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A solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy (KtK_t) as well as rotational kinetic energy (KrK_r) simultaneously. The ratio Kt:(Kt+Kr)K_t : (K_t + K_r) for the sphere is

1

10:710 : 7

2

5:75 : 7

3

7:107 : 10

4

2:52 : 5

Step-by-Step Solution

Kt=12mv2K_t = \frac{1}{2}mv^2. Kt+Kr=12mv2+12Iω2=12mv2+12(25mr2)(vr)2=710mv2K_t + K_r = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{7}{10}mv^2. Therefore, KtKt+Kr=1/27/10=57\frac{K_t}{K_t + K_r} = \frac{1/2}{7/10} = \frac{5}{7}.

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