Question

A circular disc of M.I. 10 $kg-m^2$ rotates about its own axis at a constant speed of 60 r.p.m. under the action of an electric motor of power 31.4 W. If the motor is switched off, how many rotations will it cover before coming to rest?

• A. $3.14$
• B. $31.4$
• C. $314$
• D. $6.28$

Answer 1

The correct option is D $6.28$

The moment of inetia of the disc is $10kg-m^2$

The speedof the disc is $60rpm$ i.e. $60\times \frac{2\pi }{60}rad/s$

The power supplied by the motor is $31.4W$

The relation between the power and the torque developed is:

$P=\tau \omega$

$\tau =\frac{P}{\omega }=\frac{31.4}{2\pi }$

$\tau =5Nm$

The angular acceleration of motor due to torque is :

$\alpha =\frac{\tau }{I}$

$\alpha =\frac{5}{10}=0.5rad/s^2$

Now, when the motor is switched off, the angular displacement of the disc to come to rest is given as:

${\omega }_f^2={\omega }_0^2+2\alpha \theta$

$0-(2\pi {)}^2=2\times 0.5\times \theta$

$\theta =4{\pi }^2$

In one rotation, displacement is $2\pi$. SO, total number of rotations is:

$n=\frac{\theta }{2\pi }=\frac{4{\pi }^2}{2\pi }$

$n=2\pi =6.28rotations$