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A uniform rod of mass m and length l res...

A uniform rod of mass `m` and length `l` rests on a smooth horizontal surface. One of the ends of the rod is struck in a horizontal direction at right angles to the rod. As a result the rod obtains velocity `v_(0)`. Find the force with which one-half of the rod will act ont he other in the process of motion.

Text Solution

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Let `J` be the linear impulse applied at B and `omega` and angular speed of rod
`J=mv_(0)` ..(i) ,brgt `J((l)/(2))=(ml^(2))/(12).omega` .(ii)
solving these two equations
`omega=(6v_(0))/(l)`
Linear speed of D (mid-point of CB) relative C
`v=omega((l)/(4))=(3)/(2)v_(0)`
`therefore` force exerted by upper half on the lower half
`F=(((m)/(2))v^(2))/(((l)/(4)))`
substituting `v=(3)/(2)v_(0)`, we get
`F=(9)/(2)(mv_(0)^(2))/(l)`
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