Instantanoeus axis of rotation

[Saitama]

Homework Statement

A uniform rod of length l is given an impulse at right angles to its length as shown. Find the distance of instantaneous centre of rotation from the centre of rod.

Homework Equations

The Attempt at a Solution

Impulse=change in momentum
or Impulse=mv_CM (Am i right here?)

I am not sure about this but i think i need to conserve angular momentum. But how should i make the equation for initial angular momentum. Should it be mv_CMx or mv_CM(x+d), d is the distance of instantaneous centre of rotation of rod from the CM.

Any help is appreciated!

 

[ehild]

Homework Statement

A uniform rod of length l is given an impulse at right angles to its length as shown. Find the distance of instantaneous centre of rotation from the centre of rod.

The Attempt at a Solution

Impulse=change in momentum
or Impulse=mv_CM (Am i right here?)

Impulse is FΔt,force multiplied by its time of action. And it is equal to the change of momentum. So the equation mv_CM=Impulse is right.

I am not sure about this but i think i need to conserve angular momentum. But how should i make the equation for initial angular momentum. Should it be mv_CMx or mv_CM(x+d), d is the distance of instantaneous centre of rotation of rod from the CM.

The applied impulse also means torque for time Δt: τΔt=Impulse*(x+d) The applied torque changes the angular momentum. You can calculate both the torque and angular momentum with respect to the instantaneous axis, considering the motion of the rod a pure rotation. Use the relation between torque and change of angular momentum:τ =I(Δω)/(Δt).You know ω from the initial speed of the CM: The CM will move around the instantaneous axis with angular speed ω=v_CM/d.


ehild

 

[Saitama]

Thanks ehild for your reply!

The applied impulse also means torque for time Δt: τΔt=Impulse*(x+d) The applied torque changes the angular momentum. You can calculate both the torque and angular momentum with respect to the instantaneous axis, considering the motion of the rod a pure rotation. Use the relation between torque and change of angular momentum:τ =I(Δω)/(Δt).You know ω from the initial speed of the CM: The CM will move around the instantaneous axis with angular speed ω=v_CM/d.

I have got the answer using the relations you posted but i still don't understand what do you mean by "pure rotation"?

 

[ehild]

How is the instantaneous axis defined?

See picture. It shows a rod moved from position 1 to position 2 and rotated by 90°. You can do it by translating the rod from 1 to 2 and then rotating about the CM, or by rotating the whole rod about the instantaneous axis O.

ehild

 

[Saitama]

How is the instantaneous axis defined?

See picture. It shows a rod moved from position 1 to position 2 and rotated by 90°. You can do it by translating the rod from 1 to 2 and then rotating about the CM, or by rotating the whole rod about the instantaneous axis O.

ehild

Thank you ehild for the explanation!