Question

Consider two spheres A (solid sphere) and B (hollow sphere) each of radius \(R\), kept on a perfectly rough surface. Mass of sphere A is 5 m and that of B is 2 m. If the same force \(F\) is applied at the top tangentially, then find the ratio of acceleration of A to acceleration of B.

• A. \(\frac{5}{17}\)
• B. \(\frac{10}{21}\)
• C. \(\frac{3}{5}\)
• D. \(\frac{2}{7}\)

Hint:

In problems involving rolling motion, use the relationship \(a = R \alpha\) and consider the moments of inertia for different objects (solid sphere or hollow sphere) to find the accelerations.

Answer 1

The Correct Option is B

Let the acceleration of sphere A be \(a_A\) and the acceleration of sphere B be \(a_B\). For rolling motion, we know that the force applied causes both linear and rotational acceleration. The relation between the linear and rotational acceleration for rolling objects is given by: \[ F = ma + I \alpha \] where \(m\) is the mass, \(a\) is the linear acceleration, \(I\) is the moment of inertia, and \(\alpha\) is the angular acceleration. For both spheres, the condition for rolling without slipping is: \[ a = R \alpha \] For solid sphere (A): - Moment of inertia \(I_A = \frac{2}{5}mR^2\) - Using \(F = ma_A + I_A \alpha_A\) and \(a_A = R \alpha_A\), we get: \[ F = m a_A + \frac{2}{5} m R^2 \frac{a_A}{R} \] \[ F = m a_A \left(1 + \frac{2}{5}\right) \] \[ F = \frac{7}{5} m a_A \] \[ a_A = \frac{5}{7} \frac{F}{m} \] For hollow sphere (B): - Moment of inertia \(I_B = \frac{2}{3}mR^2\) - Using \(F = ma_B + I_B \alpha_B\) and \(a_B = R \alpha_B\), we get: \[ F = m a_B + \frac{2}{3} m R^2 \frac{a_B}{R} \] \[ F = m a_B \left(1 + \frac{2}{3}\right) \] \[ F = \frac{5}{3} m a_B \] \[ a_B = \frac{3}{5} \frac{F}{m} \] Step 3: Find the ratio of accelerations: \[ \frac{a_A}{a_B} = \frac{\frac{5}{7} \frac{F}{m}}{\frac{3}{5} \frac{F}{m}} = \frac{5}{7} \times \frac{5}{3} = \frac{25}{21} \approx \frac{10}{21} \] Thus, the ratio of the accelerations is \(\frac{10}{21}\). Final Answer: Option (B) \(\frac{10}{21}\).