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Question: A flywheel rotating at \(420\,rpm\) slows at a constant rate of \(2\,rad\,{s^{ - 2}}\) . Find the ti...

A flywheel rotating at 420 rpm420\,rpm slows at a constant rate of 2 rad s−22\,rad\,{s^{ - 2}} . Find the time required to stop the flywheel.
A) 22 s22\,s
B) 11 s11\,s
C) 44 s44\,s
D) 12 s12\,s

Explanation

Solution

First we need to convert rate per minute to radian per second. Then we need to apply the first equation of rotational motion to get the answer.

Complete step by step solution:
Angular velocity is the rate of change in the angular direction of the moving body. The angular velocity of a particle can be defined as the rate at which the particle rotates around the centre point, i.e. the time rate of shift of its angular displacement relative to the origin.

Angular acceleration is the rate of angular velocity transition. In SI units, it is measured in radians per second squared and is generally denoted by the Greek letter alpha (α)\left( \alpha \right) .

The fusion of rotation and translational motion of a rigid body is known as rolling motion. According to the law of preservation of angular momentum, if there is no external pair operating, the total angular momentum of a solid body or a particle structure shall be preserved.
Given,
Initial angular velocity,
ω∘=420 rpm =42060 =7 rps =7×2π =14π rad s−1  {\omega _ \circ } = 420\,rpm \\\ = \dfrac{{420}}{{60}} \\\ = 7\,rps \\\ = 7 \times 2\pi \\\ = 14\pi \,rad\,{s^{ - 1}} \\\

Final angular velocity,
ω=0\omega = 0

Acceleration, α=−2 rad s−2\alpha = - 2\,rad\,{s^{ - 2}} (negative since the flywheel is slowing down)
According to first equation of kinematics:
ω=ω∘+αt ⇒t=ω−ω∘α ⇒t=−14π−2 ⇒t=7π ⇒t=22 s  \omega = {\omega _ \circ } + \alpha t \\\ \Rightarrow t = \dfrac{{\omega - {\omega _ \circ }}} {\alpha } \\\ \Rightarrow t = \dfrac{{ - 14\pi }} {{ - 2}} \\\ \Rightarrow t = 7\pi \\\ \Rightarrow t = 22\,s \\\

Hence, option A is correct.

Additional information:
The earlier the angular velocity transition happens, the larger the angular acceleration is.
Linear acceleration at the point of interest is tangent to the globe of circular motion, which is termed tangential acceleration.
Centripetal acceleration refers to changes in the direction of the velocity but not its frequency in circular motion. Centripetal acceleration is experienced by an object undergoing circular motion.

Note: Here we have to be careful while writing the acceleration since it may be negative or positive. Also, we have to check the unit of the angular velocity and see whether it is standard or not.