Question

A flywheel of moment of inertia 5.0 kg.m² is rotated at a speed of 60 rad/s. Because of the friction at the axle, it comes to rest in 5.0 minutes.
Find (a) the average torque of the friction (b) the total work done by the friction and (c) the angular momentum of the wheel 1 minute before it stops rotating.?

Answer 1

Given in the question :-

Moment of Inertia = 5kg m

speed of rotation = 60 rad/s.

(a)Now we have to calculate the angular retardation (α)

Angular speed (initial) ⍵ =60 rad/s

Angular speed (final) ⍵' =0

Time, t = 5 min or 300 s.

Now , we know the formula

$\omega' = \omega - \alpha t$

$0 = 60 - 300\alpha$

$\alpha =60/300$

α = 0.20 rad/s²

Now we have

Moment of Inertia = 5.0 kg.m²

Hence, Average frictional torque = Iα

=5.0 × 0.20

[tex] \boxed{1 Nm } [/tex]

(b) Let the angle rotated

$\theta = \omega \frac{1}{2}\alpha t^2$

θ =60 × 300 - 0.50 × 0.20 × 300²

θ =18000 - 0.10 × 90000

θ =18000-9000

θ =9000 rad

Now , Work done is given as

W.D = Tθ

= 1.0 × 9000

[tex] \boxed{ W =9000 J } [/tex]

(c) Assume that the angular speed at time 240 s will be ω''

$\omega'' = \omega - \alpha t$

ω'' = 60 - 0.20 × 240

ω'' =60 - 48

ω'' =12 rad/s

1 minute before stopping the wheel , now the angular momentum will be

= Iω''

= 5.0 × 12

$\boxed{ 60 kg.m^2/s}$

Hope it Helps :-)