Center of mass and angular momentum

  • #1
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Homework Statement


On a frictionless table, a glob of clay of mass 0.240kg strikes a bar of mass 1.560kg perpendicularly at a point 0.390m from the center of the bar and sticks to it.

a) If the bar is 1.260m long and the clay is moving at 9.300m/s before striking the bar, what is the final speed of the center of mass?
b) At what angular speed does the bar/clay system rotate about its center off mass after the impact (in rad/s)?

Homework Equations


-

The Attempt at a Solution


I have found a) which is 1.24m/s.

In the case of b), I'm not sure on where to start. I have calculated that the center of mass is 0.682. I'm not sure on how to continue by using conservation of angular momentum, because we will have:

Icwc + Irwr = Ifwf

I'm not sure on how to continue. Isn't it possible to solve this problem by taking as a reference the center of the rod? (angular momentum will be conserved or not?)

Thank you.
 

Answers and Replies

  • #2
Simon Bridge
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I have found a) which is 1.24m/s.
So you reckon it makes no difference that the blob struck some distance from the center of mass of the rod?

Which center of mass are they tracking - that for the rod alone, or that of the combined rod+blob system?

angular momentum will be conserved or not
Angular momentum is conserved, yes.
The center you are looking for is the center of mass of the combined blob+rod system.
 
  • #3
haruspex
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Isn't it possible to solve this problem by taking as a reference the center of the rod?
You do need to be careful in picking your reference point for angular momentum conservation. There are two safe choices: any fixed point (which will not be any part of the rod or mass, of course, but might be where some part of them is at a particular point in time); the centre of mass of the system whose angular momentum you are calculating.
 
  • #4
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So you reckon it makes no difference that the blob struck some distance from the center of mass of the rod?

Which center of mass are they tracking - that for the rod alone, or that of the combined rod+blob system?


Angular momentum is conserved, yes.
The center you are looking for is the center of mass of the combined blob+rod system.

Isn't the velocity obtained in the first question the one of the center of mass before the collision? (Or at least whst this part asks for)

I was thinking of the problem and came up woth an idea. I can first find the center of mass position, and then calculate the distance of the objects with respect to the center of mass (which in turn gives us the angular velocity). For this part i should be using the velocity obtained in a) right?

Then, the final angular momentum will be the final angular velocity and the new moment of inertia (which should be the mass of the combined system multiplied by the square of the distance relative to the center of mass, which is the new rotating point).

Is my reasoning acceptable?
 
  • #5
Simon Bridge
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To me? You are the one doing the problem ;)

Consider though: after the collision you are expecting the combined system to rotate right?
Are you not expecting it to rotate about the center of mass of the combined systrem?

Does't the blob+rod have a combined center of mass all through the motion?
Also read post #3 carefully.
 
  • #6
haruspex
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Also read post #3 carefully.
Thanks Simon ... and I particularly stress this clause: any fixed point (which will not be any part of the rod or mass, but might be where some part of one of them is at a particular point in time).
 
  • #7
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Thanks Simon ... and I particularly stress this clause: any fixed point (which will not be any part of the rod or mass, but might be where some part of one of them is at a particular point in time).
I still cannot get my head around this problem. I am now trying to analyze it around the center of mass. I have found the center of mass and the distances of the bodies relative to this center (and then calculated the angular velocity by the combined velocity over the distance to the center of mass). Then, for final angular momentum, i found the combined moment of inertia (which should be of the two masses with respect to point b) and then cleared out for the angular velocity final.

What am i doing wrong? I cannot understand this problem. The main thing is that I don't get how the moments of inertia work after and before the collision.
 
  • #8
haruspex
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I still cannot get my head around this problem. I am now trying to analyze it around the center of mass. I have found the center of mass and the distances of the bodies relative to this center (and then calculated the angular velocity by the combined velocity over the distance to the center of mass). Then, for final angular momentum, i found the combined moment of inertia (which should be of the two masses with respect to point b) and then cleared out for the angular velocity final.

What am i doing wrong? I cannot understand this problem. The main thing is that I don't get how the moments of inertia work after and before the collision.

We can only tell what you are doing wrong if you post your working.
The hint I was giving you was to take moments about the point where the bar's centre of mass is before the impact. That is, not about the bar's centre of mass, since that moves, but about the point where it lay before impact. This simplifies the equations, and avoids having to work out where the combined mass centre is. But whichever way you do it, please post that working.
 
  • #9
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Thank you for your help guys. I finally got the right answer.

My method was the following:

Find the center of mass of the system:

[itex]x_{cm}=\displaystyle\frac{m_{c}(d+L/2)+m_{r}(L/2)}{m_{c}+m_{r}}[/itex], which gives [itex]x_{cm}=0.682m[/itex].

Then we can find the distance of the ball of clay to this center of mass and the distance of the rod to this center of mass:

[itex]d_{ccm}=0.338m, d_{rcm}=0.052m[/itex]

Now we can find the angular velocity from this information, note that the angular velocity for the rod will be zero.

[itex]w_{cm}=\displaystyle\frac{v}{d_{cm}}=27.5rad/s[/itex]

Then, note that angular momentum will be conserved and given by the following expression:

[itex]I_{c}w_{c}+I_{r}w{r}=I_{t}w_{f}[/itex]

Where the moments of inertia of the rod and clay are:

[itex]I_{c}=m_{c}d_{rcm}^{2}[/itex] and [itex]I_{r}=1/12 m_{r}L^{2}+m_{r}d_{rcm}^{2}[/itex]

Then you can clear out those expressions and you will find that:

[itex]w_{f}=\displaystyle\frac{I_{c}w_{c}}{I_{r}+I_{c}}=3.17rad/s[/itex]
 
  • #10
haruspex
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Thank you for your help guys. I finally got the right answer.

My method was the following:

Find the center of mass of the system:

[itex]x_{cm}=\displaystyle\frac{m_{c}(d+L/2)+m_{r}(L/2)}{m_{c}+m_{r}}[/itex], which gives [itex]x_{cm}=0.682m[/itex].

Then we can find the distance of the ball of clay to this center of mass and the distance of the rod to this center of mass:

[itex]d_{ccm}=0.338m, d_{rcm}=0.052m[/itex]

Now we can find the angular velocity from this information, note that the angular velocity for the rod will be zero.

[itex]w_{cm}=\displaystyle\frac{v}{d_{cm}}=27.5rad/s[/itex]

Then, note that angular momentum will be conserved and given by the following expression:

[itex]I_{c}w_{c}+I_{r}w{r}=I_{t}w_{f}[/itex]

Where the moments of inertia of the rod and clay are:

[itex]I_{c}=m_{c}d_{rcm}^{2}[/itex] and [itex]I_{r}=1/12 m_{r}L^{2}+m_{r}d_{rcm}^{2}[/itex]

Then you can clear out those expressions and you will find that:

[itex]w_{f}=\displaystyle\frac{I_{c}w_{c}}{I_{r}+I_{c}}=3.17rad/s[/itex]
Good job.
Fwiw, here's my way, taking moments about the point where the centre of the bar is initially:
Moment of clay, prior = mcud
Moment of clay after impact: mcωd2
Moment of bar after impact: Irω
mcud = mcωd2 + Irω
ω = mcud/(mcd2 + Ir)
 

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