Question

A uniform rod length l and mass is 4m keep on horizontel surface it is free for move. m Mass ball is strike at a corner of rod with v velocity.if after colligen ball is rest so find the angular velocity after colligen

Answer 1

Solution:

Some information to be added:

Ball strikes at an angle of 45°. Hence velocity would be v.cos45°

Therefore the velocity with which ball causes angular momentum is (v/√2).

This question can be solved by using the knowledge of "Conservation of Angular Momentum".

Let us assume the angular momentum before collision to be Li and the angular momentum after collision to be Lf.

By principle of conservation of angular momentum,

Li = Lf

Calculating initial angular momentum, we get:

→ Li = m ( v/√2 ) ( L/2 )

→ Li =  (mvL / 2√2 )

Calculating final angular momentum, we get:

→ Lf = Iω

→ Lf = ( I of rod + I of ball ) ω

I of rod is given as: 1/12 ML²

According to question, M = 4m. Therefore substituting the values we get:

→ I of rod = ( 1/12 (4m) L² ) = mL²/3   ... ( 1 )

I of ball is given as MR², where R is the distance between axis of rotation and Point of application of force.

→ I of ball = m ( L/2 )²

→ I of ball = mL²/4   .... ( 2 )

Combining (1) and (2) we get:

→ ( mL²/4 + mL²/3 ) ω

→ Lf = ( 7mL²/12 ) ω

Equating Li and Lf we get:

→ (mvL / 2√2 ) =  ( 7mL²/12 ) ω

→ v / 2√2 = 7ωL/12

→ 12v/14√2 L = ω

→ 6v / 7√2 L = ω

Rationalising the denominator we get:

→ 3√2 v / 7L = ω

Hence the angular velocity after collision is ( 3√2 v / 7 L )

Answer 2

Answer:

A uniform rod of length L and mass 4m lies on a frictionless horizontal surface on which it is free to move anyway. A ball of mass m moving with speed v as shown in figure collides with the rod at one of the ends. If ball comes to rest immediately after collision with the rod then find out angular velocity ω of rod just after collision.