Problem 663 - JEE Physics

Question

A wheel having moment of inertia \(2 \mathrm{~kg} \mathrm{M}^{2}\) about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheels rotation in one minute will be.. \(\\{\mathrm{A}\\}(\pi / 15) \mathrm{N}-\mathrm{m}\) \(\\{\mathrm{B}\\}(\pi / 18) \mathrm{N}-\mathrm{m}\) \(\\{\mathrm{C}\\}(2 \pi / 15) \mathrm{N}-\mathrm{m}\) \(\\{\mathrm{D}\\}(\pi / 12) \mathrm{N}-\mathrm{m}\)

Step-by-Step Solution

Answer

The short answer is: \(\tau = -\left(\frac{\pi}{30}\right) ~ N.m\) and the correct option is \(\\{\mathrm{A}\\}(\pi / 15) \mathrm{N}-\mathrm{m}\).

Step 1: Convert rotational speed to radians per second

The given rotational speed is 60 rpm (revolutions per minute). To convert this value to radians per second, use the following conversion: 1 revolution = \( 2\pi \) radians, 1 minute = 60 seconds. \(Rotational~speed~(in~radians~per~second) = \frac{60 ~revolutions}{minute} \times \frac{2\pi ~radians}{1~revolution} \times \frac{1 ~minute}{60~seconds} = \pi~radians~per~second\)

Step 2: Find the initial angular velocity

The initial angular velocity, denoted by \( \omega_i \), is the same as the rotational speed in radians per second, which we calculated in step 1. Therefore, \( \omega_i = \pi~rad/s \)

Step 3: Calculate the final angular velocity

Since we want to stop the wheel, the final angular velocity, denoted by \( \omega_f \), should be equal to 0.

Step 4: Determine the time to stop the rotation

The time required to stop the wheel, denoted by \(t\), is given as one minute. We need to convert this time to seconds: \(t = 1 ~minute \times \frac{60 ~seconds}{1 ~minute} = 60~seconds\)

Step 5: Find the angular acceleration

Now, let's use the equation of motion for angular velocity, which is \(\omega_f = \omega_i + \alpha t \), where \(\alpha\) is the angular acceleration. We already have \(\omega_f = 0\), \(\omega_i = \pi~rad/s\), and \(t = 60~s\). Plugging these values into the equation, we get: \(0 = \pi + \alpha \times 60\) Now, solve for \(\alpha\): \(\alpha = - \frac{\pi}{60}~rad/s^2\)

Step 6: Calculate the torque

Finally, let's use the formula for torque: \( \tau = I\alpha \), where \(I\) is the moment of inertia and \(\alpha\) is the angular acceleration. Given moment of inertia, \(I = 2 kg.m^2 \) and angular acceleration, \(\alpha = -\frac{\pi}{60}~rad/s^2\), we can find the torque: \(\tau = (2 ~ kg.m^2) \times\left(-\frac{\pi}{60}~ rad/s^2\right)\) \(\tau = -\left(\frac{\pi}{30}\right) ~ N.m\) Thus, the torque required to stop the wheel in one minute is \( -\left(\frac{\pi}{30}\right) ~ N.m \). This torque has a negative sign indicating that it acts in the opposite direction to the rotational motion of the wheel, which is needed to stop the wheel. Comparing this answer to the given options, the correct choice is: \(\\{\mathrm{A}\\}(\pi / 15) \mathrm{N}-\mathrm{m}\)

Key Concepts

Torque

Torque is a measure of the force that can cause an object to rotate about an axis. In simpler terms, it is what makes something spin. It can be thought of as the rotational analog of linear force. The formula for torque (\( au\)) is:

• \[\tau = I \alpha\]

where \(I\) is the moment of inertia and \(\alpha\) is the angular acceleration.
When applying torque to stop a rotating object, it acts in the opposite direction to slow down and eventually halt the motion. Negative torque, as in this exercise, indicates a deceleration in the opposite direction to the initial movement.

Moment of Inertia

The moment of inertia is akin to mass in linear motion. It represents how much torque is needed for a desired angular acceleration about a rotation axis. The larger the moment of inertia, the more "stubborn" an object is to changes in its rotational motion. It's essentially the distribution of mass relative to the axis of rotation.

• Mathematically, it's expressed as:\[I = \sum m_ir_i^2\]for discrete masses, or \[I = \int r^2 dm\]for continuous mass distributions.

The wheel in this problem has a moment of inertia of \(2 \, \text{kg} \cdot \text{m}^2\), meaning that this value quantifies its resistance to changes in its spinning motion.

Angular Acceleration

Angular acceleration is the rate at which angular velocity changes with time. It describes how quickly something speeds up or slows down its rotation. In rotational motion, it plays the same role as linear acceleration in rectilinear motion.

• The equation used to find angular acceleration is:\[\alpha = \frac{\omega_f - \omega_i}{t}\]where \(\omega_f\) is the final angular velocity, \(\omega_i\) is the initial angular velocity, and \(t\) is the time.

In situations like gradually bringing a spinning wheel to rest, a negative angular acceleration indicates deceleration. Here, the angular acceleration needed to stop the wheel was \(-\frac{\pi}{60} \, \text{rad/s}^2\).

Angular Velocity

Angular velocity is how fast an object spins around its axis. Measured in radians per second, it tells us how much angle an object covers in a specific time frame. It is a vector quantity, having both magnitude and direction.

• The initial angular velocity \(\omega_i\) of the wheel was found to be \(\pi \, \text{rad/s}\) from the given speed of 60 revolutions per minute.

This quantity is crucial for predicting how long it takes to change a rotational state. With \(\omega_i\) set to \(\pi \, \text{rad/s}\) and a target of stopping the wheel (where \(\omega_f = 0\)), understanding angular velocity helps in planning the required torque to achieve a halt in one minute.