A horizontal force F is applied at the center of mass of a cylindrical object of mass m and radius R

Q: Why does friction act backwards? Shouldn't it oppose the motion?

Friction opposes *relative* motion at the point of contact. The force $F$ tries to pull the cylinder forward, which would cause the bottom point to slide forward. To prevent this slipping, static friction must act backwards.

Q: Why isn't the frictional force just equal to $\mu mg$?

$\mu mg$ is the *maximum* possible value of static friction (limiting friction). In rolling without slipping, static friction only provides whatever force is necessary to maintain the condition $a = \alpha R$. It's a self-adjusting force.

Q: Why does the hollow cylinder have a different acceleration than the solid one?

A hollow cylinder has a larger moment of inertia ($mR^2$) compared to a solid one ($\frac{1}{2}mR^2$). This means it's harder to rotate. More of the applied force $F$ is 'used up' to generate rotation, leaving less for translational acceleration.

Q: What happens if the applied force $F$ is very large?

If $F$ is too large, the required static friction to maintain rolling will exceed the maximum limit ($\mu mg$). At that point, the cylinder will start to slip while rolling.

Q: Can we solve this using torque about the bottom contact point?

Yes! Taking torque about the instantaneous center of rotation (the contact point) eliminates friction from the equation. $\tau_{\text{contact}} = F \cdot R = I_{\text{contact}} \cdot \alpha$. Using the parallel axis theorem, $I_{\text{contact}} = I_{\text{cm}} + mR^2$. This gives the acceleration directly!

Question

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 as shown in the figure. The coefficient of friction between the object and the ground is $\mu$. The center of mass of the object has an acceleration $a$. The acceleration due to gravity is $g$. Given that the object rolls without slipping, which of the following statement(s) is(are) correct ?

Options:

(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 $2\mu g$

(C) The magnitude of the frictional force on the object due to the ground is always $\mu mg$

(D) For a thin-walled hollow cylinder, $a = \frac{F}{2m}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are pushing a heavy cylindrical roller across the floor. You apply a horizontal force $F$ directly at its center of mass.

Because the cylinder is resting on a rough surface, it doesn't just slide; it rolls. This rolling motion is a beautiful interplay between pure translation and pure rotation.

To prevent the bottom point of the cylinder from sliding forward under the influence of $F$, a static frictional force $f$ must act backwards at the point of contact.

### The Master Equations

Let's break down the motion using Newton's laws. For the translational motion of the center of mass, the net force is the applied force minus friction.

$$F - f = ma$$

For the rotational motion, the applied force $F$ passes right through the center of mass, so it creates zero torque. The only force creating a torque is the static friction $f$.

$$fR = I\alpha$$

Since the problem states the cylinder rolls **without slipping**, the translational acceleration $a$ and angular acceleration $\alpha$ are perfectly locked together by the constraint equation.

$$a = \alpha R$$

### Solving for Acceleration

Now, let's do some algebra to find the acceleration. We can substitute $\alpha = \frac{a}{R}$ into our torque equation to express friction in terms of acceleration.

$$f = \frac{Ia}{R^2}$$

Substituting this expression for $f$ back into our translational force equation gives us a master equation for acceleration.

$$F - \frac{Ia}{R^2} = ma$$

Factoring out $a$, we arrive at a highly revealing result.

$$a = \frac{F}{m + \frac{I}{R^2}}$$

Notice how the acceleration intrinsically depends on the moment of inertia $I$! This means that a solid cylinder and a hollow cylinder of the same mass will accelerate differently under the same force. Therefore, option (A) is incorrect.

### The Hollow Cylinder Case

Let's test option (D) by applying our master equation to a thin-walled hollow cylinder. For a hollow cylinder, all the mass is concentrated at the rim, so its moment of inertia is $I = mR^2$.

Substituting this into our acceleration formula, the $R^2$ terms cancel out beautifully.

$$a = \frac{F}{m + \frac{mR^2}{R^2}} = \frac{F}{2m}$$

The acceleration is exactly $\frac{F}{2m}$. This confirms that option (D) is absolutely correct!

### The Nature of Static Friction

What about the frictional force itself? Let's look back at our expression for $f$.

$$f = \frac{I}{R^2} a = \frac{F}{\frac{mR^2}{I} + 1}$$

This equation tells us a profound truth: **static friction is a self-adjusting force**. It only provides the exact amount of force necessary to maintain rolling without slipping.

It depends on the applied force $F$ and the mass distribution $I$. It is *not* always equal to its maximum limiting value $\mu mg$. Therefore, option (C) is incorrect.

### Pushing to the Limit

Finally, let's find the maximum possible acceleration before the cylinder starts to slip. For rolling to be maintained, the required static friction $f$ must not exceed the limiting friction $\mu mg$.

$$f \le \mu mg$$

Substituting our expression $f = \frac{Ia}{R^2}$, we get a condition for the maximum acceleration.

$$\frac{Ia}{R^2} \le \mu mg \implies a \le \frac{\mu mg R^2}{I}$$

Let's apply this to a solid cylinder, where the moment of inertia is $I = \frac{1}{2}mR^2$.

$$a \le \frac{\mu mg R^2}{\frac{1}{2}mR^2} = 2\mu g$$

The maximum possible acceleration for a solid cylinder is exactly $2\mu g$. This confirms that option (B) is correct!

### Final Conclusion

By systematically applying the principles of rolling dynamics, we have evaluated every statement. The correct options are (B) and (D).

Select Answer:

Visualized Solution

Free Body Diagram

Equations of Motion

Solving for Acceleration $a$

Thin-Walled Hollow Cylinder

Frictional Force $f$

Maximum Acceleration for Solid Cylinder

Conclusion

The Sigma Insight: Rolling Motion

Analyzing the Setup

The Master Equations

Solving for Acceleration

The Hollow Cylinder Case

The Nature of Static Friction

Pushing to the Limit

Final Conclusion

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