Angular Velocity of Center of Mass

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A 0.50 kg clay glob strikes a 1.58 kg bar on a frictionless table, sticking to it and creating a combined system. The center of mass speed was calculated to be 1.587 m/s, while the angular velocities for the clay and bar were initially found to be 22.84 rad/s and 72.53 rad/s, respectively. The discussion focused on how to relate these angular velocities and the conservation of angular momentum and linear momentum to find the final angular speed of the bar/clay system. After several calculations and clarifications, the final angular speed was determined to be 3.433 rad/s. The conversation emphasized the importance of systematic problem-solving and understanding the physical principles involved.
baubletop
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Homework Statement


On a frictionless table, a 0.50 kg glob of clay strikes a uniform 1.58 kg bar perpendicularly at a point 0.38 m from the center of the bar and sticks to it. If the bar is 1.30 m long and the clay is moving at 6.60 m/s before striking the bar, what is the final speed of the center of mass?
At what angular speed does the bar/clay system rotate about its center of mass after the impact?


Homework Equations


xcm = Ʃmixi / Mtotal
ω = v/r

The Attempt at a Solution


I was able to find the speed of the center of mass, which was 1.587 m/s.
I also found ω for both the clay and the bar, where r = 0.289 m. For the clay it's 22.84 rad/s, and for the bar it's 72.53 rad/s. I just don't know how to relate them into one "bar/clay system". Adding, subtracting, dividing, and averaging didn't work, and I know there's something bigger I'm missing.
 
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baubletop said:
I was able to find the speed of the center of mass, which was 1.587 m/s.
OK.

I also found ω for both the clay and the bar, where r = 0.289 m. For the clay it's 22.84 rad/s, and for the bar it's 72.53 rad/s.
Not sure what you did here.

I just don't know how to relate them into one "bar/clay system". Adding, subtracting, dividing, and averaging didn't work, and I know there's something bigger I'm missing.
Do it systematically. Where's the center of mass of the system? In the center of mass frame, what's the angular momentum of each piece (rod and clay) about the cm before the collision? So what's the total angular momentum? What's the rotational inertia of the system?
 
For ω of the clay and bar I did this:
v = 6.6 m/s
r = (d+D/2) - xcm
xcm clay = [mclay*(d+D/2)+Mbar*(D/2)]/Mtotal = 0.741 m
D = 1.30 m (length of bar)
d = distance of clay = 0.38 m
so r = 0.289 m
then ωclay = 22.84 m/s

For the bar:
d = xcm - D/2 = 0.091 m
ωbar = 72.53 m/sThe center of mass of the system would be the 0.741 m as above.
Before the collision the rod is at rest, so its angular momentum is 0, but for the clay would it be:
ω = v/r = 6.6/(.741-.38) = 6.6/.361 = 18.27 rad/s ? If r is the distance from the center of mass.
For the rotational inertia, I'm not sure.
 
baubletop said:
For ω of the clay and bar I did this:
v = 6.6 m/s
r = (d+D/2) - xcm
xcm clay = [mclay*(d+D/2)+Mbar*(D/2)]/Mtotal = 0.741 m
D = 1.30 m (length of bar)
d = distance of clay = 0.38 m
so r = 0.289 m
then ωclay = 22.84 m/s
Ok, that's the angular velocity of the clay about the point which will be the common mass centre, just before impact - right?
For the bar:
d = xcm - D/2 = 0.091 m
ωbar = 72.53 m/s
I don't understand what you have calculated there.
What is the angular momentum of the clay about the point which will be the common mass centre, just before impact? What is the moment of inertia of the combined system about its common mass centre?
 
haruspex said:
Ok, that's the angular velocity of the clay about the point which will be the common mass centre, just before impact - right?
I believe so--although now I'm getting a bit confused.

I don't understand what you have calculated there.
What is the angular momentum of the clay about the point which will be the common mass centre, just before impact? What is the moment of inertia of the combined system about its common mass centre?
Sorry, the way I looked at it was:
v = 6.6 m/s, the velocity of the clay
The center of mass of the entire system is at .741 m, and the center of mass of the bar itself is at half the length, so 0.65 m. Then r would be the distance from the center of mass of the whole system to the center of the bar. The velocity divided by that gives 72.53 rad/s.
The moment of inertia would be 0.0548 kg m2. Do I need to look at kinetic energy to be able to find ω?
The way I found the moment of inertia was:
mclay = 0.5 kg
mbar = 1.58 kg
rclay = (1.3m/2)+0.38m = 1.03m, then this minus 0.741m gives the distance from the cm, which is 0.289 m
rbar = (1.3m/2) = 0.65m, subtracting this from 0.741m gives its distance, which is .091 m
Then using sum(miri^2) gives 0.0548
 
