A body rotating with an angular velocity of 3 | vedclass

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(A) The rotational kinetic energy $(K_{rot})$ of the rotating body is given by $K_{rot} = \frac{1}{2} I \omega^2$. Given $I = 3 \ kg \cdot m^2$ and $\

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(A) The rotational kinetic energy $(K_{rot})$ of the rotating body is given by $K_{rot} = \frac{1}{2} I \omega^2$. Given $I = 3 \ kg \cdot m^2$ and $\omega = 3 \ rad/s$,we have: $K_{rot} = \frac{1}{2} 	imes 3 	imes (3)^2 = \frac{1}{2} 	imes 3 	imes 9 = 13.5 \ J$. The translational kinetic energy $(K_{trans})$ of the second body is given by $K_{trans} = \frac{1}{2} m v^2$. Given $m = 27 \ kg$ and $K_{trans} = K_{rot} = 13.5 \ J$,we have: $\frac{1}{2} 	imes 27 	imes v^2 = 13.5$. $v^2 = \frac{13.5 	imes 2}{27} = \frac{27}{27} = 1$. Therefore,$v = 1 \ m/s$.

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