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Kinetic energy involving angular momentum and inertia

  • #1

Homework Statement


Problem with diagram here

Homework Equations


I=(1/3)MR2 (I believe the formula given in the problem is incorrect since it is pivoting on one end rather than the center of mass)
KE=(1/2)Iω2+(1/2)Mv2
L=Iω=MvR

The Attempt at a Solution



a.
True, the velocity is still going in the same direction at the instant of the collision.
False, it is a perfectly inelastic collision, which means kinetic energy is not conserved.
True, no real reason why. I just think it's true? No forces seem to be working on the center of mass during the collision.
True, no external forces/torque are working on the collision at that moment.

b.
(M/3)Lv=(1/3)ML2ω+(M/3)ML2ω
(M/3)Lv=(2/3)ML2ω
v/(2L)=ω

c.

KE=(1/2)Mv2+(1/2)Icmω2
I have no idea how to approach this correctly other than the possible use of the parallel axis theorem. This was as far as I got

The answer is given: (5/6)MgL+(1/12)Mv2
Can someone please explain to me the steps to get to that answer?
 

Answers and Replies

  • #2
ehild
Homework Helper
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Homework Equations


I=(1/3)MR2 (I believe the formula given in the problem is incorrect since it is pivoting on one end rather than the center of mass)
KE=(1/2)Iω2+(1/2)Mv2
L=Iω=MvR
These equations are valid only if you explain the meaning of the notations.
The formula for the moment of inertia is correct: I=1/12 ML2with respect to the CM. With respect to the pivot, it is 1/3ML2

The Attempt at a Solution



a.
True, the velocity is still going in the same direction at the instant of the collision.
No, the rod is not free to move, the force at the pivot acts during the collision, it can be of any magnitude, so can not be ignored.
False, it is a perfectly inelastic collision, which means kinetic energy is not conserved.
That is right.
True, no real reason why. I just think it's true? No forces seem to be working on the center of mass during the collision.
False, the change of angular momentum depends on the external torque. With respect of the CM, there is an external torque from the force at the pivot. The angular momentum of the whole system conserves, and its includes both the angular momentum of the rotation about the cm and the angular momentum of the cm orbiting about the pivot.
True, no external forces/torque are working on the collision at that moment.
The angular momentum conserves with respect to the pivot, as the torque of the external force at the pivot is zero, and also the torque of gravity is zero with respect to the pivot.
b.
(M/3)Lv=(1/3)ML2ω+(M/3)ML2ω
(M/3)Lv=(2/3)ML2ω
v/(2L)=ω
Correct.
c.

KE=(1/2)Mv2+(1/2)Icmω2
I have no idea how to approach this correctly other than the possible use of the parallel axis theorem. This was as far as I got

The answer is given: (5/6)MgL+(1/12)Mv2
Can someone please explain to me the steps to get to that answer?
After the collision, the energy is conserved. When the rod starts to move, it has both kinetic and potential energy. What is its initial potential energy?
The KE of a rigid body rotating about a fixed axis is 1/2 Iω2. What is the moment of inertia of the system putty+rod with respect to the pivot?

ehild
 
  • #3
After the collision, the energy is conserved. When the rod starts to move, it has both kinetic and potential energy. What is its initial potential energy?
The KE of a rigid body rotating about a fixed axis is 1/2 Iω2. What is the moment of inertia of the system putty+rod with respect to the pivot?

ehild
I think the PE is:

PE=mgL

since the height fallen with respect to the putty and rod is the entire length of the rod.

As for the moment of inertia, is it:

I=(1/3)(M/2)L2 + (M/3)L2?

The mass of the rod being halved because not all of the rod's mass is (because I'm unable to think of a better word at the moment) "used"?

Or since the masses are combined now and are rotating about the pivot point

I=(1/3)(M/2 + M/3)L2?
 
  • #4
ehild
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I think the PE is:

PE=mgL


since the height fallen with respect to the putty and rod is the entire length of the rod.
consider the potential energy with respect to the ground. When the whole system fell down, it has zero PE. Initially, the putty had M/3 gL potential energy, but the PE of the rod was? .... The pieces of the rod are not at equal height, so the potential energy of a bottom piece is about zero and a top piece is at height L...
As for the moment of inertia, is it:

I=(1/3)(M/2)L2 + (M/3)L2?

