A string is wound around a hollow cylinder of mass 5 kg and radius 0.5 m. If the string is now pulle

Q: Why is the moment of inertia $MR^2$ and not $\frac{1}{2}MR^2$?

The problem specifies a *hollow* cylinder. In a hollow cylinder, all the mass is concentrated at the outer rim, at a distance $R$ from the center. Therefore, the moment of inertia is $MR^2$. A solid cylinder would have its mass distributed throughout, giving $\frac{1}{2}MR^2$.

Q: Why did we ignore the torque due to friction?

In this highly specific scenario—a hollow cylinder pulled exactly at its top edge—the static friction required to maintain pure rolling is mathematically zero! Since $f=0$, it produces no torque. If it were a solid cylinder, friction would not be zero, and we would have to account for it.

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

Absolutely! Taking torque about the Instantaneous Center of Rotation (ICR) at the bottom is a fantastic shortcut. The torque is $F \times 2R$, and the moment of inertia about the ICR is $I_{COM} + MR^2 = 2MR^2$. Equating them gives $F(2R) = (2MR^2)\alpha$, which simplifies directly to $\alpha = \frac{F}{MR}$, completely bypassing the need to calculate friction!

Q: Does the force $F$ do work in pure rolling?

Yes, the applied force $F$ does work because the point of application (the top of the cylinder) is moving. However, the static friction at the bottom does *no* work because the instantaneous point of contact is always at rest relative to the ground.

Q: What does "rolling without slipping" actually mean mathematically?

It means the velocity of the point of contact with the ground is zero. Mathematically, this enforces a strict relationship between the translational acceleration of the center of mass ($a$) and the angular acceleration ($\alpha$), given by the constraint equation $a = \alpha R$.

Question

A string is wound around a hollow cylinder of mass 5 kg and radius 0.5 m. If the string is now pulled with a horizontal force of 40 N and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string)

Options:

(A) 10 rad /s$^2$

(B) 16 rad /s$^2$

(C) 20 rad /s$^2$

(D) 12 rad /s$^2$

Sigma Percentile · Accepted Answer

### The Symphony of Pure Rolling

Imagine a hollow cylinder resting peacefully on a horizontal surface.

Suddenly, a string wrapped around its outer edge is pulled with a crisp, horizontal force.

This isn't just a simple push; it's the beginning of a beautiful physical symphony called **pure rolling**.

In pure rolling, the object translates forward while simultaneously spinning, perfectly synchronized so that the bottom point never slips against the ground.

Our mission is to find the **angular acceleration** of this hollow cylinder.

### The Friction Paradox

When we think of rolling without slipping, we immediately think of static friction.

Friction is usually the unsung hero that grips the ground and forces the object to spin instead of just sliding.

But here is a mind-blowing catch for this specific setup!

If you rigorously solve the equations of motion for a **hollow cylinder** pulled exactly at its top edge, the required static friction turns out to be exactly zero.

Because all the mass is concentrated at the rim, the force at the top provides exactly the right amount of torque to match the translational acceleration.

Since the static friction is zero, we can confidently calculate the torque about the center of mass using only the applied force.

### Setting Up the Torque

Let's calculate the torque generated by our applied force.

Torque is the rotational equivalent of force, calculated as the force multiplied by the perpendicular distance from the axis of rotation.

$$ 	au = F 	imes R $$

We are given a force of $40	ext{ N}$ and a radius of $0.5	ext{ m}$.

Substituting these values gives us the total torque acting on the cylinder.

$$ 	au = 40 	imes 0.5 = 20	ext{ N-m} $$

### The Resistance to Spin

Now, we must consider how much the cylinder resists this rotational push.

This resistance is quantified by the **moment of inertia**.

For a hollow cylinder, all of its mass is distributed as far away from the center as possible—right at the rim.

This makes it harder to spin compared to a solid cylinder.

The moment of inertia for a hollow cylinder is simply its mass times the square of its radius.

$$ I = mR^2 $$

### Newton's Second Law for Rotation

We are ready to bring it all together using **Newton's Second Law for Rotation**.

This fundamental law states that the net torque on an object equals its moment of inertia multiplied by its angular acceleration.

$$ 	au = I\alpha $$

We already know the torque is $20	ext{ N-m}$, and we have our expression for the moment of inertia.

Let's equate them to solve for the angular acceleration, $\alpha$.

$$ mR^2 \alpha = 20 $$

### The Final Calculation

It is time to plug in our final numbers and execute the math.

The mass $m$ is $5	ext{ kg}$, and the radius $R$ is $0.5	ext{ m}$.

$$ 5 	imes (0.5)^2 	imes \alpha = 20 $$

Be careful with the decimals here!

The square of $0.5$ is $0.25$.

$$ 5 	imes 0.25 	imes \alpha = 20 $$

Multiplying $5$ by $0.25$ gives us $1.25$.

$$ 1.25 \alpha = 20 $$

Finally, dividing $20$ by $1.25$ reveals our answer.

$$ \alpha = 16	ext{ rad/s}^2 $$

The hollow cylinder will accelerate rotationally at **$16	ext{ rad/s}^2$**.

### The Way Forward

We solved this elegantly, but physics always has more layers to explore.

What if the object was a **solid cylinder** instead of a hollow one?

Because its mass is distributed differently, its moment of inertia would be $\frac{1}{2}mR^2$.

If you pull a solid cylinder from the top, the static friction is no longer zero!

Try solving that variation using the **Instantaneous Center of Rotation** at the bottom contact point.

It is a fantastic shortcut that bypasses friction entirely!

Select Answer:

Visualized Solution

$Analyzing the Setup$

$Torque about Center of Mass$

$Calculating Torque$

$Calculating Torque$

$Newton’s Second Law for Rotation$

$Equating Torques$

$Substituting Values$

$Final Angular Acceleration$

$The Way Forward$

The Sigma Insight: Dynamics of Rigid Body Rotation

The Symphony of Pure Rolling

The Friction Paradox

Setting Up the Torque

The Resistance to Spin

Newton's Second Law for Rotation

The Final Calculation

The Way Forward

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A rigid massless rod of length $3 l$ has two masses attached at each end as shown in the figure. The rod is pivoted at point $P$ on the horizontal axis (see figure). When released from initial horizontal position, its instantaneous angular acceleration will be

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Torque about Center of Mass

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Torque

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Torque

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Newton’s Second Law for Rotation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating Torques

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Angular Acceleration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward

The Sigma Insight: Dynamics of Rigid Body Rotation

The Symphony of Pure Rolling

The Friction Paradox

Setting Up the Torque

The Resistance to Spin

Newton's Second Law for Rotation

The Final Calculation

The Way Forward

Similar Questions

$Analyzing the Setup$

$Torque about Center of Mass$

$Calculating Torque$

$Calculating Torque$

$Newton’s Second Law for Rotation$

$Equating Torques$

$Substituting Values$

$Final Angular Acceleration$

$The Way Forward$