Question

A constant torque of magnitude 5000Nm acting on a body increases its angular velocity from 4rad /s to 20rad /s in 8 second calculate the MI of the body about the axis of rotation

Answer 1

Answer:

MI of the body about the axis of rotation is 2500 kgm^2.

Explanation:

Given,

T = 5000 Nm

Wi = 4 rad/s

Wf = 20 rad/s

t = 8 seconds

a = (Wf - Wi)/t

a = (20 - 4)/8 rad/s^2

a = 16/8 rad/s^2

a = 2 rad/s^2

T = I a

I = T/a

I = 5000/2 kgm^2

I = 2500 kgm^2

Here,

T is the torque .

Wi is the initial angular velocity.

Wf is the final angular velocity.

t is the time .

a is the angular acceleration.

I is the moment of inertia.

Hence,

MI of the body about the axis of rotation is 2500 kgm^2.