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

A uniform rod of mass M and length `l` is dropped on horizontal smooth surface from the position shown. What is the velocity (in `ms^(-1)`) of centre of mass of rod immediately after elastic collision?
`(g= 10 ms^(-2))`

Text Solution

Verified by Experts

The correct Answer is:
`02.00`
Impulse will be in vertical direction, also apply equation for coefficient of restitution.
`v+(omegal)/(2) cos theta =v_(0)`

`intN dt = m(v_(0) +v)`
`int N (l)/(2) cos theta dt = (ml^(2))/(12) omega`
`m(v_(0) +v) (l)/(2) cos theta =(ml^(2))/(12) omega`
`6(v_(0) +v) = (omega l)/(cos theta)`
`3(v_(0) +v) = ((v_(0)-v))/(cos^(2) theta)`
`3v_(0) +3v=4(v_(0)-v)`
`7v =v_(0)`
`v=(v_(0))/(7)`
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