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Kelvi

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

medium

Kinetic Theory

2018

physics

10:7

5:7

7:10

2:5

Explanation

To solve this problem, we need to find the ratio of translational kinetic energy $K_t$ to the total kinetic energy $(K_t + K_r)$

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for a solid sphere in rolling motion.

\\

For a solid sphere, the moment of inertia

I

about its center is

\frac{2}{5} m r^2,

where

m

is the mass and

r

is the radius.

\\

The translational kinetic energy

K_t

is given by:

\\ K_t = \frac{1}{2} m v^2 \\

The rotational kinetic energy

K_r

is given by:

\\ K_r = \frac{1}{2} I \omega^2 \\

Since the sphere is rolling without slipping, the angular velocity

\omega

is related to the linear velocity

v

by

\omega = \frac{v}{r}. \\

Substitute

I = \frac{2}{5} m r^2

and

\omega = \frac{v}{r}

into the expression for

K_r: \\ K_r = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v}{r}\right)^2 \\ K_r = \frac{1}{2} \cdot \frac{2}{5} m r^2 \cdot \frac{v^2}{r^2} \\ K_r = \frac{1}{5} m v^2 \\

Now, the total kinetic energy

(K_t + K_r)

is:

\\ K_t + K_r = \frac{1}{2} m v^2 + \frac{1}{5} m v^2 \\ K_t + K_r = \left(\frac{1}{2} + \frac{1}{5}\right) m v^2 \\ K_t + K_r = \left(\frac{5}{10} + \frac{2}{10}\right) m v^2 \\ K_t + K_r = \frac{7}{10} m v^2 \\

The ratio

K_t : (K_t + K_r)

is:

\\ \frac{K_t}{K_t + K_r} = \frac{\frac{1}{2} m v^2}{\frac{7}{10} m v^2} \\ \frac{K_t}{K_t + K_r} = \frac{\frac{1}{2}}{\frac{7}{10}} \\ \frac{K_t}{K_t + K_r} = \frac{1}{2} \times \frac{10}{7} \\ \frac{K_t}{K_t + K_r} = \frac{10}{14} \\ \frac{K_t}{K_t + K_r} = \frac{5}{7} \\

Therefore, the ratio

K_t : (K_t + K_r)

for the sphere is

5:7. \\

This corresponds to Option 2.