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

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(B) Given: Moment of inertia $I = 1.2 ~kg \cdot m^2$,Rotational Kinetic Energy $K.E. = 1500 ~J$,Angular acceleration $\alpha = 25 ~rad/s^2$,Initial an

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$2$

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(B) Given: Moment of inertia $I = 1.2 ~kg \cdot m^2$,Rotational Kinetic Energy $K.E. = 1500 ~J$,Angular acceleration $\alpha = 25 ~rad/s^2$,Initial angular velocity $\omega_0 = 0$. Step $1$: Calculate the final angular velocity $\omega$ using the rotational kinetic energy formula: $K.E. = \frac{1}{2} I \omega^2$ $1500 = \frac{1}{2} 	imes 1.2 	imes \omega^2$ $1500 = 0.6 	imes \omega^2$ $\omega^2 = \frac{1500}{0.6} = 2500$ $\omega = 50 ~rad/s$. Step $2$: Calculate the time $t$ using the kinematic equation for rotation: $\omega = \omega_0 + \alpha t$ $50 = 0 + 25 	imes t$ $t = \frac{50}{25} = 2 ~s$.

The moment of inertia of a body about a given axis is $1.2 ~kg \cdot m^2$. Initially,the body is at rest. In order to produce a rotational kinetic energy of $1500 ~J$,an angular acceleration of $25 ~rad/s^2$ must be applied about the axis for a time duration of: (in $~s$)

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