Step 1: Understanding the given data
A horizontal force \( F \) is applied at the center of mass of a cylindrical object of mass \( m \) and radius \( R \), perpendicular to its axis. The object is rolling without slipping, which implies that the object is undergoing both translation and rotation.
The coefficient of friction between the object and the ground is \( \mu \). The center of mass of the object has an acceleration \( a \), and the acceleration due to gravity is \( g \).
Step 2: Translational and Rotational Motion
The object is rolling without slipping, which implies that the following condition holds between the translational velocity \( v \) and the angular velocity \( \omega \):
\[
v = R \omega
\]
Similarly, for rolling motion without slipping, the following equation for the accelerations holds:
\[
a = R \alpha
\]
where \( \alpha \) is the angular acceleration.
The forces acting on the object are:
- The applied horizontal force \( F \),
- The frictional force \( f \),
- The gravitational force \( mg \), acting downward,
- The normal force \( N \), acting upward.
The net force causes translational acceleration, and the friction force causes rotational acceleration.
Step 3: Equations of Motion
Using Newton's second law for translation and rotation:
Translational equation:
\[
F - f = ma
\]
Rotational equation (for torque about the center of mass):
\[
fR = I \alpha
\]
where \( I \) is the moment of inertia of the object.
For a solid cylinder, the moment of inertia is:
\[
I = \frac{1}{2} m R^2
\]
For a thin-walled hollow cylinder, the moment of inertia is:
\[
I = m R^2
\]
Step 4: Solving for the acceleration \( a \)
For a solid cylinder, using \( I = \frac{1}{2} m R^2 \) and \( \alpha = a / R \), we substitute into the rotational equation:
\[
f R = \frac{1}{2} m R^2 \cdot \frac{a}{R}
\]
Simplifying this, we get:
\[
f = \frac{1}{2} ma
\]
Now, using the translational equation \( F - f = ma \), we substitute \( f = \frac{1}{2} ma \) into it:
\[
F - \frac{1}{2} ma = ma
\]
Solving for \( a \), we get:
\[
a = \frac{2F}{3m}
\]
The maximum possible value of acceleration \( a \) occurs when the frictional force is at its maximum, \( f_{\text{max}} = \mu mg \), which gives:
\[
a_{\text{max}} = 2 \mu g
\]
This corresponds to option (B).
Step 5: For a thin-walled hollow cylinder
For a thin-walled hollow cylinder, the moment of inertia is \( I = m R^2 \). Using the same procedure as above, we substitute into the rotational equation:
\[
fR = mR^2 \cdot \frac{a}{R}
\]
Simplifying this, we get:
\[
f = ma
\]
Substituting into the translational equation \( F - f = ma \), we get:
\[
F - ma = ma
\]
Solving for \( a \), we get:
\[
a = \frac{F}{2m}
\]
This corresponds to option (D).
Step 6: For the same \( F \), the value of \( a \) does not depend on whether the cylinder is solid or hollow
The acceleration \( a \) depends on the force applied and the mass of the object. However, the moment of inertia of the object affects the distribution of the applied force and its effect on rotational motion. Despite this, for a given force, the resulting acceleration is the same for both solid and hollow cylinders.
This corresponds to option (A).
Step 7: Conclusion
Therefore, the correct answers are:
(A): For the same \( F \), the value of \( a \) does not depend on whether the cylinder is solid or hollow.
(B): For a solid cylinder, the maximum possible value of \( a \) is \( 2 \mu g \).
(D): For a thin-walled hollow cylinder, \( a = \frac{F}{2m} \).