Question

A grinding wheel is initially at rest. A constant external torque of 50N*n is applied to the wheel for 20 s, giving the wheel an angular speed of 600 rev/min. The external torque is then removed, and the wheel comes to rest 120 s later.

a) Find the amount of inertia of the wheel

b)Find the frictional torque, which is assumed to be constant

Answer 1

Solution) Text = 50 Nm

t1 = 20 s

w = 600 rev/min

w = 600×(2(pi))/(60)

w = 62.83 rad/s

t2 = 120 s

(a) Moment of inertia , I = ?

Net torque , Tnet = I(angular acceleration on speeding)

Tnet = Text - Tf

Tf is frictional torque which occurs while slowing

Tf = I(angular acceleration while slowing)

Text = I(angular acceleration on speeding) + I(angular acceleration while slowing)

Angular acceleration on speeding = w/t1 = (62.83/20) = 3.14 rad/s^2

Angular acceleration while slowing =- w/t2 = - (62.83/120) = -0.523 rad/s^2

Text = I( 3.14 - 0.523)

I = (Text)/(3.14 - 0.523)

I = (50)/(3.14 - 0.523)

I = 19.1 kgm^2

(b) Frictional torque , Tf = ?

Tf = I(angular acceleration while slowing)

Tf = 19.1×(- 0.523)

Tf = - 9.98 Nm

Tf = - 10 Nm ( approximately)