A uniform rod AB of length 1 m and mass 4 k | vedclass

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(B, C) Let the rod make an angle $	heta$ with the $OX$ axis. The velocity of end $A$ is $v_A = 3 \ m/s$ along $OX$ and the velocity of end $B$ is $v_

Vedclass · Accepted Answer

The angular velocity of the rod is $5 \ rad/s$ anticlockwise.

Vedclass · Answer

(B, C) Let the rod make an angle $	heta$ with the $OX$ axis. The velocity of end $A$ is $v_A = 3 \ m/s$ along $OX$ and the velocity of end $B$ is $v_B = 4 \ m/s$ along $OY$ (downwards).Using the condition that the length of the rod is constant,the component of velocity along the rod must be the same for both ends:$v_A \cos 	heta = v_B \sin 	heta \implies 3 \cos 	heta = 4 \sin 	heta \implies 	an 	heta = \frac{3}{4}$.Thus,$\sin 	heta = \frac{3}{5}$ and $\cos 	heta = \frac{4}{5}$.The angular velocity $\omega$ is given by $\omega = \frac{v_A \sin 	heta + v_B \cos 	heta}{\ell} = \frac{3(3/5) + 4(4/5)}{1} = \frac{9/5 + 16/5}{1} = \frac{25}{5} = 5 \ rad/s$.Since the rod is rotating such that $A$ moves right and $B$ moves down,the rotation is clockwise.The rotational kinetic energy is $(KE)_{rot} = \frac{1}{2} I \omega^2 = \frac{1}{2} (\frac{1}{12} m \ell^2) \omega^2 = \frac{1}{2} 	imes \frac{1}{12} 	imes 4 	imes (1)^2 	imes (5)^2 = \frac{1}{24} 	imes 4 	imes 25 = \frac{25}{6} \ J$.Both statements $B$ and $C$ are correct.

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