Tutorial Questions for Week Nov All taken from text book with permission

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Branden Hunter
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1 Tutorial Questions for Week Nov -5. All taken from text book with permission Text 9.6 Ricardo, of mass 80 kg, and Carmelita, who is lighter, are enjoying Lake Merced at dusk in a 30 kg canoe. When the canoe is at rest in the placid water, they exchange seats, which are 3.0 m apart and symmetrically located with respect to the canoe's center. If the canoe moves 40 cm horizontally relative to a pier post, what is Carmelita's mass? (Ignore effects of friction). 6. When they exchange places, the centre of mass will remain constant (in the absence of friction) Hence if the exchange of Ricardo and Carmelita moves mass in one direction, the canoe will move backwards in the opposite direction to keep the canter of mass constant. Suppose the pier post is at the exact center of the canoe initially, and take this to be 0 in the x direction. Suppose Ricardo is initially on the left. Then the center of mass of Ricardo, Carmelita and the canoe is given by. MR (.5 m) MC (.5 m) Mcanoe(0) xcm M M M R C canoe Now after they exchange places, the canoe will have moved.4 m to the left, so will be at -0.4 m with respect to the post. Ricardo will be.5 m to the right of that, so at x =. m, and Carmelita will be.5 m to the left of the center of the canoe, so at x = -.9 m. Then the center of mass is given by: MR(. m) MC(.9 m) Mcano e (.4) xcm M M M R C canoe Since the center of mass doesn t change, these two expressions are equal. Ricardo has mass 80 kg, and the canoe is 30, so one obtains:.5(80).5mc.(80).9 MC.4(30) Solving this for the unknown mass gives: 3.4M C 96 or M 57.6 kg C Text 0.37 Calculate the rotational inertia of a meter stick, with mass 0.56 kg, about an axis perpendicular to the stick and located at the 0 cm mark. (Treat the stick as a thin rod.) 37. We use the parallel axis theorem: I = I com + Mh, where I com is the rotational inertia about the center of mass (see Table 0-(d)), M is the mass, and h is the distance between the center of mass and the chosen rotation axis. The center of mass is at the center of the meter stick, which implies h = 0.50 m 0.0 m = 0.30 m. We find

2 I com b gb g. ML kg 0. m kgm Consequently, the parallel axis theorem yields b gb g I kgm 056. kg 030. m kgm. Text 0.4 In Fig. 0.34, two particles, each with mass m 0.85 kg, are fastened to each other, and to a rotation axis at O by two thin rods, each with length d 5.6 cm and mass M. kg. The combination rotates around the rotation axis with the angular speed 0.30 rad/s. Measured about O, what are the combination's (a) rotational inertia and (b) kinetic energy? Figure 0.34 Problem The particles are treated point-like in the sense that Eq yields their rotational inertia, and the rotational inertia for the rods is figured using Table 0-(e) and the parallel-axis theorem (Eq. 0-36). (a) With subscript standing for the rod nearest the axis and 4 for the particle farthest from it, we have 3 I II I3 I4 Md M d md Md M d m( d) 8 8 Md 5 md (. kg)(0.056 m) +5(0.85 kg)(0.056 m) 3 3 =0.03 kg m. (b) Using def of rotational kinetic energy we have: K I 3 (0.03)(0.30).040 J

3 Text 0.43 The uniform solid block in Fig has mass 0.7 kg and edge lengths a 3.5 cm, b 8.4 cm, and c.4 cm. Calculate its rotational inertia about an axis through one corner and perpendicular to the large faces. Figure 0.35 Problem Since the rotation axis does not pass through the center of the block, we use the parallel-axis theorem to calculate the rotational inertia. According to Tables, the rotational inertia of a uniform slab about an axis through the center and perpendicular to M the large faces is given by I com a b c h. A parallel axis through the corner is a b g b g distance h a / b / from the center. Therefore, M M M I I Mh a b a b a b 4 3 With M 0.7 kg, a 3.5 cm, and com. b 8.4 cm, we have M 0.7 kg 4 [(0.035 m) (0.084 m) ] I a b kg m. 3 3 Text 0.5 In Fig. 0.38, block has mass, m 460 g, block has mass m 500 g, and the pulley, which is mounted on a horizontal axle with negligible friction, has radius R 5.00 cm. When released from rest, block falls 75.0 cm in 5.00 s without the cord slipping on the pulley. (a) What is the magnitude of the acceleration of the blocks? What are (b) tension T and (c) tension T? (d) What is the magnitude of the pulley's angular acceleration? (e) What is its rotational inertia?

