Velocity of COM after collision

Click For Summary
The velocity of the head before the collision is calculated as sqrt(2gh), indicating complete conversion of gravitational potential energy to kinetic energy. After the collision, the velocity at point A is determined by the formula vf=e•sqrt(2gh), where e is the coefficient of restitution. The head is treated as a rod with a moment of inertia of (ml^2)/12. The discussion emphasizes the importance of the impulse-momentum theorem and angular momentum conservation, particularly about a chosen origin. Understanding these principles helps in achieving a zero velocity for the center of mass after the collision.
Mooy
Messages
2
Reaction score
0
Homework Statement
Show that the velocity of the centre mass of the body is zero immediately after impact if the following equation holds.
b^2=(e•l^2)/12
Note: The head is released from rest and the only force acting on the head during impact is at point A
Relevant Equations
KE=0.5mv^2
GPE=mgh
V=rω
I=(ml^2)/12
244206

I calculated that the velocity of the head prior to the collision is sqrt(2gh), as all of the gravitational potential energy is converted to kinetic energy.

And I believe the velocity at point A after the collision is given by the formula vf=e•sqrt(2gh), with e representing the coefficient of restitution. As the ground does not change in velocity during the collision.

I believe that the head is meant to be treated as a rod with a moment of inertia of (ml^2)/12.

I’m unsure how to go about getting the centre of mass to have zero velocity, any help would be greatly appreciated
 
Physics news on Phys.org
Have you covered the "impulse-momentum theorem" and the corresponding "angular impulse-angular momentum theorem"?

Recall that angular momentum is always calculated relative to some point ("origin"). What would be a good choice for the origin in this problem?
 
TSny said:
Have you covered the "impulse-momentum theorem" and the corresponding "angular impulse-angular momentum theorem"?

Recall that angular momentum is always calculated relative to some point ("origin"). What would be a good choice for the origin in this problem?
Thanks heaps think I get it now.

Would it be to conserve angular momentum about A, and the COM has an initial angular momentum of sqrt(2gh)mb prior to the collision.
I forgot that an object without angular velocity still has angular momentum
 
Mooy said:
Thanks heaps think I get it now.

Would it be to conserve angular momentum about A, and the COM has an initial angular momentum of sqrt(2gh)mb prior to the collision.
I forgot that an object without angular velocity still has angular momentum
That will give the desired answer, when combined with your two beliefs in post #1.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 4 ·
Replies
4
Views
897
Replies
34
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
Replies
20
Views
3K
  • · Replies 15 ·
Replies
15
Views
4K
Replies
10
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K