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

The correct answer is "0.66 kg.m²".

Step-by-step explanation:

The given values are:

K.E = 500 J

w₁ = 65 rad/s

w₂ = 52 rad/s

As we know,

⇒
`K.E=(1)/(2)I[w_1^2-w_2^2]`

then,

⇒
`I=(2(K.E))/(w_1^2-w_2^2)`

On putting the given values, we get

⇒
`=(2* 500)/((65)^2-(52)^2)`

⇒
`=(1000)/(4225-2704)`

⇒
`=(1000)/(1521)`

⇒
`=0.66 \ kg.m^2`