The velocity of the centre of mass of a rigid | vedclass

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(B) The velocity of point $A$ with respect to the observer $O$ is given by the vector sum of the velocity of the centre of mass and the velocity of po

Vedclass · Accepted Answer

$8\hat i + 11\hat j - 6\hat k$

Vedclass · Answer

(B) The velocity of point $A$ with respect to the observer $O$ is given by the vector sum of the velocity of the centre of mass and the velocity of point $A$ relative to the centre of mass:$\vec v_A = \vec v_{cm} + \vec v_{A/cm}$The velocity of point $A$ relative to the centre of mass is given by the cross product of angular velocity and position vector:$\vec v_{A/cm} = \vec \omega 	imes \vec r = \begin{vmatrix} \hat i & \hat j & \hat k \ 0 & 3 & 4 \ 2 & 0 & 2 \end{vmatrix}$$= \hat i(3 	imes 2 - 4 	imes 0) - \hat j(0 	imes 2 - 4 	imes 2) + \hat k(0 	imes 0 - 3 	imes 2)$$= 6\hat i + 8\hat j - 6\hat k 	ext{ m/s}$Now,adding the velocity of the centre of mass:$\vec v_A = (2\hat i + 3\hat j) + (6\hat i + 8\hat j - 6\hat k)$$\vec v_A = (2+6)\hat i + (3+8)\hat j - 6\hat k = 8\hat i + 11\hat j - 6\hat k 	ext{ m/s}$

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