The moment of inertia of a body about a given | vedclass

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Question

(C) The rotational kinetic energy $(K)$ of a body is given by the formula: $K = \frac{1}{2} I \omega^2$. Given that the angular velocity $\omega = 1 \

Vedclass · Accepted Answer

$2$

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(C) The rotational kinetic energy $(K)$ of a body is given by the formula: $K = \frac{1}{2} I \omega^2$. Given that the angular velocity $\omega = 1 \ rad/s$,we substitute this into the formula: $K = \frac{1}{2} I (1)^2 = \frac{1}{2} I$. According to the problem,the moment of inertia $(I)$ is numerically equal to '$P$' times the rotational kinetic energy $(K)$: $I = P \cdot K$. Substituting the expression for $K$: $I = P \cdot (\frac{1}{2} I)$. Dividing both sides by $I$ (assuming $I 
eq 0$): $1 = P \cdot \frac{1}{2}$. Therefore,$P = 2$.

The moment of inertia of a body about a given axis,rotating with an angular velocity of $1 \ rad/s$,is numerically equal to '$P$' times its rotational kinetic energy. The value of '$P$' is:

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