Step-by-Step Derivation

1. Analyze the Forces and Motion Let the mass of the cylindrical object be $m$ and its radius be $R$. A horizontal force $F$ is applied at the center of mass (COM). The object rolls without slipping, which means there is a static friction force $f$ between the object and the ground. Let the linear acceleration of the COM be $a$ and the angular acceleration about the COM be $α$.

   • Vertical forces: The normal force $N$ balances the gravitational force $mg$. So, $N = mg$.

   • Horizontal forces: The applied force $F$ acts in the forward direction. The force $F$ tends to cause the bottom point of the cylinder to slide forward relative to the ground. Therefore, the static friction force $f$ must act in the backward direction to oppose this tendency and provide the torque for rotation. The net horizontal force is given by Newton's second law for linear motion: $F - f = ma \quad \text{(Equation 1)}$

   • Torque: The torque is calculated about the center of mass. The applied force $F$ and the normal force $N$ pass through the COM, so they produce zero torque. The frictional force $f$ creates a torque that causes the object to rotate. Assuming the object rolls to the right, the rotation is clockwise. The net torque is given by Newton's second law for rotational motion: $τ_{COM} = fR = Iα \quad \text{(Equation 2)}$ where $I$ is the moment of inertia of the object about its axis.

2. Apply the Rolling Without Slipping Condition For an object rolling without slipping, the linear acceleration of the COM and the angular acceleration are related by: $a = αR \quad \implies \quad α = \frac{a}{R} \quad \text{(Equation 3)}$

3. Solve the System of Equations Substitute $α$ from Equation 3 into Equation 2: $fR = I \left(\frac{a}{R}\right) \implies f = \frac{I}{R^2}a \quad \text{(Equation 4)}$ Now substitute this expression for $f$ into Equation 1: $F - \frac{I}{R^2}a = ma$ $F = ma + \frac{I}{R^2}a = a\left(m + \frac{I}{R^2}\right)$ Solving for acceleration $a$: $a = \frac{F}{m + I/R^2} \quad \text{(Equation 5)}$ The moment of inertia for a general cylindrical object can be written as $I = kmR^2$, where $k$ is a dimensionless constant that depends on the mass distribution. Substituting $I = kmR^2$ into Equation 5: $a = \frac{F}{m + (kmR^2)/R^2} = \frac{F}{m + km} = \frac{F}{m(1+k)}$ And the friction force is: $f = kma$

4. Evaluate Each Statement

   A: For the same F, the value of a does not depend on whether the cylinder is solid or hollow

   • For a solid cylinder, $I = \frac{1}{2}mR^2$, so $k = 1/2$. The acceleration is $a_{solid} = \frac{F}{m(1+1/2)} = \frac{F}{1.5m} = \frac{2F}{3m}$.
   • For a thin-walled hollow cylinder, $I = mR^2$, so $k=1$. The acceleration is $a_{hollow} = \frac{F}{m(1+1)} = \frac{F}{2m}$. Since $a_{solid} \neq a_{hollow}$, the acceleration $a$ depends on whether the cylinder is solid or hollow. Thus, statement A is incorrect.

   B: For a solid cylinder, the maximum possible value of a is 2 μ g

   • The condition for rolling without slipping is that the static friction force $f$ must be less than or equal to the maximum possible static friction, $f_{max} = μN = μmg$. $f ≤ μmg$
   • For a solid cylinder, $k=1/2$, so the friction force is $f = kma = \frac{1}{2}ma$.
   • Substituting this into the inequality: $\frac{1}{2}ma ≤ μmg$ $a ≤ 2μg$ The maximum possible value of acceleration for a solid cylinder to roll without slipping is $a_{max} = 2μg$. Thus, statement B is correct.

   C: The magnitude of the frictional force on the object due to the ground is always μ mg

   • The frictional force is $f = kma$. Its value depends on the acceleration $a$ (which in turn depends on the applied force $F$).
   • The value $μmg$ represents the maximum available static friction. The actual static friction $f$ is an adjustable force and is only equal to $μmg$ when the object is on the verge of slipping ($a = a_{max}$). In general, $f ≤ μmg$. Thus, statement C is incorrect.

   D: For a thin-walled hollow cylinder, $a = {F \over {2m}}$

   • For a thin-walled hollow cylinder, the moment of inertia is $I = mR^2$, which corresponds to $k=1$.
   • Using the general formula for acceleration, $a = \frac{F}{m(1+k)}$: $a = \frac{F}{m(1+1)} = \frac{F}{2m}$ The problem states that the object rolls without slipping, so we assume the conditions for this are met (i.e., $f ≤ μmg$). Under this condition, the acceleration is indeed $F/(2m)$. Thus, statement D is correct.