(B) The rotational kinetic energy $(K_r)$ is given by the formula $K_r = \frac{1}{2} I \omega^2$.Given $I = 2.4 \ kg \cdot m^2$ and $K_r = 750 \ J$,we

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(B) The rotational kinetic energy $(K_r)$ is given by the formula $K_r = \frac{1}{2} I \omega^2$.Given $I = 2.4 \ kg \cdot m^2$ and $K_r = 750 \ J$,we have:$750 = \frac{1}{2} \times 2.4 \times \omega^2$$750 = 1.2 \times \omega^2$$\omega^2 = \frac{750}{1.2} = 625$$\omega = \sqrt{625} = 25 \ rad/s$.Assuming the body starts from rest,the initial angular velocity $\omega_0 = 0$.Using the kinematic equation $\omega = \omega_0 + \alpha t$:$25 = 0 + 5 \times t$$t = \frac{25}{5} = 5 \ s$.

Class 11 Physics System of Particles and Rotational Motion Energy conservation, Kinetic Energy, Work and Power of Rotational Motion

Medium

The moment of inertia of a body about a given axis is $2.4 \ kg \cdot m^2$. To produce a rotational kinetic energy of $750 \ J$,an angular acceleration of $5 \ rad/s^2$ must be applied about that axis for how many seconds?

A
$6$
B
$5$
C
$4$
D
$3$

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Class 11 Questions

Class 11

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Chapter

System of Particles and Rotational Motion

Subtopic

Energy conservation, Kinetic Energy, Work and Power of Rotational Motion

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The moment of inertia of a body about a given axis is $2.4 \ kg \cdot m^2$. To produce a rotational kinetic energy of $750 \ J$,an angular acceleration of $5 \ rad/s^2$ must be applied about that axis for how many seconds?

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