Ratio of Acceleration of Rolling Solid Sphere and Spherical Shell

To find the ratio of the linear accelerations $a_A$ and $a_B$ of the solid sphere and the spherical shell respectively, we analyze the dynamics of rolling without slipping for a general body subjected to a tangential force applied at its highest point.

General Derivation for Rolling Acceleration

Consider a body of mass $M$, radius $R$, and moment of inertia about its center of mass $I_{\text{cm}} = k M R^2$, where $k$ is a dimensionless constant depending on the geometric shape of the body.

A horizontal force $F$ is applied tangentially at the top of the body. Let $f$ be the force of friction acting at the contact point with the rough surface.

1. Translational Motion: Applying Newton's second law for the center of mass motion: $F + f = M a$

2. Rotational Motion: Taking torque about the center of mass: $\tau = F \cdot R - f \cdot R = I_{\text{cm}} \alpha$ $F - f = \frac{I_{\text{cm}} \alpha}{R}$

3. Condition for Pure Rolling: Since the body rolls without slipping: $a = \alpha R \implies \alpha = \frac{a}{R}$

Substituting $\alpha$ into the rotational equation gives: $F - f = \frac{I_{\text{cm}}}{R^2} a$

Adding the translational and rotational equations eliminates the friction force $f$: $(F + f) + (F - f) = M a + \frac{I_{\text{cm}}}{R^2} a$ $2F = \left(M + \frac{I_{\text{cm}}}{R^2}\right) a$

Since $I_{\text{cm}} = k M R^2$, this simplifies to: $2F = M(1 + k)a \implies a = \frac{2F}{M(1 + k)}$

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Acceleration of Solid Sphere ($A$)

For a solid sphere of mass $M_A = 5m$:

• Moment of inertia $I_A = \frac{2}{5} M_A R^2 \implies k_A = \frac{2}{5}$

Substituting the values into the acceleration formula: $a_A = \frac{2F}{5m \left(1 + \frac{2}{5}\right)} = \frac{2F}{5m \left(\frac{7}{5}\right)} = \frac{2F}{7m}$

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Acceleration of Spherical Shell ($B$)

For a spherical shell of mass $M_B = m$:

• Moment of inertia $I_B = \frac{2}{3} M_B R^2 \implies k_B = \frac{2}{3}$

Substituting the values into the acceleration formula: $a_B = \frac{2F}{m \left(1 + \frac{2}{3}\right)} = \frac{2F}{m \left(\frac{5}{3}\right)} = \frac{6F}{5m}$

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Ratio of Accelerations

Taking the ratio of $a_A$ to $a_B$: $\frac{a_A}{a_B} = \frac{\frac{2F}{7m}}{\frac{6F}{5m}} = \frac{2}{7} \times \frac{5}{6} = \frac{10}{42} = \frac{5}{21}$

Thus, the ratio $a_A : a_B$ is $5 : 21$.