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

Step 1: Understanding the Question
We have two objects, a solid sphere and a spherical shell, with different masses but the same radius. The same force is applied to the top of each, causing them to roll without slipping. We need to find the ratio of their linear accelerations.
Step 2: Key Formula or Approach
We apply Newton's second law for both linear and rotational motion. 1. Linear motion: \( F_{net} = m_{total}a \)
2. Rotational motion: \( \tau_{net} = I\alpha \), where \( I \) is the moment of inertia and \( \alpha \) is the angular acceleration.
3. Rolling without slipping condition: \( a = R\alpha \).
Moment of inertia for a solid sphere: \( I_{solid} = \frac{2}{5}MR^2 \).
Moment of inertia for a spherical shell: \( I_{shell} = \frac{2}{3}MR^2 \).
Step 3: Detailed Explanation
Let \(F\) be the applied force, \(f\) be the force of friction, \(M\) be the mass of the object, \(a\) be the linear acceleration, and \(R\) be the radius. The force \(F\) is applied at the top, and friction \(f\) acts at the bottom, opposing the tendency to slip. The equations of motion for a general rolling body are:
Linear motion (horizontal direction): \( F + f = Ma \) (Both F and f act forward to produce linear acceleration)
Rotational motion (about the center of mass): The applied force F and friction f both produce a torque. \( F \cdot R - f \cdot R = I\alpha \)
Using the no-slip condition \( \alpha = a/R \): \[ (F - f)R = I(a/R) \implies F - f = \frac{Ia}{R^2} \] We have a system of two linear equations for \(a\) and \(f\): 1) \( F + f = Ma \) 2) \( F - f = \frac{Ia}{R^2} \) Adding the two equations to eliminate \(f\): \[ 2F = Ma + \frac{Ia}{R^2} = a\left(M + \frac{I}{R^2}\right) \] \[ a = \frac{2F}{M + I/R^2} \] For the solid sphere (A): Mass \( M_A = 5m \), \( I_A = \frac{2}{5}(5m)R^2 = 2mR^2 \). \[ a_A = \frac{2F}{5m + (2mR^2)/R^2} = \frac{2F}{5m + 2m} = \frac{2F}{7m} \] For the spherical shell (B): Mass \( M_B = m \), \( I_B = \frac{2}{3}mR^2 \). \[ a_B = \frac{2F}{m + ( (2/3)mR^2 )/R^2} = \frac{2F}{m + 2m/3} = \frac{2F}{5m/3} = \frac{6F}{5m} \] Ratio of accelerations: \[ \frac{a_A}{a_B} = \frac{2F/7m}{6F/5m} = \frac{2F}{7m} \times \frac{5m}{6F} = \frac{10}{42} = \frac{5}{21} \] The ratio \( a_A : a_B \) is 5 : 21.
Step 4: Final Answer
The ratio of \( a_A \) and \( a_B \) is 5:21.