Question

500 J of work is used to decrease the angular velocity of a disk from 65 rad/s to 52 rad/s.What is the rotational inertia of the disk?

Answer 1

Answer:

The correct answer is "0.66 kg.m²".

Explanation:

The given values are:

K.E = 500 J

w₁ = 65 rad/s

w₂ = 52 rad/s

As we know,

⇒  $K.E=\frac{1}{2}I[w_1^2-w_2^2]$

then,

⇒  $I=\frac{2(K.E)}{w_1^2-w_2^2}$

On putting the given values, we get

⇒     $=\frac{2\times 500}{(65)^2-(52)^2}$

⇒     $=\frac{1000}{4225-2704}$

⇒     $=\frac{1000}{1521}$

⇒     $=0.66 \ kg.m^2$