1please answer
A uniform rod of mass=2kg and length l=1m rests upon a smooth horizontal surface.One of the ends of the rod is struck with impulse=2Ns in horizontal direction perpendicular to the rod.As a result the rod obtains a momentum=2Ns.The force exerted by one half of the rod on the other during the process of motion is?
Options :
(A)6N (B)9N (C)12N (D)15N
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5 Answers
1well this is a small problem ..
for the lower mass we have (if the force exerted by the other mass be F)
F=(m/2)w2L/4 as after the impulse it moves in a circle of radius L/4 w.r.t the mid point ..
further
J=Iw
2*1/2=2/12(w) w=6
thus we get answer as = 2/2* 36/4= 9