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

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(B) Given: Moment of inertia $I = 1.2 \; kg \cdot m^2$,initial angular velocity $\omega_0 = 0$,rotational kinetic energy $K_r = 1500 \; J$,and angular

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(B) Given: Moment of inertia $I = 1.2 \; kg \cdot m^2$,initial angular velocity $\omega_0 = 0$,rotational kinetic energy $K_r = 1500 \; J$,and angular acceleration $\alpha = 25 \; rad/s^2$.The formula for rotational kinetic energy is $K_r = \frac{1}{2} I \omega^2$.Substituting the values: $1500 = \frac{1}{2} 	imes 1.2 	imes \omega^2$.$1500 = 0.6 	imes \omega^2 \Rightarrow \omega^2 = \frac{1500}{0.6} = 2500$.Thus,the final angular velocity $\omega = \sqrt{2500} = 50 \; rad/s$.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 that axis for a duration of: (in $; s$)

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