Question

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

• A. 7:11
• B. 10:7
• C. 5:7
• D. 2:5

Answer 1

The Correct Option is C

To solve this problem, we need to understand the concept of kinetic energy in rolling motion. A rolling object, such as a solid sphere, has both translational kinetic energy (\(K_t\)) due to the movement of its center of mass and rotational kinetic energy (\(K_r\)) due to its rotation around the center of mass.

The translational kinetic energy is given by:

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

where:

• \(m\) is the mass of the sphere
• \(v\) is the velocity of the center of mass

The rotational kinetic energy for a solid sphere is given by:

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

where:

• \(I\) is the moment of inertia of the sphere, which is \(\frac{2}{5} m r^2\) for a solid sphere
• \(\omega\) is the angular velocity

Since the sphere is in rolling motion without slipping, the velocity of the center of mass and angular velocity are related by:

v = r \omega

Substituting this relationship into the formula for rotational kinetic energy gives:

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

Thus, the total kinetic energy (\(K_{total}\)) is:

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

Now, let's find the ratio of translational kinetic energy to the total kinetic energy:

\text{Ratio} = \frac{K_t}{K_{total}} = \frac{\frac{1}{2} m v^2}{\frac{7}{10} m v^2} = \frac{5}{7}

Therefore, the ratio \(K_t : (K_t + K_r)\) is \(5:7\), which matches option 5:7.