Moment of inertia of a body about a given axi | vedclass

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(A) Given: Moment of inertia $I = 1.5\, kg\, m^2$,Rotational kinetic energy $K = 1200\, J$,Angular acceleration $\alpha = 20\, rad/s^2$,Initial angula

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(A) Given: Moment of inertia $I = 1.5\, kg\, m^2$,Rotational kinetic energy $K = 1200\, J$,Angular acceleration $\alpha = 20\, rad/s^2$,Initial angular velocity $\omega_0 = 0$.Formula for rotational kinetic energy is $K = \frac{1}{2}I\omega^2$.Substituting the values: $1200 = \frac{1}{2} 	imes 1.5 	imes \omega^2$.$1200 = 0.75 	imes \omega^2 \Rightarrow \omega^2 = \frac{1200}{0.75} = 1600$.Thus,final angular velocity $\omega = \sqrt{1600} = 40\, rad/s$.Using the equation of rotational motion: $\omega = \omega_0 + \alpha t$.$40 = 0 + 20 	imes t$.$t = \frac{40}{20} = 2\, s$.

Moment of inertia of a body about a given axis is $1.5\, kg\, m^2$. Initially,the body is at rest. In order to produce a rotational kinetic energy of $1200\, J$,an angular acceleration of $20\, rad/s^2$ must be applied about the axis of rotation for a duration of ......... $\sec$.

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