Question

A uniform cylindrical grinding wheel of mass 50.0 kg and diameter 1.0 m is turned on by an electric motor. The friction in the bearings is negligible. What torque must be applied to the wheel to bring it from rest to 120 rev/min in 20 revolutions?

Answer 1

Answer:

The torque must be applied to the wheel is 15.7 N-m.  

Explanation:

Given that,

Mass of the wheel of cylinder, m = 50 kg

Diameter of the wheel, d = 1 m

Radius, r = 0.5 m      

Initial speed off the wheel is 0

Final angular speed of the wheel, 120 rev/min = 12.56 rad/s

Angular displacement, $\theta=20\ rev=125.6\ rad$

The torque is given by :

$\tau=I\alpha \\\\\tau=\dfrac{mr^2}{2}\times (\dfrac{\omega_f^2}{2\theta})\\\\\tau=\dfrac{50\times1^{2}}{2}\times(\dfrac{12.56^{2}}{2\times125.6})\\\\\tau=15.7\ N-m$

So, the torque must be applied to the wheel is 15.7 N-m.