The mass of the rod being halved because not all of the rod's mass is (because I'm unable to think of a better word at the moment) "used"?

Or since the masses are combined now and are rotating about the pivot point

I=(1/3)(M/2 + M/3)L2?
Why do you think that not all the mass of the rod is "used"?
The moment of inertia of the rod with respect to the pivot is (M/3)L2. The moment of inertia of the putty (of mass M/3) is (M/3) L2. The momenta add up.

ehild
 
  • #5
consider the potential energy with respect to the ground. When the whole system fell down, it has zero PE. Initially, the putty had M/3 gL potential energy, but the PE of the rod was? .... The pieces of the rod are not at equal height, so the potential energy of a bottom piece is about zero and a top piece is at height L...


Why do you think that not all the mass of the rod is "used"?
The moment of inertia of the rod with respect to the pivot is (M/3)L2. The moment of inertia of the putty (of mass M/3) is (M/3) L2. The momenta add up.

ehild
Oh I thought because not all the mass is at the far end of the rod, the mass should be halved. My mistake.

So Itot=(2/3)ML2
KE=(1/2)(2/3)ML2ω=(1/3)ML2ω=(1/3)MLv2
PE=(M/3)gL+MgL=(4/3)MgL

(1/3)MLv2=(4/3)MgL
v2=4g ?

Then do I plug this into the KE equation so it is

(1/3)ML4g=(4/3)MLg?

I think I figured out the (1/12)Mv2 portion as well. Could it be:

ω=v/2L
(M/3)(L2)(v2/4L2)=(1/12)Mv2 ?
 
  • #6
ehild
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KE=(1/2)(2/3)ML2ω=(1/3)ML2ω=(1/3)MLv2
PE=(M/3)gL+MgL=(4/3)MgL
ω has to be squared in the equation for the KE.
The potential energy of the rod is not mgL as not all the mass is at height L.
The potential energy of the rod is the same as if all its mass concentrated at the centre of mass.

ehild
 
  • #7
ω has to be squared in the equation for the KE.
The potential energy of the rod is not mgL as not all the mass is at height L.
The potential energy of the rod is the same as if all its mass concentrated at the centre of mass.

ehild
So PE=(M/3)gL+Mg(L/2)=(5/6)MgL?

KE=PE
(1/3)MLv2=(5/6)MgL
v2=(5/2)g

ω=v/4L (from part b)

So KEf= (M/3)L2ω2+(M/3)L2(v2/4L2)

KEf=(1/3)ML(5/2)g+(1/12)Mv2=(5/6)MgL+(1/12)Mv2

Is that all right? If so, what happened to the (1/2) in front of both of the MLω2?
 
  • #8
ehild
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So PE=(M/3)gL+Mg(L/2)=(5/6)MgL?

KE=PE
It is not true. After the collision, the system rod+putty has bot KE and PE.
At the final state, there is only KE.

ehild
 
  • #9
It is not true. After the collision, the system rod+putty has bot KE and PE.
At the final state, there is only KE.

ehild
Ahh! I see. I think I got it now! So once the rod falls all the way down, all the PE is transformed (is that the right word to use? I feel like it isn't.) into KE. So the we can just use PE for part of the final KE.
Then we also have the KE from the putty and rod during the collision, right? So it'd be:
(1/2)Iω2+(1/2)(M/3)L2ω2
=(1/2)(M/3)L2ω2+(1/2)(M/3)L2ω2
=(M/3)L2ω2
=(M/3)L2(v2/4L2)
=(1/12)Mv2
Yes?

Then you add them together to get
(5/6)MgL+(1/12)Mv2

Thank you so much for all your help! I can't believe I forgot about how PE turns into KE as the rod falls down... Thank you!
 
  • #10
ehild
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Ahh! I see. I think I got it now! So once the rod falls all the way down, all the PE is transformed (is that the right word to use? I feel like it isn't.) into KE.

It is perfect... :smile:

And the solution is correct. Good job!

ehild
 

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