4 Figure 0.38 Problem (a) We use constant acceleration kinematics. If down is taken to be positive and a is the acceleration of the heavier block m, then its coordinate is given by y at, so y ( m) a m/s. t (. 500s) Block has an acceleration of m/s upward. (b) Newton s second law for block is mg T ma, where m is its mass and T is the tension force on the block. Thus, T m g a ( ) (0.500 kg) 9.8 m/s m/s 4.87 N. (c) Newton s second law for block is mg, T ma where T is the tension force on the block. Thus, T m g a ( ) (0.460 kg) 9.8 m/s m/s 4.54 N. (d) Since the cord does not slip on the pulley, the tangential acceleration of a point on the rim of the pulley must be the same as the acceleration of the blocks, so a R m/s m 0. rad / s. (e) The net torque acting on the pulley is (T T) R. Equating this to I we solve for the rotational inertia:

5 I T T R 4.87 N 4.54 N m.38 0 kg m..0 rad/s Text 0.66 A uniform spherical shell of mass M 4.5 kg and radius R 8.5 cm can rotate about a vertical axis on frictionless bearings (Fig. 0.44). A massless cord passes around the 3 equator of the shell, over a pulley of rotational inertia I 3.00 kg m and radius r 5.0 cm, and is attached to a small object of mass m 0.6 kg. There is no friction on the pulley's axle; the cord does not slip on the pulley. What is the speed of the object when it has fallen 8 cm after being released from rest? Use energy considerations. Figure 0.44 Problem From Tables, the rotational inertia of the spherical shell is MR /3, so the kinetic energy (after the object has descended distance h) is I F K MR I m H G sphere pulley v. 3 KJ Since it started from rest, then this energy must be equal (in the absence of friction) to the potential energy mgh with which the system started. We substitute v/r for the pulley s angular speed and v/r for that of the sphere and solve for v. v mgh gh m I mr M m I M r 3 ( / ) ( /3 ) (9.8)(0.8) /((0.60)(0.050) ) (4.5)/3(0.60).4 m/s.

6 Text 0.7 In Fig. 0.47, two 6.0 kg blocks are connected by a massless string over a pulley of radius.40 cm and rotational inertia kg m. The string does not slip on the pulley; it is not known whether there is friction between the table and the sliding block; the pulley's axis is frictionless. When this system is released from rest, the pulley turns through 0.3 rad in 9.0 ms and the acceleration of the blocks is constant. What are (a) the magnitude of the pulley's angular acceleration, (b) the magnitude of either block's acceleration, (c) string tension, and (d) string tension T? T Figure 0.47 Problem We choose positive coordinate directions (different choices for each item) so that each is accelerating positively, which will allow us to set a = a = R (for simplicity, we denote this as a). Thus, we choose rightward positive for m = M (the block on the table), downward positive for m = M (the block at the end of the string) and clockwise for positive sense of disk rotation. This means that we interpret given in the problem as a positive-valued quantity. Applying Newton s second law to m, m and to M, respectively, we arrive at the following three equations (where we allow for the possibility of friction f acting on m ). mg T ma T f ma TRTR I (a) From Eq. 0-3 (with 0 = 0) we find (0.30 rad) 0t t 3.4 rad/s. t (0.090 s) (b) From the fact that a = R (noted above), we obtain

7 R (0.04 m)(0.30 rad) a m/s. t (0.090 s) (c) From the first of the above equations, we find R (6.0 kg) 9.80 m/s (0.04 m)(0.30 rad) T m ga M g t (0.090 s) 56. N. (d) From the last of the above equations, we obtain the second tension: 4 I (7.400 kg m )(3.4 rad/s ) T T 56. N 55. N. R 0.04 m

PHYSICS 111 HOMEWORK SOLUTION #10. April 8, 2013

PHYSICS 111 HOMEWORK SOLUTION #10. April 8, 2013 PHYSICS HOMEWORK SOLUTION #0 April 8, 203 0. Find the net torque on the wheel in the figure below about the axle through O, taking a = 6.0 cm and b = 30.0 cm. A torque that s produced by a force can be