Problem 660 - JEE Physics

Question

A circular disc of radius \(R\) is free to oscillate about an axis passing through a point on its rim and perpendicular to its plane. The disc is turned through an angle of \(60 ?\) and released. Its angular velocity when it reaches the equilibrium position will be \(\\{\mathrm{A}\\} \sqrt{(\mathrm{g} / 3 \mathrm{R})}\) \(\\{\mathrm{B}\\} \sqrt{(2 \mathrm{~g} / 3 \mathrm{R})}\) \(\\{\mathrm{C}\\} \sqrt{(2 \mathrm{~g} / \mathrm{R})}\) \(\\{\mathrm{D}\\} 2 \sqrt{(2 \mathrm{~g} / \mathrm{R})}\)

Step-by-Step Solution

Answer

The correct answer is the angular velocity ω = \(\\{\mathrm{B}\\} \sqrt{(2 \mathrm{~g} / 3 \mathrm{R})}\), which is obtained by using the conservation of mechanical energy principle and solving for ω.

Step 1: Calculate the Initial Gravitational Potential Energy

First, let's calculate the initial gravitational potential energy when the disc is turned through an angle of 60°. The center of mass of the disc is located at a distance R/2 from the axis of rotation. In the initial position, the height of the center of mass above the equilibrium position can be calculated using trigonometry: Height = (R/2) * (1 - cos(60°)) Now, we can calculate the gravitational potential energy (PE_initial): PE_initial = m * g * Height, where m is the mass of the disc.

Step 2: Calculate the Final Rotational Kinetic Energy

When the disc reaches the equilibrium position, its gravitational potential energy is converted to rotational kinetic energy (KE_final): KE_final = (1/2) * I * ω^2, where ω is the angular velocity, and I is the moment of inertia of the disc. The moment of inertia of the circular disc about the given axis can be calculated with the parallel axis theorem: I = (1/2) * m * R^2 + m * (R/2)^2 = (3/4) * m * R^2

Step 3: Equate the Initial Gravitational Potential Energy to the Final Rotational Kinetic Energy

By conservation of energy, the initial gravitational potential energy equals the final rotational kinetic energy: PE_initial = KE_final m * g * Height = (1/2) * (3/4) * m * R^2 * ω^2

Step 4: Solve for the Angular Velocity ω

Now, we can solve for the angular velocity ω: ω^2 = (2 * g * Height) / ((3/2) * R^2) Since Height = (R/2) * (1 - cos(60°)) = R/4, we can substitute it into the equation: ω^2 = (2 * g * (R/4)) / ((3/2) * R^2) ω^2 = (2 * g) / (3 * R) Now, we can find the angular velocity ω by taking the square root: ω = √((2 * g) / (3 * R)) So the correct answer is: \(\\{\mathrm{B}\\} \sqrt{(2 \mathrm{~g} / 3 \mathrm{R})}\).

Key Concepts

Rotational Kinetic Energy

When discussing rotational motions, it's important to understand rotational kinetic energy, which is the energy due to the rotation of an object around an axis. For a rotating disc, this energy can be expressed as follows:

• KE_final = \(\frac{1}{2} I \omega^2\)

where \(I\) is the moment of inertia, and \(\omega\) is the angular velocity.
Rotational kinetic energy is analogous to translational kinetic energy (\(\frac{1}{2} mv^2\)), where mass is replaced by moment of inertia and linear velocity by angular velocity.
In the case of the oscillating disc, once it is released and it reaches the equilibrium position, all of its gravitational potential energy has been converted into rotational kinetic energy.

Gravitational Potential Energy

Gravitational potential energy refers to the energy an object possesses because of its position in a gravitational field. When the disc is turned through an angle of 60°, its center of mass is elevated, resulting in gravitational potential energy as given by:

• PE_initial = \(m \cdot g \cdot \text{Height}\)

where \(m\) is the mass of the disc, \(g\) is the acceleration due to gravity, and Height is the vertical elevation of the center of mass.
For a circular disc, the height is determined with the equation:

• Height = \(\frac{R}{2} (1 - \cos 60°) = \frac{R}{4}\)

Understanding how potential energy is transformed as the disc moves is essential, especially in calculating its conversion into rotational kinetic energy.

Conservation of Energy

The principle of conservation of energy is fundamental in physics and states that energy can neither be created nor destroyed; it can only be transformed from one form to another. For the oscillating disc, this principle implies that the initial gravitational potential energy must be equal to the final rotational kinetic energy:

• \(m \cdot g \cdot \text{Height} = \frac{1}{2} I \omega^2\)

This relationship allows the equation setup where we equate both forms of energy to solve for unknown variables like the angular velocity.
This principle enables us to connect the height and angular velocity by translating the change in energy states.

Moment of Inertia

Moment of inertia is a measure of an object's resistance to changes in its rotational motion and depends on mass distribution relative to the axis of rotation. For a circular disc oscillating about an axis at its rim, the moment of inertia is calculated by the parallel axis theorem:

• \(I = \frac{1}{2} m R^2 + m \left(\frac{R}{2}\right)^2 = \frac{3}{4} m R^2\)

The formula takes into account both the inherent rotational inertia of the disc and its added inertia due to its offset by a distance of \(R/2\) from its center of mass.
Moment of inertia plays a crucial role in determining how fast the disc will spin once released, affecting the angular velocity calculation.