What is the formula for torque in rotational motion?

Torque is given by $\tau = I\alpha$, where $I$ is the moment of inertia and $\alpha$ is the angular acceleration.

How do you calculate angular acceleration?

Angular acceleration $\alpha$ is the rate of change of angular velocity, calculated as $\alpha = \frac{\omega_f - \omega_i}{t}$.

What is the moment of inertia of a solid disk about its diameter?

The moment of inertia of a solid disk of mass $M$ and radius $R$ about an axis passing through its diameter is $I = \frac{1}{4}MR^2$.

How do you convert rpm to rad/s?

To convert revolutions per minute (rpm) to radians per second (rad/s), multiply by $\frac{2\pi}{60}$.

What is the relation between diameter and radius?

The diameter $D$ is twice the radius ($D = 2R$).

What are the initial and final angular velocities in this problem?

Initial $\omega_i = 60\pi$ rad/s and final $\omega_f = 70\pi$ rad/s.

Is mass used in finding the radius?

Yes, the mass $M=1$ kg is a required parameter in the moment of inertia formula to solve for $R$.

Why is the moment of inertia $\frac{1}{4}MR^2$ and not $\frac{1}{2}MR^2$?

$\frac{1}{2}MR^2$ is for the axis perpendicular to the disk through the center. For the diameter axis, we use the perpendicular axis theorem which results in $\frac{1}{4}MR^2$.

What is the SI unit of torque?

The SI unit of torque is Newton-meter (Nm).

How does torque affect angular speed?

A constant external torque produces constant angular acceleration, which linearly increases the angular speed over time.

A thin solid disk of 1 kg is rotating along its diameter axis at the speed of 1800 rpm. By applying an external torque of 25π Nm for 40 s, the speed increases to 2100 rpm. The diameter of the disk is ___

A thin solid disk of 1 kg is rotating along its diameter axis at the speed of 1800 rpm. By applying an external torque of 25π Nm for 40 s, the speed increases to 2100 rpm. The diameter of the disk is ____ m. | JEE Main Mathematics

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PhysicsRotational Dynamics

QNumerical

A thin solid disk of $1$ kg is rotating along its diameter axis at the speed of $1800$ rpm. By applying an external torque of $25\pi$ Nm for $40$ s, the speed increases to $2100$ rpm. The diameter of the disk is ____ m.

✅ Correct Answer

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Solution Steps

Convert Speeds to Angular Velocity ($\omega$)

Initial angular velocity, $\omega_i = 1800 \text{ rpm} = \frac{1800 \times 2\pi}{60} = 60\pi \text{ rad/s}$.

Final angular velocity, $\omega_f = 2100 \text{ rpm} = \frac{2100 \times 2\pi}{60} = 70\pi \text{ rad/s}$.

Calculate Angular Acceleration ($\alpha$)

Using the kinematic equation $\omega_f = \omega_i + \alpha t$:

$$70\pi = 60\pi + \alpha(40)$$

$$10\pi = 40\alpha \implies \alpha = \frac{\pi}{4} \text{ rad/s}^2$$

Relate Torque to Moment of Inertia

Using $\tau = I\alpha$, where $\tau = 25\pi$ Nm:

$$25\pi = I \left(\frac{\pi}{4}\right)$$

$$I = 25 \times 4 = 100 \text{ kg}\cdot\text{m}^2$$

Find Radius using Moment of Inertia Formula

For a disk about its diameter, $I = \frac{1}{4}MR^2$. Given $M = 1$ kg:

$$100 = \frac{1}{4}(1)R^2 \implies R^2 = 400$$

$$R = \sqrt{400} = 20 \text{ m}$$

Calculate Diameter

Diameter $D = 2R = 2 \times 20 = 40$ m.

Final Answer: 40

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Theory

1. Moment of Inertia of a Disk

The moment of inertia (MOI) represents an object’s resistance to rotational acceleration. For a uniform thin solid disk of mass $M$ and radius $R$, the MOI depends on the axis of rotation. About the central longitudinal axis (perpendicular to the plane), $I = \frac{1}{2}MR^2$. However, about any diameter axis (lying in the plane of the disk), we apply the Perpendicular Axis Theorem ($I_z = I_x + I_y$), which yields $I_{dia} = \frac{1}{4}MR^2$. Understanding the specific axis is critical for JEE problems.

2. Rotational Kinematics

Rotational kinematics describes the motion of rotating objects using angular displacement ($\theta$), angular velocity ($\omega$), and angular acceleration ($\alpha$). These variables are analogous to linear displacement, velocity, and acceleration. Under constant torque, the angular acceleration is constant, allowing the use of equations like $\omega_f = \omega_i + \alpha t$ and $\theta = \omega_i t + \frac{1}{2}\alpha t^2$. Always ensure angular velocity is converted from RPM to rad/s ($1 \text{ rpm} = \pi/30 \text{ rad/s}$) for SI consistency.

3. Torque and Newton’s Second Law

Torque ($\tau$) is the rotational equivalent of force. According to Newton’s Second Law for rotation, the net external torque applied to a rigid body is equal to the product of its moment of inertia and its angular acceleration ($\tau = I\alpha$). This relationship is valid when $I$ is measured about the same axis as the torque. In this problem, a constant torque over a period of time causes a linear change in the angular momentum and angular velocity of the disk.

4. Work and Energy in Rotation

Applying torque over an angular displacement results in work done: $W = \tau \theta$. This work increases the rotational kinetic energy of the system, given by $K_{rot} = \frac{1}{2}I\omega^2$. The Work-Energy Theorem for rotation states that the net work done by external torques equals the change in rotational kinetic energy. While not directly required to find the diameter here, energy methods are often faster alternatives for complex JEE problems involving varying torques or speeds.

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FAQs

Why did we use $I = \frac{1}{4}MR^2$?

Because the problem specifies the rotation is “along its diameter axis” rather than its center perpendicular axis.

How do I convert 1800 rpm to rad/s?

$\omega = \frac{1800 \times 2\pi}{60} = 30 \times 2\pi = 60\pi \text{ rad/s}$.

What if the disk was a ring?

For a ring about its diameter, $I = \frac{1}{2}MR^2$. The resulting diameter would be different.

Is the “thin” disk assumption important?

Yes, “thin” implies we can treat it as a 2D object and ignore the height (thickness) in MOI calculations.

What is the value of $\alpha$ in this question?

The angular acceleration $\alpha$ is $\pi/4$ rad/s².

Can the diameter be negative?

No, diameter is a physical length and must be positive ($+40$ m).

What is the SI unit of Moment of Inertia?

The unit is $\text{kg}\cdot\text{m}^2$.

Does the time (40s) affect the MOI?

No, MOI is a property of mass distribution; time only affects how much the speed changes for a given torque.

What if torque was not constant?

Then $\alpha$ would vary, and we would need to use integration: $\int \tau dt = I(\omega_f – \omega_i)$.

How do I calculate $\pi$ in the final answer?

Notice that $\pi$ cancels out during the $\tau = I\alpha$ calculation, leaving a purely numerical result.

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