Problem 13 - Master Resource Book in JEE Main Physics

Question

A string is would round the rim of a mounted fly wheel of mass \(20 \mathrm{~kg}\) and radius \(20 \mathrm{~cm}\). A steady ball of \(25 \mathrm{~N}\) is applied on the cord. Neglecting friction and mass of the string, the angular acceleration of the wheel is (a) \(50 \mathrm{rad} \mathrm{s}^{-2}\) (b) \(25 \mathrm{rad} \mathrm{s}^{-2}\) (c) \(6.25 \mathrm{rad} \mathrm{s}^{-2}\) (d) \(12.5 \mathrm{rad} \mathrm{s}^{-2}\)

Step-by-Step Solution

Answer

The angular acceleration of the wheel is \(12.5\, \text{rad/s}^2\) (Option d).

Step 1: Understand the problem

The problem is about finding the angular acceleration of a flywheel when a force is applied to a string wrapped around it. The flywheel has a mass of \(20\, \text{kg}\) and a radius of \(20\, \text{cm}\). An effective weight of \(25\, \text{N}\) is pulling on the string.

Step 2: Calculate the torque

Torque \(\tau\) is given by the formula \(\tau = F \times r\), where \(F\) is the force applied (\(25\, \text{N}\)) and \(r\) is the radius of the wheel (\(0.2\, \text{m}\)). Thus, the torque is \(\tau = 25\, \text{N} \times 0.2\, \text{m} = 5\, \text{Nm}\).

Step 3: Determine the moment of inertia

For a solid disk or wheel, the moment of inertia \(I\) is given by \(\frac{1}{2}mr^2\), where \(m\) is the mass and \(r\) is the radius. Substituting the given values, \(I = \frac{1}{2} \times 20\, \text{kg} \times (0.2\, \text{m})^2 = 0.4\, \text{kg} \cdot \text{m}^2\).

Step 4: Use Newton's Second Law for Rotation

Newton's second law for rotation states that \(\tau = I\alpha\), where \(\alpha\) is the angular acceleration. Substituting the values from steps 2 and 3, we have \(5\, \text{Nm} = 0.4\, \text{kg} \cdot \text{m}^2 \times \alpha\).

Step 5: Solve for Angular Acceleration

Rearrange the equation from Step 4 to solve for \(\alpha\). This gives \(\alpha = \frac{5\, \text{Nm}}{0.4\, \text{kg} \cdot \text{m}^2} = 12.5\, \text{rad/s}^2\). Therefore, the angular acceleration of the wheel is \(12.5\, \text{rad/s}^2\).

Key Concepts

Torque Calculation

In simple terms, torque can be thought of as a measure of how much force acts on an object to make it rotate. It's very similar to how force makes an object move in a straight line. But with torque, we are dealing with rotation. The formula for calculating torque \( \tau \) is:\[ \tau = F \times r\]

• \( F \) is the force applied, measured in newtons (N).
• \( r \) is the radius or distance from the center of rotation to where the force is applied, measured in meters (m).

In our problem, the force \( F \) applied is \( 25 \text{ N} \), and the radius \( r \) of the wheel's rim is \( 0.2 \text{ m} \). So, the torque \( \tau \) is calculated by multiplying the force and the radius:\[ \tau = 25 \text{ N} \times 0.2 \text{ m} = 5 \text{ Nm}\]This tells us that a torque of \( 5 \text{ Nm} \) is acting on the wheel, trying to rotate it around its axis.

Moment of Inertia

The moment of inertia is essentially the rotational equivalent of mass in linear motion. It determines how easy or difficult it is to change an object's rotational speed. The greater the moment of inertia, the harder it is to spin the object or stop it from spinning.For a solid wheel or disk, such as the flywheel in our problem, the moment of inertia \( I \) is calculated using the formula:\[ I = \frac{1}{2} m r^2\]

• \( m \) is the mass of the object, measured in kilograms (kg).
• \( r \) is the radius, measured in meters (m).

Substituting the values for the mass of the wheel \( 20 \text{ kg} \) and its radius \( 0.2 \text{ m} \), we find:\[ I = \frac{1}{2} \times 20 \text{ kg} \times (0.2 \text{ m})^2 = 0.4 \text{ kg} \cdot \text{m}^2\]This means the wheel has a moment of inertia of \( 0.4 \text{ kg} \cdot \text{m}^2 \), indicating how much rotational resistance it has.

Newton's Second Law for Rotation

Newton's second law for rotation bridges the gap between torque, moment of inertia, and angular acceleration. It tells us how the torque applied to an object relates to its moment of inertia and the angular acceleration it experiences.The law is expressed as:\[ \tau = I \alpha\]

• \( \tau \) represents torque, measured in newton-meters (Nm).
• \( I \) stands for the moment of inertia in \( \text{kg} \cdot \text{m}^2 \).
• \( \alpha \) is the angular acceleration, measured in radians per second squared \((\text{rad/s}^2)\).

From our earlier calculations, we know the torque \( \tau = 5 \text{ Nm} \) and the moment of inertia \( I = 0.4 \text{ kg} \cdot \text{m}^2 \). Rearranging the equation to solve for angular acceleration \( \alpha \):\[ \alpha = \frac{\tau}{I} = \frac{5 \text{ Nm}}{0.4 \text{ kg} \cdot \text{m}^2} = 12.5 \text{ rad/s}^2\]Thus, the angular acceleration of the wheel is \( 12.5 \text{ rad/s}^2 \). This demonstrates how quickly the wheel speeds up due to the applied force.