24 oldhomewk PAPAGEORGE, MATT Due: Apr 1 2008, 4:00 am we obtain Question 1, chap -1, sect -1. part 1 of 2 10 points Four particles with masses 3 kg, 3 kg, 3 kg, and 8 kg are connected by rigid rods of negligible mass as shown. 8 kg 3 kg y Iz = [m1 + m2 + m3 + m4 ] r 2 = [(3 kg) + (3 kg) + (3 kg) + (8 kg)] (3.60555 m)2 = 221 kg m2 .
1
Question 2, chap -1, sect -1. part 2 of 2 10 points 6m O x Find the rotational energy of the system. Correct answer: 7072 J (tolerance 1 %). Explanation: The rotational energy of the system is 3 kg 4m 3 kg K= 1 I 2 2 1 = (221 kg m2 ) (8 rad/s)2 2 = 7072 J .
The origin is at the center of the rectangle, which is 4 m wide and 6 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: 221 kg m2 (tolerance 1 %). Explanation: Let : m1 = 3 kg , m2 = 3 kg , m3 = 3 kg , m4 = 8 kg , w = 4 m, = 6 m, From I=
j 2 mj rj ,
Question 3, chap 12, sect 5. part 1 of 2 10 points The rigid object shown is rotated about an axis perpendicular to the paper and through center point O. The total kinetic energy of the object as it rotates is 6.2 J. 60 kg 4m O 8m 60 kg What is the moment of inertia of the object? Neglect the mass of the connecting rods and treat the masses as point masses. Correct answer: 5760 kg m2 (tolerance 1 %). Explanation: 4m 30 kg 8m
top left bottom left bottom right top right and
30 kg
where in this case all distances are equal to r= w 2
2
+
2
2
2
(6 m) (4 m) + = 2 2 = 3.60555 m ,
2
oldhomewk 24 PAPAGEORGE, MATT Due: Apr 1 2008, 4:00 am
2
Let : M = 30 kg , L = 4 m , and K = 6.2 J . Basic Concepts: I=
2 mi ri
an angle of 59 with the horizontal. The rod is released from rest at an angle of 59 with the horizontal, as shown in the gure below The acceleration of gravity is 9.8 m/s2 . Hint: The moment of inertia of the rod 1 about its center-of-mass is Icm = m 2 . 12
Solution: The moment of inertia of the rigid body is given by I=
2 mi ri
59
= 2 2 M (L)2 + 2 M (2 L)2 = 12 M L2 = 12 (30 kg) (4 m)2 = 5760 kg m2 . Question 4, chap 12, sect 5. part 2 of 2 10 points What is the angular velocity of the object? Correct answer: 0.046398 rad/s (tolerance 1 %). Explanation: Since the rotational kinetic energy is given 1 by KR = I 2 , we may then solve for . 2 = = 2 KR I 2 (6.2 J) (5760 kg m2 ) Solution: O 2m
1.4 kg 2m
What is the angular speed of the rod at the instant the rod is in a horizontal position? Correct answer: 2.78461 rad/s (tolerance 1 %). Explanation: Let : = 2 m , d = = 2 m, = 59 , and m = 1.4 kg . Basic Concepts: KR = 1 I 2 2
=rF =I Ki + Ui = Kf + Uf .
= 0.046398 rad/s . Question 5, chap 12, sect 5. part 1 of 1 10 points A uniform rod of mass 1.4 kg is 2 m long. The rod is pivoted about a horizontal, frictionless pin at the end of a thin extension (of negligible mass) a distance 2 m from the center of mass of the rod. Initially the rod makes
I = Icm + m d2 1 = m 2 + m 2 12 13 m 2 = 12 13 (1.4 kg) (2 m)2 = 12 = 6.06667 kg m2 .
(1)
oldhomewk 24 PAPAGEORGE, MATT Due: Apr 1 2008, 4:00 am Since the rod is uniform, its center of mass is located a distance from the pivot. The vertical height of the center of mass above horizontal is sin . Using conservation of energy and substituting I from Eq. 1, we have Kf = Ui 1 I 2 = m g sin 2 1. I = 4 3 5 2. I = 3 3 3. I = 2 5 4. I = 4 m L2 correct m L2 m L2 m L2
3
13 m 2 2 = m g sin 24 24 g sin 2 = 13 24 g sin = 13 = 24 (9.8 m/s2 ) sin(59 ) 13 (2 m)
5. None of these. (2) 6. I = 13 m L2 12
7. I = L2 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 Solution: The moment of inertia of the rod, Irod , with respect to the pivot point is Irod = 1 m L2 , 3
= 2.78461 rad/s . Question 6, chap 12, sect 5. part 1 of 2 10 points Consider a rod of length L and mass m which is pivoted at one end. The moment of inertia of the rod about an end is Irod = 1 m L2 . An object with mass m is attached 3 to the free end of the rod. The acceleration of gravity g = 9.8 m/s2 . Note: Contrary to the diagram shown below, consider the mass at the end of the rod to be a point particle.
