KE Conservation: Calculating Final Velocities

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Homework Help Overview

The discussion revolves around a problem involving the conservation of kinetic energy and angular momentum in a perfectly elastic collision scenario, specifically focusing on a beam and a block. The participants explore the relationships between translational and rotational motion, as well as the implications of conservation laws in this context.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the conservation of kinetic energy and angular momentum, questioning the correct application of these principles in the context of a rotating beam and a colliding block. There is exploration of the moment of inertia and its relevance to the problem.

Discussion Status

The conversation is ongoing, with participants attempting to clarify their understanding of angular momentum and its conservation. Some guidance has been offered regarding the equations needed to solve for the final velocities, but confusion remains about the relationships between the variables involved.

Contextual Notes

There is uncertainty regarding the moment of inertia for the beam and its application in the equations. Participants are also grappling with the implications of the initial conditions and the direction of velocities post-collision.

goonking
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Homework Statement


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Homework Equations


KEi = KEf

The Attempt at a Solution



since it is perfectly elastic, kinetic energy before = kinetic energy after

(1/2) (m) (20m/s^2) = (1/2) (2m) (VBeamf^2) + (1/2) (m) (VBlockf^2)

is this the right approach? [/B]
 
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goonking said:

Homework Statement



Homework Equations


KEi = KEf

The Attempt at a Solution



since it is perfectly elastic, kinetic energy before = kinetic energy after

(1/2) (m) (20m/s^2) = (1/2) (2m) (VBeamf^2) + (1/2) (m) (VBlockf^2)

is this the right approach? [/B]

Not quite. The beam can only rotate about the axis. What is its kinetic energy?
 
ehild said:
Not quite. The beam can only rotate about the axis. What is its kinetic energy?
(1/2) (I) (ω^2) ?
 
goonking said:
(1/2) (I) (ω^2) ?
Yes. And you need one more equation. What else is conserved?
 
ehild said:
Yes. And you need one more equation. What else is conserved?
i have no idea, momentum?
 
Momentum is conserved when translation is involved. But the beam can not translate. What can be conserved?
 
ehild said:
Momentum is conserved when translation is involved. But the beam can not translate. What can be conserved?
I can't think of anything else besides momentum and kinetic energy. any hints?
 
For rotation, you have quantities analogous to translational ones. For position, it is the angle of rotation. For velocity, it is angular velocity. For mass, it is moment of inertia. What do you have for momentum?
 
ehild said:
For rotation, you have quantities analogous to translational ones. For position, it is the angle of rotation. For velocity, it is angular velocity. For mass, it is moment of inertia. What do you have for momentum?
angular momentum?

but if the initial angular momentum is 0, then the final has to be 0 too? since it was at rest
 
  • #10
The angular momentum of the beam is zero, but the little mass has initial and final angular momentums with respect to the axis. How is the angular momentum defined for a point-like body, moving in the plane normal to an axis, traveling with velocity v, at distance D from an axis?
 
  • #11
ehild said:
The angular momentum of the beam is zero, but the little mass has initial and final angular momentums with respect to the axis. How is the angular momentum defined for a point-like body, moving in the plane normal to an axis, traveling with velocity v, at distance D from an axis?
L = mvr ?
 
  • #12
goonking said:
L = mvr ?
yes. What is r in this problem?
 
  • #13
ehild said:
yes. What is r in this problem?
distance traveled before the block hit the beam? I have no idea
 
  • #14
goonking said:
distance traveled before the block hit the beam? I have no idea
No, it is the distance of the line of track of the little mass from the axis of rotation. It is the same as the distance from the axis of the point where the small mass hits the beam .
 
  • #15
ehild said:
No, it is the distance of the line of track of the little mass from the axis of rotation. It is the same as the distance from the axis of the point where the small mass hits the beam .
so it is just D/2?
 
  • #16
goonking said:
so it is just D/2?
Yes. Now you can write the equation for conservation of angular momentum.
 
  • #17
ehild said:
Yes. Now you can write the equation for conservation of angular momentum.
i don't understand the use of this equation here.

m vi D/2 = m vf D/2 ?
 
  • #18
The beam also has got angular momentum, you need to include it.
 
  • #19
ehild said:
The beam also has got angular momentum, you need to include it.

m vi D/2 = (m vf D/2) + ( I ω)

?
 
  • #20
goonking said:
m vi D/2 = (m vf D/2) + ( I ω)

?
Yes, it is. You know the expression for the moment of inertia for the beam. And you have the equation for the energies.

Eliminate ω and solve for vf.
 
  • #21
ehild said:
Yes, it is. You know the expression for the moment of inertia for the beam. And you have the equation for the energies.

Eliminate ω and solve for vf.
what is the moment of inertia for the beam? isn't there a whole list of moment of inertia for each shape?
 
  • #22
You can use the formula valid for a thin rod of length D wit respect of the centre.
 
  • #23
goonking said:
what is the moment of inertia for the beam? isn't there a whole list of moment of inertia for each shape?
You can use the formula valid for a thin rod of length D with respect of the centre.
 
  • #24
ehild said:
You can use the formula valid for a thin rod of length D with respect of the centre.
hmm, i didn't know you could do that, since the beam doesn't look like a thin rod.

anyways, it is 1/12*M* L^2

L being D/2 correct?
 
  • #25
goonking said:
hmm, i didn't know you could do that, since the beam doesn't look like a thin rod.

anyways, it is 1/12*M* L^2

L being D/2 correct?
No the length L is D.
It does not look like a thin rod, but nothing is given about its width.
 
  • #26
ehild said:
No the length L is D.
It does not look like a thin rod, but nothing is given about its width.
so basically, we have 2 equations.

the kinetic energy and the angular momentum.

do we equate them?
 
  • #27
You can not equate angular momentum with energy. You have a system of two equations for the unknowns, final velocity vf and angular velocity ω. Solve.
 
  • #28
1/2 mv^2 = (1/2) (I) (ω^2)

correct?

we use that to solve for ω?
 
  • #29
goonking said:
1/2 mv^2 = (1/2) (I) (ω^2)

correct?

we use that to solve for ω?
It is wrong.
 
  • #30
ehild said:
It is wrong.
ok, I'm so lost right now.

lets try to clear things up a bit.

the block is moving, hits into the beam.

I'm guessing the beam makes a full revolution and comes back down to the initial position?

then the block moves backwards with a velocity that was less than what it initially was. and since it was going backwards, the velocity is negative.

is everything correct so far?
 

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