Question

A wheel rotating at an angular speed of 20 rads–1 is brought to rest by a
constant torque in 4.0 seconds. If the moment of inertia of the wheel about the axis of rotation is 0.20 kg m2
, calculate the work done by the torque in
the first two seconds.

Answer 1

Dear Student,

◆ Answer -

W = -30 J

● Explaination -

# Given -

w = 20 rad/s

w' = 0 rad/s

t = 4 s

I = 0.2 kgm^2

# Solution -

According to 1st law of rotational motion,

w' = w + α.t

0 = 20 + α×4

α = -20/4

α = -5 rad/s^2

Angular speed after 2 s is calculated as -

w" = w + α.t

w" = 20 + (-5)×2

w" = 10 rad/s

Work done by the torque in first 2 s is calculated by -

W = 1/2 I(w"²-w²)

Substitute, w = 20 rad/s and w" = 10 rad/s,

W = 1/2 × 0.2 × (10² - 20²)

W = 0.1 × (100 - 400)

W = -30 J

Therefore, calculate the work done by the torque in the first two seconds is -30 J.

Thanks dear. Hope this helps you...