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,
⇒ [tex]K.E=\frac{1}{2}I[w_1^2-w_2^2][/tex]
then,
⇒ [tex]I=\frac{2(K.E)}{w_1^2-w_2^2}[/tex]
On putting the given values, we get
⇒ [tex]=\frac{2\times 500}{(65)^2-(52)^2}[/tex]
⇒ [tex]=\frac{1000}{4225-2704}[/tex]
⇒ [tex]=\frac{1000}{1521}[/tex]
⇒ [tex]=0.66 \ kg.m^2[/tex]
