Question

A rod of mass M and length L lies on frictionless horizontal table.It is free to move on any way on table.A small body of mass m moving with velocity u collides in elastically with rod. (a)find velocity of center of mass of rod? (b)find angular velocity of rod about center of mass?

Answer 1

Answer:

:

Linear momentum angular momentum and kinetic energy are conserved in the process. From conservation of linear momentum

`Mv'=mv` ltbr. Or `v'=(m)/(M)v` ..(i)

Conservation of angular momentum gives,

`mvd=sqrt((Ml^(2))/(12))omega`

or `omega=((12mvd)/(Ml^(2)))` ...(ii)

Collision is elastic, Hence

`e=1`

or relative speed of approach

`=` relative speed of separation

`thereforev=v'+domega`

Subsitituting the values, we have

`v=(m)/(M)v+(12mvd^(2))/(Ml^(2))`

Solving it we get

`m=(Ml^(2))/(12d^(2)+l^(2))`

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Explanation:

Answer 2

Vcm = mu ÷ M

ω = Vcm / r

Explanation:

Given: A rod of mass M and length L lies on frictionless horizontal table.It is free to move on any way on table.A small body of mass m moving with velocity u collides in elastically with rod.

Find: Velocity of center of mass of rod and angular velocity of rod about center of mass.

Solution:

a) The center of mass velocity equation is the sum of each particle's momentum (mass times velocity) divided by the total mass of the system.

Vcm = mu ÷ M

m - mass of particle, u - velocity of of particle and M - total mass of the system

b) Angular velocity = Vcm ÷ radius

v = rω

Therefore ω = Vcm / r