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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?

Sagot :

Cricetus

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]

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