Question

A torque of 50 Nm is applied to a grinding wheel with moment of inertia 20 kg m2 for 10 s. If it starts from rest, what is the angular velocity of the wheel after the torque is removed?

Answer 1

Answer:

Final angular velocity, $\omega_f=25\ rad/s$

Explanation:

Given that,

Torque, $\tau=50\ N-m$

Moment of inertia, $I=20\ kg m^2$

Time, t = 10 s

Initial angular velocity, $\omega_i=0$

The relation between the torque and the moment of inertia is :

$\tau=I\times \alpha$

$\alpha$ is the angular acceleration

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

$\alpha =\dfrac{50}{20}$

$\alpha =2.5\ rad/s^2$

Using first equation of kinematics to find the final angular velocity of the wheel. It is given by :

$\omega_f=\omega_i+\alpha t$

$\omega_f=0+2.5\times 10$

$\omega_f=25\ rad/s$

So, the final angular velocity of the wheel is 25 rad/s. Hence, this is the required solution.