 2 7 3 7 g sin . g cos . g cos . g cos . g sin . g sin . correct g sin . g cos .
where I is obtained using the parallel-axis theorem 2 7 I = M R2 + M R2 = M R2 . (2) 5 5 If the sphere rolls without slipping, a (3) = . R Now substituting I from Eq. 2 and from Eq. 3 into Eq. 1, we have 7 a M g R sin = M R2 , so 5 R 5 a = g sin . 7 Question 22 part 2 of 2 10 points The minimum coecient of friction such that the sphere rolls without slipping is 3 sin . 7 2 2. = tan . correct 7 2 3. = sin . 7 5 4. = tan . 7 2 5. = cos . 7 3 6. = tan . 7 3 7. = cos . 5 5 8. = cos . 7 Explanation: 1. =
Explanation: Basic Concepts: F =ma
= I .
homework 08 BAUTISTA, ALDO Due: Oct 23 2006, 4:00 am The net force along the direction of the incline is F = M g sin f = M 5 g sin 7 ,
11
where f N = M g cos is the minimum no slipping criterion. Then M g sin M g cos = M cos = = 5 g sin 7 5 1 sin , 7
so
2 tan . 7