Question

A grinding wheel of mass 5.0 kg and diameter 0.4m is rotating with an angular speed of 2

rev s-1. Calculate the torque which will increase its angular speed to 8 revolution per

second in 2s.

Answer 1

Your final answer is 683.672Nm

Answer 2

Answer: The torque is 0.03 N-m.

Explanation:

Given that,

Mass of wheel = 5.0 kg

Diameter of wheel = 0.4 m

Initial angular speed $\omega_{1} = 2\ rev/s$

Final angular speed$\omega_{2} = 8\rev/s$

The angular acceleration is

$\alpha = \dfrac{\omega_{2}-\omega_{1}}{t}$

$\alpha = \dfrac{8-2}{2}$

$\alpha = 3 rad/s^{2}$

The torque is the product of the moment of inertia and angular acceleration.

The torque is

$\tau = I \alpha$

Where, I = moment of inertia

$\alpha$ = angular acceleration

$\tau = \dfrac{MR^{2}}{2}\times\alpha$

$\tau = \dfrac{5.0\ kg\times0.04m^{2}\times3rad/sec^{2}}{2}$

$\tau = 0.03 N-m$

Hence, the torque is 0.03 N-m.