DYNAMICS OF ROTATIONAL MOTION 10 10.9. IDENTIFY: Apply I. rad/rev SET UP: 0 0. (400 rev/min) 419 rad/ 60 /min EXECUTE: 0 419 rad/ I I (0 kg m ) 11 N m. t 800 EVALUATE: In I, mut be in rad/. 10.. IDENTIFY: Apply F ma to the tranlational motion of the center of ma I to the rotation about the center of ma. SET UP: Let be down the incline let the hell be turning in the poitive direction. The free-body diagram for the hell i given in Figure 10.. From Table 9., EXECUTE: (a) F ma give mg in f ma. I / I mr. give fr mr. With a R thi become f ma. Combining the equation give mg in ma ma g in (980 m/ )(in80 ) a 6 m/. f ( 00 kg)( 6 m/ ) 4 8 N. ma The friction i tatic ince there i no lipping at the point of contact. n mg co 1 4 N. f 48 N 0 1. n 14 N (b) The acceleration i independent of m doen t change. The friction force i proportional to m o will double; f 9 66 N. The normal force will alo double, o the minimum required for no lipping wouldn t change. EVALUATE: If there i no friction the object lide without rolling, the acceleration i g in. Friction rolling without lipping reduce a to 0.60 time thi value.
10.. IDENTIFY: Apply Fet ma I to the motion of the ball. (a) SET UP: The free-body diagram i given in Figure 10.a. Figure 10.a F ma EXECUTE: y y n mg co f mg co F ma mg in mg co ma g(in co ) a (eq. 1) SET UP: Conider Figure 10.b. n mg act at the center of the ball provide no torque. Figure 10.b EXECUTE: f mg co R; I I give mg co R mr mr No lipping mean ar /, o g co a (eq.) We have two equation in the two unknown a. Solving give a gin 7 7 tan tan 60 0 61. (b) Repeat the calculation of part (a), but now tan tan 60 0 88 I mr. a gin The value of calculated in part (a) i not large enough to prevent lipping for the hollow ball. (c) EVALUATE: There i no lipping at the point of contact. More friction i required for a hollow ball ince for a given m R it ha a larger I more torque i needed to provide the ame. Note that the required i independent of the ma or radiu of the ball only depend on how that ma i ditributed. 7
10.9. IDENTIFY: A the ball roll up the hill, it kinetic energy (tranlational rotational) i tranformed into gravitational potential energy. Since there i no lipping, it mechanical energy i conerved. SET UP: The ball ha moment of inertia coordinate where ball kinetic energy i K = I mr. Rolling without lipping mean v R. Ue y i upward y 0 at the bottom of the hill, o y1 0 y h 00 m. The 1 1 I it potential energy i U mgh. EXECUTE: (a) Conervation of energy give K 1 U 1 K U. U1 0, K 0 (the ball top). Therefore K1 U 1 1 I mgh. 1 1 v 1 I mr R 6gh 6(980 m/ )(00 m) 6 mgh. Therefore v 7 67 m/ v 767 m/ 679 rad/. R 011 m (b) rot 1 1 1 K I (0 46 kg)(7 67 m/) 8 J. EVALUATE: It tranlational kinetic energy at the bae of the hill i kinetic energy i 0.9 J, which equal it final potential energy: mgh (046 kg)(980 m/ )(00 m) 0 9 J. 1 rot o K 1 J. It total 10.. IDENTIFY: The power output of the motor i related to the torque it produce to it angular velocity by P, where mut be in rad/. SET UP: The work output of the motor in 60.0 i (9 00 kj) 6 00 kj, o 600 kj P 100 W. 600 00 rev/min 6 rad/. EXECUTE: P 100 W 0 8 N m 6 rad/ EVALUATE: For a contant power output, the torque developed decreae when the rotation peed of the motor increae. 10.9. IDENTIFY SET UP: Ue L I. EXECUTE: The econd h make 1 revolution in 1 minute, o (1 00 rev/min)( rad/1 rev)(1 min/60 ) 0 1047 rad/. For a lender rod, with the ai about one end, I 1 ML 1 (600 10 kg)(010 m) 40 10 kg m. Then 6 L I (40 10 kg m )(01047 rad/) 47110 kg m /. EVALUATE: L i clockwie. 10.4. IDENTIFY: Apply conervation of angular momentum to the motion of the kater. SET UP: For a thin-walled hollow cylinder center, I 1 1 ml. L L o I i i I f f. EXECUTE: i f 1 Ii 1 0 40 kg m (8 0 kg)(1 8 m) 6 kg m. Ii f i I f 090 kg m 6 kg m (0 40 rev/) 1 14 rev/. EVALUATE: 1 1 I mr. For a lender rod rotating about an ai through it If 040 kg m (80 kg)(0 m) 090 kg m. K I L. increae L i contant, o K increae. The increae in kinetic energy come from the work done by the kater when he pull in hi h.
EQUILIBRIUM AND ELASTICITY 11 11.1. IDENTIFY: Ue Eq. (11.) to calculate. The center of gravity of the bar i at it center it can be treated a a point ma at that point. SET UP: Ue coordinate with the origin at the left end of the bar the ai along the bar. m1 0 10 kg, m 0 0 kg, m 0110 kg. m1 1 m m (0.10 kg)(0.0 m) 0 (0.110 kg)(0.00 m) EXECUTE: 0.98 m. m m m 0.10 kg 0.0 kg 0.110 kg 1 fulcrum hould be placed 9.8 to the right of the left-h end. EVALUATE: The ma at the right-h end i greater than the ma at the left-h end. So the center of gravity i to the right of the center of the bar. 11.. IDENTIFY: Ue Eq. (11.) to calculate of the compoite object. SET UP: Ue coordinate where the origin i at the original center of gravity of the object i to the right. With the 1.0 g ma added, 0, m1 00 g m 1 0 g. 1 0. m m EXECUTE:. 1 m 00 g 10 g ( 0 ) 9. m1 m m 10 g The additional ma hould be attached 9. to the left of the original center of gravity. EVALUATE: The new center of gravity i omewhere between the added ma the original center of gravity. 11.14. IDENTIFY: Apply the firt econd condition of equilibrium to the beam. SET UP: The free-body diagram for the beam i given in Figure 11.14. H v H h are the vertical horiontal component of the force eerted on the beam at the wall (by the hinge). Since the beam i uniform, it center of gravity i.00 m from each end. The angle ha co 0 800 in 0.600. The tenion T ha been replaced by it y component. EXECUTE: (a) H v, H T Tco all produce ero torque. 0 give h (10 N)(00 m) (00 N)(400 m) w(00 m) wload (400 m) T in (400 m) 0 T 6 N. (400 m)(0600) (b) F 0 give Hh Tco 0 Hh (6 N)(0800) 00 N. 0 give Hv w wload T in 0 Hv w wload T in 10 N 00 N (6 N)(0600) 7 N. EVALUATE: For an ai at the right-h end of the beam, only w to w i counterclockwie o the torque due to H v H mut be upward, in agreement with our reult from 0. v H v F y The produce torque. The torque due mut be clockwie. To produce a clockwie torque, F y
Figure 11.14