A carpet of mass m made of inextensible material is rolled along its length in the form of a cyllinder 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 cyllindrical part of carpet when its radius reduces to r/2. Ans: √(14rg/3) Can you explain this answer?

Question

Answer

Problem
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 carpet when its radius reduces to r/2.

Solution
To solve this problem, we need to consider the conservation of angular momentum and the work-energy theorem.

Conservation of Angular Momentum
When the carpet starts unrolling without sliding, the angular momentum about its axis of rotation is conserved. Initially, the angular momentum is given by:

L_initial = I_initial * ω_initial

where:
- L_initial is the initial angular momentum
- I_initial is the moment of inertia of the carpet about its axis of rotation
- ω_initial is the initial angular velocity

As the carpet unrolls, its radius reduces to r/2. At this point, the angular momentum is given by:

L_final = I_final * ω_final

where:
- L_final is the final angular momentum
- I_final is the moment of inertia of the carpet about its axis of rotation when the radius is r/2
- ω_final is the final angular velocity

Since angular momentum is conserved, we can equate the initial and final angular momenta:

L_initial = L_final

I_initial * ω_initial = I_final * ω_final

Moment of Inertia of the Carpet
The moment of inertia of a cylindrical object can be calculated using the formula:

I = (1/2) * m * r^2

where:
- I is the moment of inertia
- m is the mass of the object
- r is the radius of the object

Initially, the moment of inertia of the carpet is:

I_initial = (1/2) * m * r^2

When the radius reduces to r/2, the moment of inertia becomes:

I_final = (1/2) * m * (r/2)^2 = (1/2) * m * (r^2/4)

Substituting into the Conservation Equation
Now, we can substitute the expressions for I_initial and I_final into the conservation equation:

(1/2) * m * r^2 * ω_initial = (1/2) * m * (r^2/4) * ω_final

Simplifying the equation:

r^2 * ω_initial = (r^2/4) * ω_final

ω_final = 4 * ω_initial

Relationship between Linear and Angular Velocity
The relationship between linear and angular velocity is given by:

v = r * ω

where:
- v is the linear velocity
- r is the radius of rotation
- ω is the angular velocity

Using this relationship, we can express the final linear velocity of the axis of the cylindrical part of the carpet:

v_final = (r/2) * ω_final

Substituting the value of ω_final:

v_final = (r/2) * 4 * ω_initial = 2 * r * ω_initial

Final Result
Since the initial velocity is not given, we

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