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

A uniform rod of mass m and length l is at rest on smooth horizontal surface. An impulse P is applied to end B as shown.

Distance travelled by the centre of the rod, while rod rotates by `90^(@)`

Text Solution

Verified by Experts

Let force Facts at the end of the rod for a very brief period `Deltat` and as a result centre of mass of rod acquires velocity v and angular velocity about its centre of mass. Linear impulse - change in linear momentum `rArr F Delta t= mv rArr v=(F Delta t)/(m) " "...(1)`
In above equation mis mass of the rod. Angular impulse about the centre - Angular momentum of rod about the centre.
`rArr F(L)/(2) Deltat=I omega`
`rArr F(L)/(2) Deltat=(ML^(2))/(12) omega rArr omega=(6FDelta)/(mL) " "...(2)`
Distance travelled by the centre of the rod by the time it rotates through an angle `pi//2` can be written as follows:
`x=vxx (pi)/(2omega)`
Substituting values from equations (1) and (2) we get the following:
`x=(F Deltat)/(m)xx(pi)/(2(6F Delta t)/(mL))=(piL)/(12)`
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