A) 0.25 m/s
B) 0.64 m/s
C) 0.5 m/s
D) 1.2 m/s
Correct Answer: C
Solution :
Let the magnitude of the impulse delivered by the blow be represented by p. From the translational and rotational forms of Newton's second law\[{{v}_{c}}=\frac{p}{M}\]and \[{{I}_{c}}{{\omega }_{c}}=\frac{pl}{2}.\] Here, \[{{v}_{c}}\]is the rightward velocity of the centre of mass after the blow and \[{{\omega }_{c}}\]is the (positive clockwise) angular velocity of the rod after the blow about the point C. Using the fact that \[{{I}_{c}}=\frac{{{(Ml)}^{2}}}{12},\] we find that \[{{\omega }_{c}}l=\frac{6p}{M}.\]Therefore,\[{{v}_{c}}=\frac{({{\omega }_{c}}l)}{6}\] Since, \[{{\omega }_{c}}=3.00\,\text{rad/s}\]and l = 1.00 m, we find\[{{v}_{c}}=0.5\,\text{m/s}\].
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