Question

A solid sphere of mass 2 kg and radius 0.5 m is rolling without slipping on a horizontal surface. The ratio of the rotational and translational kinetic energies of the sphere is

• A. 3:5
• B. 2:5
• C. 4:5
• D. 7:5

Hint:

For any object rolling without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is a constant that depends only on its shape (through the moment of inertia formula). It can be written as $KE_R/KE_T = I/(mr^2)$, where $I=kmr^2$, so the ratio is just $k$. For a solid sphere, $k=2/5$.

Answer 1

The Correct Option is B

Step 1: Write the formulas for translational and rotational kinetic energy.
The translational kinetic energy ($KE_T$) of an object with mass $m$ and center-of-mass velocity $v$ is: \[ KE_T = \frac{1}{2}mv^2. \] The rotational kinetic energy ($KE_R$) of an object with moment of inertia $I$ and angular velocity $\omega$ is: \[ KE_R = \frac{1}{2}I\omega^2. \]

Step 2: Use the properties of a solid sphere and rolling without slipping.
For a solid sphere, the moment of inertia about its center is $I = \frac{2}{5}mr^2$. The condition for rolling without slipping is $v = \omega r$, which implies $\omega = v/r$.

Step 3: Express the rotational kinetic energy in terms of m and v.
Substitute the expressions for $I$ and $\omega$ into the formula for $KE_R$: \[ KE_R = \frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{v}{r}\right)^2. \] \[ KE_R = \frac{1}{2} \cdot \frac{2}{5}mr^2 \cdot \frac{v^2}{r^2} = \frac{1}{5}mv^2. \]

Step 4: Calculate the required ratio.
The ratio of rotational to translational kinetic energy is $\frac{KE_R}{KE_T}$. \[ \frac{KE_R}{KE_T} = \frac{\frac{1}{5}mv^2}{\frac{1}{2}mv^2}. \] The $mv^2$ terms cancel out. \[ \frac{KE_R}{KE_T} = \frac{1/5}{1/2} = \frac{1}{5} \times 2 = \frac{2}{5}. \] The ratio is 2:5. Note that the mass (2 kg) and radius (0.5 m) are not needed for this calculation as the ratio is independent of these values.