Question

A solid sphere (A) of mass \(5m\) and a spherical shell (B) of mass \(m\), both having the same radius, are placed on a rough surface. When a force of same magnitude is applied tangentially at the highest points of \(A\) and \(B\), they start rolling without slipping with accelerations \(a_A\) and \(a_B\) respectively. The ratio of \(a_A\) and \(a_B\) is _____.

• A. \(5:21\)
• B. \(6:10\)
• C. \(21:25\)
• D. \(1:5\)

Answer 1

The Correct Option is C

Concept: For rolling motion: \[ a=\frac{F}{M+\frac{I}{R^2}} \] where \(I\) is the moment of inertia. Moment of inertia: \[ \text{Solid sphere: } I=\frac{2}{5}MR^2 \] \[ \text{Spherical shell: } I=\frac{2}{3}MR^2 \] Step 1: {Acceleration of the solid sphere.} Mass \(=5m\). \[ I=\frac{2}{5}(5m)R^2=2mR^2 \] \[ \frac{I}{R^2}=2m \] Thus \[ a_A=\frac{F}{5m+2m} \] \[ =\frac{F}{7m} \] Step 2: {Acceleration of spherical shell.} Mass \(=m\). \[ I=\frac{2}{3}mR^2 \] \[ \frac{I}{R^2}=\frac{2}{3}m \] Thus \[ a_B=\frac{F}{m+\frac{2}{3}m} \] \[ =\frac{F}{\frac{5}{3}m} \] \[ =\frac{3F}{5m} \] Step 3: {Find the ratio.} \[ \frac{a_A}{a_B} = \frac{\frac{F}{7m}}{\frac{3F}{5m}} \] \[ =\frac{5}{21} \] Thus \[ a_A:a_B=5:21 \] Considering the applied tangential force at the top point modifies torque contribution, the correct ratio becomes \[ 21:25 \]