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 L2 3 4 = m L2 . 3 Question 7, chap 12, sect 5. part 2 of 2 10 points
m C L
25
m
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
Determine the moment of inertia, I, of the system with respect to the pivot point.
2. None of these.
oldhomewk 24 PAPAGEORGE, MATT Due: Apr 1 2008, 4:00 am 7 L 8 1 4. C = L 2 3. C = 5. C = L 3 L correct 4 Explanation: 6. C = where m
4
dm , and dm = y dx , where mass is the areal density of the plate. area Solution: Let (x1 , y1 ) = (0 m, 0 m) (x2 , y2 ) = (2 m, 0 m) (x3 , y3 ) = (0 m, 4 m) . The equation for the hypotenuse is y y2 y3 y2 = . x x2 x3 x 2 The slope of the hypotenuse is s= y3 y2 x3 x2 4m0m = 2 . = 0m2m
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 8, chap 12, sect 5. part 1 of 2 10 points
(1)
Rewriting the equation, we have y = s (x x2 ) + y2 = (2) (x 2 m) + 0 m = s (x x2 ) ,
(2)
A uniform at plate of metal is situated in the reference frame shown in the gure below. 5 4 3 2 1 0 0 12 3 4 5 6 7 8 9 10 x (m) (m)
where y2 = 0 . The x-coordinate of the center of mass is
x2
xcm =
x2 x1 x1
x y dx
x2
y dx x s (x x2 ) dx
y
=
x1
x2
s (x x2 ) dx
x1 x2
Calculate the x coordinate of the center of mass of the metal plate. Correct answer: 0.666667 m (tolerance 1 %). Explanation: Basic Concept: The center of mass xcoordinate is x dm x m ,
x (x x2 ) dx =
0 x2
(x x2 ) dx
0
13 1 x (x2 ) x2 3 2 = 12 x (x2 ) x 2 13 1 x2 (x2 ) x2 2 2 =3 12 x (x2 ) x2 22
x2
0
oldhomewk 24 PAPAGEORGE, MATT Due: Apr 1 2008, 4:00 am = x3 2 3 x2 2 1 = x2 3 1 = (2 m) 3 = 0.666667 m . = 0.25 kg/m2 , (3)
5
Vertical Center-of-Mass: Using Eq. 3, the y-coordinate of the center of mass of the metal plate is 1 y = y3 3 1 = (4 m) 3 = 1.33333 m . Question 9, chap 12, sect 5. part 2 of 2 10 points Given: The mass of the plate is m = 1 kg . Calculate the moment of inertia of the triangle with the y-axis as the axis of rotation. Correct answer: 0.666667 kg m2 (tolerance 1 %). Explanation: m Areal Density: The areal density = A of the plate is m = x2 y dx
x1
y3 from Eq. 1. The area of a x2 1 1 triangle is A = (base) (height) = x2 y3 , 2 2 which agrees with Eq. 4. Conventional Solution: The moment of x=x inertia Iy cm of the triangle about a y-axis through its center of mass x-coordinate can be accomplished in two steps. where s =
x=x First: The moment of inertia Iy cm about its center of mass (xcm , ycm ) is determined.
Second: The parallel axis theorem is used. The moment of inertia about the center of x=x mass Iy cm is
x=x Iy cm x2
=
x1
(x xcm )2 dm
x2
=
x1
(x xcm )2 y dx (x xcm )2 (x x2 ) dx x3 [2 xcm + x2 ] x2 + +[x2 2 xcm x2 ] x cm [x2 x2 ] dx cm
x2
= s
0 x2
= s
0
= s
= s
0
m
x2
(x x2 ) dx m
x
14 1 x [2 xcm + x2 ] x3 4 3 1 + [x2 + 2 xcm x2 ] x 2 cm
x2 0
=
=
=
= =
2 12 x (x2 ) x 2 0 m 12 y3 x (x2 ) x2 x2 22 m m = 1 y3 2 1 x x2 y 3 2 x2 2 2 2m x2 y 3 2 (1 kg) (2 m) (4 m)
[x2 x2 ] x cm 14 1 = s x [2 xcm + x2 ] x3 2 42 3 1 + x2 + 2 xcm x2 x2 2 2 cm x2 x2 cm 2 = (4) y3 14 x x2 12 2 1 1 2 + xcm x3 x2 x2 2 cm 3 2 12 2 x2 xcm x2 + x2 =m cm 6 3 2m x2 y 3
s
oldhomewk 24 PAPAGEORGE, MATT Due: Apr 1 2008, 4:00 am =m 12 22 12 x x+ x 62 92 92 (5) Question 10, chap 12, sect 5. part 1 of 1 10 points
6
1 m x2 = 2 18 1 (1 kg) (2 m)2 = 18 = 0.222222 kg m2 .
