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Grade 11Mechanics

A uniform rod of mass M and length a lies on a smooth table. A particle of mass m strikes it at distance a/4 from the centre and stops after the collision. Find the a) velocity of the centre of the rod and the b)angular velocity of the rod about its centre just after the collision.

a) They have used linear momentum formula: mv=MV. where V is the velocity of the centre of the rod.

How could they use it when the rod was undergoing rotational and not translational motion after collision?

Shouldn't velocity of the centre of rod be 0 because its stationary?

b) How do we find part b? Please explain the concept.

Gundoos 1234

15 Years agoGrade 11

1 Answer

vikas askiitian expert

Approved Tutor Answer15 Years ago

here we have to apply to concepts

first is conservation of angular momentam and second is conservation of linear momentam...

initial angular momentam of particle about the center of rod is mva/4 ....

finally particle is rest but rod is rotating with angular velocity W..

so final angular momentam=Iw=initial angular momentam (by conservation of angular momentam)

IW=mva/4

I for rod is ma²/12 about its axis passing through center perpendicular to rod...

WMa²/12 =mva/4

W= 3mv/Ma ] .......................1

now applying conservation of linear momentam

let V_com is the linear velocity of rod finally then

MV_com = mv

V_com =mv/M .......................2

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