KE Conservation: Calculating Final Velocities

In summary, the conversation revolves around a problem involving a block of mass m colliding with a beam that can rotate about an axis. The problem is to find the final velocity of the block after the collision. The approach suggested involves using the conservation of angular momentum and kinetic energy equations. The final solution requires solving a system of equations for the unknowns, final velocity vf and angular velocity ω.
  • #1
goonking
434
3

Homework Statement


5R1RjVK.png


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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  • #2
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?
 
  • #3
ehild said:
Not quite. The beam can only rotate about the axis. What is its kinetic energy?
(1/2) (I) (ω^2) ?
 
  • #4
goonking said:
(1/2) (I) (ω^2) ?
Yes. And you need one more equation. What else is conserved?
 
  • #5
ehild said:
Yes. And you need one more equation. What else is conserved?
i have no idea, momentum?
 
  • #6
Momentum is conserved when translation is involved. But the beam can not translate. What can be conserved?
 
  • #7
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?
 
  • #8
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?
 
  • #9
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?
 
  • #31
goonking said:
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?

No. You need to consider the situation just before and after collision. The block had some velocity initially, the rod was in rest.
The block has some angular momentum both before and after the collision. The beam has angular momentum after the collision. The angular momentum before collision is the same as the angular momentum after it.
Also, the initial KE of the block is the same as the KE of the block plus the rotational energy of the beam just after collision.
You do not know the direction of the final velocity of the block. Consider it positive. If it comes out to be negative at the end, it would mean that the block moves backward after the collision.
 
  • #32
it's a bit hard trying to grasp how an object moving linearly can have an angular momentum. anyways:

Kinetic energy :
1/2 m vi^2 = (1/2 m vf^2) +(1/2) (I) (ω^2)angular momentum:

m vi D/2 = (m vf D/2) + ( I ω)

I =1/12*M* D^2

=
m vi D/2 = (m vf D/2) + ( (1/12*M* D^2) ω)
 
  • #33
goonking said:
it's a bit hard trying to grasp how an object moving linearly can have an angular momentum. anyways:

Kinetic energy :
1/2 m vi^2 = (1/2 m vf^2) +(1/2) (I) (ω^2)angular momentum:

m vi D/2 = (m vf D/2) + ( I ω)

I =1/12*M* D^2

=
m vi D/2 = (m vf D/2) + ( (1/12*M* D^2) ω)

It is right now. Solve for vf.
 
  • #34
ehild said:
It is right now. Solve for vf.
is this just all algebra now?

how do we eliminate ω?

and D is unknown.
 
  • #35
ehild said:
It is right now. Solve for vf.
shouldn't the M in I =1/12*M* D^2 be 2m?

so
I =1/12*2m* D^2

?
 

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