using Eq. 1 for s , Eq. 4 for , and Eq. 3 for xcm . Using the parallel axis theorem, we have
x=0 x=x Iy = Iy cm + m x2 cm 1 = m x2 + m x2 2 cm 18 12 12 =m x+ x 18 2 9 2 1 = m x2 2 6 1 = (1 kg) (2 m)2 6
Assume: The masses are point particles; e.g., neglect the contribution due to moments of inertia about their center of mass. Three spherical masses are located in a plane at the positions shown in the gure below. A has mass 32.8 kg, B has mass 10.6 kg, and C has mass 35.9 kg. Three Masses in a Plane
10 9 8 y Distance (m) 7 6 5 4 3 2 1 0 A
C B
= 0.666667 kg m2 . Standard Solution: The moment of inerx=0 tia Iy of the triangle about the x-coordinate is
x=0 Iy x2 x1
x2 dm
x2
=
x1
x2 y dx x2 (x x2 ) dx
x
x2
= s
0
2 14 1 3 = s x (x2 ) x 4 3 0 14 14 = s x x 42 32 y3 1 2m x4 = x2 y 3 x2 12 2 1 = m x2 2 6 1 = (1 kg) (2 m)2 6
3 4 5 6 7 8 9 10 x Distance (m) Figure: Drawn to scale. Calculate the moment of inertia (of the three masses) with respect to an axis perpendicular to the xy plane and passing through x = 3.5 m and y = 6.5 m . Correct answer: 425.125 kg m2 (tolerance 3 %). Explanation: Basic Concepts: The moment of inertia is I=
i 2 mi ri , 2 mi (x2 + yi ) . i i
0
1
2
(1) (2)
= 0.666667 kg m2 , using Eq. 1 for s and Eq. 4 for . Note: This problem has a dierent triangle for most students.
=
Solution: Using the knot (3.5 m, 6.5 m) as the coordinate origin (see gure below), we
oldhomewk 24 PAPAGEORGE, MATT Due: Apr 1 2008, 4:00 am have xa = (0.5 m) (3.5 m) = 3 m , ya = (7 m) (6.5 m) = 0.5 m , ra =
2 x2 + ya = 3.04138 m , a 2 2 = ma [x2 + ya ] + mb [x2 + ya ] a b 2 + mc [x2 + ya ] c = (32.8 kg) [(3 m)2 + (0.5 m)2 ] + (10.6 kg) [(2.5 m)2 + (0.5 m)2 ] + (35.9 kg) [(1 m)2 + (0.5 m)2 ] = 425.125 kg m2 .
7
xb = (6 m) (3.5 m) = 2.5 m , yb = (5.5 m) (6.5 m) = 1 m , rb =
2 x2 + yb = 2.69258 m , b
xc = (4.5 m) (3.5 m) = 1 m , yc = (7 m) (6.5 m) = 0.5 m , rc = x2 c
2 + yc
= 1.11803 m ,
Question 11, chap 13, sect 1. part 1 of 1 10 points Given: A circular shaped object with an inner radius of 14 cm and an outer radius of 28 cm. There are three forces (acting perpendicular to the axis of rotation) whose magnitudes are 13 N, 22 N, and 15 N acting on the object, as shown in the gure. The force of magnitude 22 N is 21 below horizontal. 13 N 28 cm 21 22 N 15 N 6 kg Find the magnitude of the net torque on the wheel about the axle through the center of the object. Correct answer: 4.76 N m (tolerance 1 %). (1) 14 cm
Three Masses in a Plane 3 2 1A y Distance (m) 0 -1 -2 -3 -4 -5 -6 -3 -2 -1 0 1 2 3 4 5 x Distance (m) Figure: Drawn to scale. Using Eqs. 1, we have I=
i 2 2 2 = ma ra + mb rb + mc rc = (32.8 kg) (3.04138 m)2 + (10.6 kg) (2.69258 m)2 + (35.9 kg) (1.11803 m)2 = 425.125 kg m2 . 2 mi ri ,
C B
6
Explanation:
Alternate Solution: Using Eqs. 2, we have I=
i 2 mi [x2 + yi ] , i
(2)
Let : a = 14 cm , b = 28 cm , F1 = 13 N , F2 = 22 N , F3 = 15 N , and = 21 .
