A rectangular solid rod of length 0.3, m is h | vedclass

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(A) The angular impulse is equal to the change in angular momentum.$	au \Delta t = \Delta L$Taking the edge of the table as the pivot point,the torqu

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$0.5$

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(A) The angular impulse is equal to the change in angular momentum.$	au \Delta t = \Delta L$Taking the edge of the table as the pivot point,the torque due to gravity is $	au = mg \frac{\ell}{2}$.$mg \frac{\ell}{2} \Delta t = I \omega$Since the rod rotates about the edge,its moment of inertia is $I = \frac{m\ell^2}{3}$.$mg \frac{\ell}{2} \Delta t = \frac{m\ell^2}{3} \omega$$\omega = \frac{3g \Delta t}{2\ell} = \frac{3 	imes 10 	imes 0.01}{2 	imes 0.3} = \frac{0.3}{0.6} = 0.5\, rad/s$Time taken by the rod to hit the ground is $t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 	imes 5}{10}} = 1\, s$.The angle rotated by the rod during this time is $	heta = \omega t = 0.5 	imes 1 = 0.5\, radian$.

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