A carpet of mass M made of inextensible material is rolled along its length in the form of a cylinde

Q: Why is the work done by friction zero?

The carpet unrolls without slipping. In pure rolling, the point of contact is instantaneously at rest relative to the ground. Since there is no relative displacement at the point of application of the friction force, the work done by static friction is zero.

Q: Why does the mass of the rolled part change?

As the carpet unrolls, a portion of it lies flat on the ground. The rolled part becomes smaller. Since the carpet is uniform, its mass is proportional to its cross-sectional area ($\pi r^2$). When the radius halves, the area (and thus the mass) becomes one-fourth.

Q: Why doesn't the unwound part have kinetic energy?

The unwound part is lying flat and stationary on the floor. It is not moving, so its velocity is zero, meaning its kinetic energy is zero.

Q: How do we calculate the change in potential energy?

We track the center of mass. Initially, the entire mass $M$ is at height $R$. Finally, the rolled part of mass $M/4$ is at height $R/2$, and the unwound part of mass $3M/4$ is at height $0$. The difference between the initial and final potential energies gives the change.

Q: Why do we use the moment of inertia of a solid cylinder?

A tightly rolled carpet can be approximated as a uniform solid cylinder. The moment of inertia of a solid cylinder about its central axis is $\frac{1}{2} m r^2$.

Question

A carpet of mass $M$ made of inextensible material is rolled along its length in the form of a cylinder of radius $R$ and is kept on a rough floor. The carpet starts unrolling without sliding on the floor when a negligibly small push is given to it. Calculate the horizontal velocity of the axis of the cylindrical part of the carpet when its radius reduces to $R/2$.

Sigma Percentile · Accepted Answer

Imagine you are standing in a room with a heavy, tightly rolled carpet. You give it a tiny push, and it begins to unroll across the floor. As it unrolls, the cylindrical part gets smaller and smaller. Our goal is to find the velocity of the center of this cylinder exactly when its radius has halved from $R$ to $R/2$.

### The Mass of the Remaining Roll

The first thing we must realize is that the mass of the moving cylindrical part is not constant. As the carpet unrolls, a portion of it lies flat and stationary on the floor. Because the carpet is uniform, its mass is directly proportional to its cross-sectional area.

The initial area of the roll is $\pi R^2$, corresponding to the total mass $M$. When the radius reduces to $R/2$, the new cross-sectional area becomes:

$$A_f = \pi \left(\frac{R}{2}\right)^2 = \frac{\pi R^2}{4}$$

This is exactly one-fourth of the original area. Therefore, the mass of the remaining rolled part is:

$$M' = \frac{M}{4}$$

### Conservation of Mechanical Energy

Here is a crucial conceptual catch: the carpet unrolls **without slipping**. This means the point of contact between the roll and the floor is instantaneously at rest. Because there is no relative motion at the point of contact, the work done by friction is zero.

With no non-conservative forces doing work, we can confidently apply the principle of conservation of mechanical energy. The decrease in the system's potential energy will perfectly equal the gain in its kinetic energy.

### Tracking the Potential Energy

Let's calculate how much potential energy the system loses. Initially, the entire mass $M$ is rolled up, and its center of mass is at a height $R$ above the ground. The initial potential energy is:

$$U_i = MgR$$

In the final state, the system consists of two parts: the remaining roll and the flat unwound carpet. The flat part is on the ground, so its height (and thus its potential energy) is zero. The remaining roll has a mass of $M/4$ and its center of mass is at a height of $R/2$. The final potential energy is:

$$U_f = \left(\frac{M}{4}\right) g \left(\frac{R}{2}\right) = \frac{MgR}{8}$$

The total decrease in potential energy is the difference between the two:

$$\Delta U = U_i - U_f = MgR - \frac{MgR}{8} = \frac{7MgR}{8}$$

### The Kinetic Energy Components

Now, let's look at the kinetic energy. The unwound part of the carpet is lying dead still on the floor, so it has absolutely zero kinetic energy. All the kinetic energy resides in the moving, rotating cylindrical roll.

The roll has both translational kinetic energy (because its center of mass is moving) and rotational kinetic energy (because it is spinning). The total final kinetic energy is:

$$K_f = \frac{1}{2} M' v^2 + \frac{1}{2} I \omega^2$$

We can model the tightly rolled carpet as a solid cylinder. The moment of inertia of a solid cylinder about its central axis is $\frac{1}{2} m r^2$. Substituting our final mass and radius:

$$I = \frac{1}{2} \left(\frac{M}{4}\right) \left(\frac{R}{2}\right)^2 = \frac{MR^2}{32}$$

### The Pure Rolling Constraint

Because the carpet is in pure rolling, its translational velocity $v$ and angular velocity $\omega$ are locked together by the relation $v = \omega r$. For our final state, the radius is $R/2$, so:

$$v = \omega \left(\frac{R}{2}\right) \implies \omega = \frac{2v}{R}$$

Let's substitute the moment of inertia and angular velocity back into our kinetic energy equation:

$$K_f = \frac{1}{2} \left(\frac{M}{4}\right) v^2 + \frac{1}{2} \left(\frac{MR^2}{32}\right) \left(\frac{2v}{R}\right)^2$$

Simplifying the rotational term:

$$K_{rot} = \frac{1}{2} \left(\frac{MR^2}{32}\right) \left(\frac{4v^2}{R^2}\right) = \frac{Mv^2}{16}$$

Adding the translational and rotational parts together:

$$K_f = \frac{Mv^2}{8} + \frac{Mv^2}{16} = \frac{3Mv^2}{16}$$

### The Final Calculation

We have reached the grand finale. The energy lost by the system as it drops must equal the kinetic energy it gains. Equating the decrease in potential energy to the total kinetic energy:

$$\frac{7MgR}{8} = \frac{3Mv^2}{16}$$

We can cancel $M$ from both sides and multiply by 16:

$$14gR = 3v^2$$

Solving for the velocity $v$, we get our beautiful final result:

$$v = \sqrt{\frac{14Rg}{3}}$$

And there we have it! By carefully tracking the changing mass and applying energy conservation, we've unlocked the motion of the unrolling carpet.

Visualized Solution

$R \to \frac{R}{2}$

$M^{'} = \frac{M}{4}$

$Δ U = Δ K$

$Δ U = \frac{7 M g R}{8}$

$K_{f} = K_{t r an s} + K_{r o t}$

$ω = \frac{2 v}{R}$

$K_{f} = \frac{3 M v ^{2}}{16}$

$v = \frac{14 R g}{3}$

The Sigma Insight: Work and Energy in Rotational Motion

The Mass of the Remaining Roll

Conservation of Mechanical Energy

Tracking the Potential Energy

The Kinetic Energy Components

The Pure Rolling Constraint

The Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R→2R​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M′=4M​

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔU=87MgR​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kf​=Ktrans​+Krot​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ω=R2v​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kf​=163Mv2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v=314Rg​​