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

Concept:
Rotational kinetic energy is: \[ K=\frac12 I\omega^2 \] Moments of inertia are: \[ I_{\text{sphere}}=\frac{2}{5}mR^2 \] for a solid sphere about diameter, and \[ I_{\text{cyl}}=\frac12 mR^2 \] for a solid cylinder about its axis. ip

Step 1: Write kinetic energy of sphere.
Let angular speed of sphere be \(\omega\). Then: \[ K_s=\frac12\left(\frac{2}{5}mR^2\right)\omega^2 =\frac{1}{5}mR^2\omega^2 \] ip

Step 2: Write kinetic energy of cylinder.
Angular speed of cylinder is \(2\omega\). So, \[ K_c=\frac12\left(\frac12 mR^2\right)(2\omega)^2 \] \[ K_c=\frac12\cdot \frac12 mR^2 \cdot 4\omega^2 = mR^2\omega^2 \] ip

Step 3: Find the ratio.
\[ K_s:K_c=\frac{1}{5}:1=1:5 \] Thus the direct physics result is: \[ 1:5 \] This matches option (B), not the keyed choice implied by the source sequence. ip The direct calculation gives:
\[ \boxed{1:5} \]