Question

A solid sphere rolls without slipping on an inclined plane at an angle $\theta$. The ratio of total kinetic energy to its rotational kinetic energy is

• A. $\frac{7}{2}$
• B. $\frac{5}{2}$
• C. $\frac{7}{3}$
• D. $\frac{5}{4}$

Hint:

Logic Tip: For any rolling object, this ratio can be simplified to $1 + \frac{M R^2}{I}$.

Answer 1

The Correct Option is A

Concept:
Total kinetic energy ($E_K$) of a rolling object is the sum of translational ($\frac{1}{2}MV^2$) and rotational ($\frac{1}{2}I\omega^2$) kinetic energies.
Step 1: Identify rotational energy.
For a solid sphere, $I = \frac{2}{5}MR^2$. With no slipping, $v = R\omega$. $$E_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2} \left( \frac{2}{5}MR^2 \right) \omega^2 = \frac{1}{5}MV^2$$
Step 2: Calculate total energy.
$$E_K = E_{trans} + E_{rot} = \frac{1}{2}MV^2 + \frac{1}{5}MV^2 = \frac{7}{10}MV^2$$
Step 3: Find the ratio.
$$\frac{E_K}{E_{rot = \frac{(7/10)MV^2}{(1/5)MV^2} = \frac{7}{2}$$