Question

The motor driving a grinding wheel with a rotational inertia of 0.36 kg m2 is switched o? when the wheel has a rotational speed of 25 rad/s. After 6.6 s, the wheel has slowed down to 20 rad/s. What is the absolute value of the constant torque exerted by friction to slow the wheel down? Answer in units of Nm.

Answer 1

Answer:

Torque will be 0.2727 N -m

Explanation:

We have given rotational inertia of the motor $I=0.36kgm^2$

Initial angular speed 25 rad/sec

Final angular speed = 20 rad/sec

Time to change in speed t = 6.6 sec

From first equation of motion $\omega _f=\omega _i+\alpha t$

$20=25+\alpha \times 6.6$

$\alpha =-0.7575rad/sec^2$

Now torque $\tau =I\alpha =0.36\times 0.7575=0.2727N-m$