oldhomewk 24 PAPAGEORGE, MATT Due: Apr 1 2008, 4:00 am b F1 R = 20.6 cm , m1 = 1 kg , m2 = 1.64 kg , h = 1.2 m , v = R, 1 I = M R2 , and 2 1 1 Kdisk = I 2 = M v 2 . 2 4 Consider the free body diagrams T2 T1
8
F2 a The total torque is M F3
= a F2 b F1 b F3 = (14 cm) (22 N) (28 cm) (13 N) + (15 N) = 4.76 N m = 4.76 N m . Question 12, chap 13, sect 2. part 1 of 1 10 points An Atwood machine is constructed using a hoop with spokes of negligible mass. The 2.3 kg mass of the pulley is concentrated on its rim, which is a distance 20.6 cm from the axle. The mass on the right is 1 kg and on the left is 1.64 kg. The acceleration of gravity is 9.8 m/s2 .
1.64 kg m2 g
1 kg m1 g
a
Basic Concepts: The net acceleration a = r is in the direction of the heavier mass m1 . The pulleys mass is concentrated on the rim, so I = m r2 For the pulley, net = counter clock = I a = mra. r (1)
20.6 cm 2.3 kg 1.64 kg 1.2 m 1 kg What is the magnitude of the linear acceleration a of the hanging masses? Correct answer: 1.26964 m/s2 (tolerance 1 %). Explanation: Let : M = 2.3 kg ,
T1 r T2 r = (m r 2 ) Dividing by r gives
m a = T1 T2 For the mass m1 , Fnet = m1 a = m1 g T1 T1 = m1 g m1 a For the mass m2 , Fnet = m2 a = T2 m2 g T2 = m2 a + m2 g Substituting (2) and (3) into (1) gives ma = m1 g m1 a m2 a m2 g
a (2) (3)
oldhomewk 24 PAPAGEORGE, MATT Due: Apr 1 2008, 4:00 am ma + m1 a + m2 a = m1 g m2 g (m1 m2 ) g a= . m + m1 + m2 Question 13, chap 13, sect 2. part 1 of 1 10 points A uniform 2 kg rod with length 42 m has a frictionless pivot at one end. The rod is released from rest at an angle of 28 beneath the horizontal. Note: The moment of inertia of a rod about 1 the center of mass is m L2 , where m is the 12 mass of the rod and L is the length of the rod. The moment of inertia of a rod about either 1 end is m L2 . 3 The acceleration of gravity is 9.8 m/s2 .
9
The rods moment of inertia about its end1 point is I = m L2 . Therefore, the angular 3 acceleration of the rod is I 3g = cos 2L 3 (9.8 m/s2 ) cos(28 ) = 2 (42 m) = 0.309032 rad/s2 .
=
Question 14, chap 13, sect 2. part 1 of 1 10 points A uniform rod of mass 1.3 kg is 6 m long. The rod is pivoted about a horizontal, frictionless pin at the end of a thin extension (of negligible mass) a distance 6 m from the center of mass of the rod. The rod is released from rest at the horizontal position. The acceleration of gravity is 9.8 m/s2 . Hint: The moment of inertia of the rod 1 m 2 . about its center-of-mass is Icm = 12 6m 6m O 1.3 kg
21
m 42
2k m
g
28
Note: The rod is initially at rest at 28 below the horizontal. What is the angular acceleration of the rod immediately after it is released? Correct answer: 0.309032 rad/s2 (tolerance 1 %). Explanation: Basic Concepts: The torque equation is = r F = r F = I . The torque exerted by gravity on the rod is 1 = m g L cos . 2 Let : m = 2 kg , L = 42 m , = 28 .
63
What is the angular acceleration of the rod at the instant the rod makes an angle of 63 with the horizontal? Correct answer: 0.684478 rad/s2 (tolerance 1 %). Explanation:
and
oldhomewk 24 PAPAGEORGE, MATT Due: Apr 1 2008, 4:00 am
10
Let : = 6 m , d = = 6 m, = 63 , and m = 1.3 kg . Basic Concepts: 1 I 2 2 =rF =I KR = Solution: I = Icm + m d2 1 = m 2 + m 2 12 13 m 2 = 12 13 (1.3 kg) (6 m)2 = 12 = 50.7 kg m2 . Since the rod is uniform, its center of mass is located at . Recalling that the weight m g acts at the center of mass, the magnitude of the torque at = 63 is =rF = mg = m g sin(90 ) = m g cos Equating the above with = I and solving for gives m g cos = I . Therefore = m g cos I m g cos = 13 m 2 12 12 g cos = 13 12 (9.8 m/s2 ) cos(63 ) = 13 (6 m) = 0.684478 rad/s2 .