Problem 605 - JEE Physics

Question

A gramophone record of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) is rotating with angular speed \(\mathrm{W}\). If two pieces of wax each of mass \(\mathrm{M}\) are kept on it at a distance of \(R / 2\) from the centre on opposite side then the new angular velocity will be..... \(\\{\mathrm{A}\\}(\omega / 2)\) \(\\{B\\}[m \omega /(M+m)\) \(\\{C\\}[M \omega /(M+m)]\) \(\\{\mathrm{D}\\}[(\mathrm{M}+\mathrm{m}) / \mathrm{M} \omega]\)

Step-by-Step Solution

Answer

The new angular velocity is \(\omega' = \frac{M\omega}{M + m}\), which corresponds to option C.

Step 1: Calculate the initial angular momentum

Before the wax pieces are added, the gramophone record's angular momentum can be found using the formula: \[L_{i} = I_{record} \omega\] Where \(L_{i}\) is the initial angular momentum, \(I_{record}\) is the moment of inertia of the record, and \(\omega\) is the given angular speed. For a disc, the moment of inertia can be calculated as: \[I_{record} = \frac{1}{2} MR^2\] Now, substitute this expression into the initial angular momentum equation: \[L_{i} = \frac{1}{2} MR^2 \omega\]

Step 2: Determine the final angular momentum

After placing the wax pieces on the record, the moment of inertia of each wax piece can be calculated as: \[I_{wax} = mr^2\] Where \(m\) is the mass of each wax piece, and \(r = \frac{R}{2}\) is the distance from the center of the record. The total moment of inertia for the new system is the sum of the moments of inertia of the record and both of the wax pieces, so we can write: \[I_{total} = I_{record} + 2I_{wax}\] Substitute the expressions for the moments of inertia into this equation: \[I_{total} = \frac{1}{2} MR^2 + 2m\left(\frac{R}{2}\right)^2\] The final angular momentum of the system is: \[L_{f} = I_{total} \omega'\] Where \(\omega'\) is the new angular velocity.

Step 3: Set up the conservation of angular momentum equation

As no external torques are acting on the system, we can equate the initial and final angular momentum of the system: \[L_{i} = L_{f}\] Now, substitute the expressions for the initial and final angular momentum into this equation: \[\frac{1}{2} MR^2 \omega = I_{total} \omega'\]

Step 4: Solve for the new angular velocity

To find the new angular velocity \(\omega'\), we can divide both sides of the conservation of angular momentum equation by \(I_{total}\): \[\omega' = \frac{\frac{1}{2} MR^2 \omega}{\frac{1}{2} MR^2 + 2m\left(\frac{R}{2}\right)^2}\] Simplify the expression: \[\omega' = \frac{MR^2\omega}{MR^2 + \frac{1}{2} mR^2}\] Finally, we can see that the correct answer is: \[\omega' = \frac{M\omega}{M + m}\] Which corresponds to option C.

Key Concepts

Moment of Inertia

The moment of inertia, often referred to as the 'rotational analog' of mass in linear motion, tells us how much resistance an object has to changes in its rotational motion. For any object spinning around an axis, the moment of inertia is crucial in understanding its rotational dynamics.
In the case of our spinning gramophone record, we calculate its moment of inertia using the formula for a disc: \[ I_{record} = \frac{1}{2} MR^2 \] Here, \( M \) is the mass of the record, and \( R \) is its radius. This formula captures how the mass distribution and radius influence the moment of inertia.

When additional masses, like wax pieces in the exercise, are placed on the record at certain distances from the axis, we also consider their moment of inertia. For each wax piece, this is \( I_{wax} = mr^2 \), where \( m \) is the mass of each wax piece and \( r \) is the distance from the center of the record. Summing up the moments of inertia gives us the total moment of inertia for the entire system. This helps in determining how the extra mass influences the spinning motion.

Angular Velocity

Angular velocity refers to how fast something is rotating. It is the rate at which an object travels around a circle or part of a circle. The measure for angular velocity is usually in radians per second. In the exercise, the gramophone record initially spins at a certain angular velocity, given by \( \omega \).

Adding the wax pieces affects the angular velocity due to changes in the system's moment of inertia. Initially, the angular velocity is determined by the record's rotation alone, but when the wax is added, we must find the new angular velocity \( \omega' \). This is found by: \[ \omega' = \frac{M\omega}{M + m} \] This equation shows us how the initial angular velocity \( \omega \) decreases based on the mass and distribution of the added wax.

By using the conservation of angular momentum, we understand that even though the moment of inertia changes, the product of moment of inertia and angular velocity (angular momentum) remains constant unless external forces come into play.

Rotational Dynamics

Rotational dynamics explores how and why objects rotate—that is, the interaction between forces and rotational motion. To understand the exercise, we must tap into the principle of conservation of angular momentum, a core concept in rotational dynamics.

In the absence of external torques, the total angular momentum of a system remains constant. This principle allows us to determine how changes in moment of inertia affect angular velocity. In the provided exercise, as the moment of inertia increases with the additional wax, the angular velocity decreases to compensate and maintain the same angular momentum.

• The initial angular momentum \( L_i \) is the product of the initial moment of inertia and the initial angular velocity: \[ L_i = I_{record} \cdot \omega \]
• After the wax is added, we have a new total moment of inertia \( I_{total} \). The system's new angular momentum \( L_f \) is \[ L_f = I_{total} \cdot \omega' \]
• Since angular momentum is conserved, \( L_i = L_f \), allowing us to solve for the new angular velocity \( \omega' \).

The calculations reinforce understanding how mass distribution affects rotational dynamics and angular momentum conservation.