Question

A solid sphere of mass ' $m$ ' and radius ' $R$ ' is rotating about its diameter. A solid cylinder of the same mass and same radius is also rotating about its geometrical axis with angular speed twice that of sphere. The ratio of kinetic energy of sphere to kinetic energy of cylinder will be

• A. $2 : 3$
• B. $1 : 5$
• C. $3 : 1$
• D. $1 : 4$

Hint:

When angular speed changes, always include the square: \[ K \propto I\omega^2 \] A doubled angular speed makes the rotational KE four times, if \(I\) stays same.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Rotational kinetic energy depends on the moment of inertia and the square of the angular speed.
Step 2: Key Formula or Approach:
$KE = \frac{1}{2} I \omega^2$.
$I_{sphere} = \frac{2}{5} mR^2$.
$I_{cylinder} = \frac{1}{2} mR^2$.
Step 3: Detailed Explanation:
Let angular speed of sphere be $\omega$. Then cylinder's angular speed is $2\omega$.
\[ KE_s = \frac{1}{2} \left( \frac{2}{5} mR^2 \right) \omega^2 = \frac{1}{5} mR^2 \omega^2 \]
\[ KE_c = \frac{1}{2} \left( \frac{1}{2} mR^2 \right) (2\omega)^2 = \frac{1}{4} mR^2 (4\omega^2) = mR^2 \omega^2 \]
Ratio $KE_s : KE_c = \frac{1}{5} : 1 = 1 : 5$.
Step 4: Final Answer:
The ratio is $1 : 5$.