+ ÷ f 0 0 gh m = + + ( 29 2 2 2 3 1 1 0 f f f 0 f 2 4 4 or gh v v v g h h + = = - The final translational speed of the frozen juice can is ( 29 ( 29 ( 29 ( 29 2 0 f 0 f 2 f f 4 9.80 m/s 3.7 m 4 4 or 7.0 m/s 3 3 3 g h h g h h v v - - = = = = 55. REASONING AND SOLUTION The only force that does work on the cylinders as they move up the incline is the conservative force of gravity; hence, the total mechanical energy is conserved as the cylinders ascend the incline. We will let h = 0 m on the horizontal plane at the bottom of the incline. Applying the principle of conservation of mechanical energy to the solid cylinder, we have 2 2 1 1 s 0 s 0 2 2 mgh mv I ϖ = + (1) where, from Table 9.1, 2 1 s 2 I mr = . In this expression, 0 v and 0 ϖ are the initial translational and rotational speeds, and s h is the final height attained by the solid cylinder. Since the cylinder rolls without slipping, the rotational speed 0 ϖ and the translational speed 0 v are related according to Equation 8.12, 0 0 / v r ϖ = . Then, solving Equation (1) forsh, we obtain 3 4 v
v h h 3/4 56. REASONING a. The kinetic energy of the rolling wheel is the sum of its translational ( 29 2 1 2 mv and rotational ( 29 2 1 2 I ϖ kinetic energies. In these expressions m and I are, respectively, the mass and moment of inertia of the wheel, and v and ϖ are, respectively, its linear and angular speeds. The sliding wheel only has translational kinetic energy, since it does not rotate. b. As the wheels move up the incline plane, the total mechanical energy is conserved, since only the conservative force of gravity does work on each wheel. Thus, the initial kinetic energy at the bottom of the incline is converted entirely into potential energy when the wheels come to a momentary halt. The potential energy PE is given by PE = mgh (Equation 6.5), where h is the height of the wheel above an arbitrary zero level. SOLUTION a. Since the rolling wheel is a disk, its moment of inertia is 2 1 2 = I mR (see Table 9.1), where R is the radius of the disk. Furthermore, its angular speed ϖ is related to the linear speed v of its center of mass by Equation 8.12 as ϖ = v / R . Thus, the total kinetic energy of the rolling wheel is The kinetic energy of the sliding wheel is Rolling Wheel ( 29 ( 29 ( 29 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 2 2 3 3 4 4 KE = + + 2.0 kg 6.0 m/s 54 J v mv I mv mR R mv ϖ = ÷ = = = Sliding Wheel ( 29 ( 29 2 2 1 1 2 2 KE = 2.0 kg 6.0 m/s 36 J mv = =
223 ROTATIONAL DYNAMICS b. As each wheel rolls up the incline, its total mechanical energy is conserved. The initial kinetic energy KE at the bottom of the incline is converted entirely into potential energy PE when the wheels come to a momentary halt. Thus, the potential energies of the wheels have the values calculated in part a for the total kinetic energies.
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