Last edited:
baubletop said:
v = 6.6 m/s, the velocity of the clay
The center of mass of the entire system is at .741 m, and the center of mass of the bar itself is at half the length, so 0.65 m. Then r would be the distance from the center of mass of the whole system to the center of the bar. The velocity divided by that gives 72.53 rad/s.
Well, yes, it gives that number, but I don't understand what the physical significance is of dividing the pre-impact angular velocity of the clay by the offset of the bar's mass centre from the total system's mass centre. What physical quantity do you think that represents? What conservation law are you using?
The moment of inertia would be 0.0548 kg m2. Do I need to look at kinetic energy to be able to find ω?
No, you need to consider angular momentum and linear momentum. Those are what will be conserved through the impact.
 
Again, I recommend:

Find the angular momentum of the bar with respect to the center of mass of the system. Note that the bar is moving with respect to the center of mass, so that angular momentum is not zero.

Do the same for the angular momentum of the clay with respect to the center of mass of the system.

That will give you the total angular momentum about the center of mass, which is conserved.

Calculate the rotational inertia of the system of bar + clay about its center of mass. Use it to find ω of the system.
 
I'm sorry I'm still not getting it, I'm just going to let this problem go. Thanks for all the help but I guess it's just going over my head and confusing me.
 
baubletop said:
I'm sorry I'm still not getting it, I'm just going to let this problem go. Thanks for all the help but I guess it's just going over my head and confusing me.
Don't give up!

Answer this: With respect to the center of mass, how fast is the rod moving? (Hint: You already figured out the speed of the center of mass in the lab frame.) Using that, what is its angular momentum about the center of mass?

Same thing for the clay.
 
  • #10
Would it be
v cm = (m1v1 + m2v2)/(m1+m2)
1.587 m/s = (3.3 + 1.58v2)/(.5+1.58)
Solving for v2 this is 0.000608 m/s, correct?
Again the rod's center of mass is at 1.3/2 = 0.65 m, so its distance from the center of mass of the system is .091 m. Which gives an angular momentum of 0.0066 rad/s. But this seems way too small, so I feel like I'm getting my equations wrong or something.
(This is just for the bar, I want to make sure I'm doing the right thing before I move onto the clay.)
 
  • #11
baubletop said:
Would it be
v cm = (m1v1 + m2v2)/(m1+m2)
1.587 m/s = (3.3 + 1.58v2)/(.5+1.58)
Solving for v2 this is 0.000608 m/s, correct?
Again the rod's center of mass is at 1.3/2 = 0.65 m, so its distance from the center of mass of the system is .091 m. Which gives an angular momentum of 0.0066 rad/s. But this seems way too small, so I feel like I'm getting my equations wrong or something.
(This is just for the bar, I want to make sure I'm doing the right thing before I move onto the clay.)
Let's start again. There are many approaches to solving this. Here's what I think is the simplest.
(Pls work entirely in symbols in the equations you post. Only plug in the numbers as the final step.)
We need to pick a reference point, O, for taking angular momentum. The safest is to pick a stationary point. I'll pick the initial location of the bar's centre of mass.
1. The clay mass mc starts with speed u and strikes the bar length L distance d from its mass centre. What is the linear momentum of the clay before impact?
2. What is the angular momentum of the clay about O before impact?
3. Suppose just after impact the centre of the bar is moving at speed vb and the clay is moving at speed vc. (These will both be in the same direction as u.) The bar-clay system is rotating at rate ω. What is the relationship between ω, d, vb and vc? (This is just geometry.)
4. What is the linear momentum of the system just after impact, in terms of mb, vb, mc and vc?
5. How does the answer to (4) relate to the answer to (1)?
6. What is the angular momentum about O of the clay just after impact?
7. What is the angular momentum about O of the bar just after impact? (Since this is immediately after impact, the bar's centre is still at O.)
8. How do the answers to (6) and (7) relate to the answer to (2)?
By this point you should have an equation from each of (3), (5) and (8), and three unknowns, ω, vb and vc. Solve to find ω.
 
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  • #12
1. pc = mcu
2. ω = mc*u*d
3. I'm not sure if I'm overthinking or underthinking this one... Does this have to do with L = r x p (in this case, L = d x vc) and L = Iω?
4. psystem = mbvb + mcvc
5. Conservation of momentum, so 1 and 4 need to be equal
6. If it's the same O, would it not be different from 2?
7. ω = mb*u*d
8. These also need to be equal to 2 from conservation of angular momentum

Let me know if this stuff is wrong... I'm sorry this is taking so much work, thanks so much for your patience!
 
  • #13
baubletop said:
1. pc = mcu
Yes.
2. ω = mc*u*d
Yes, that's the angular momentum, but let's not call it ω. ω is usually reserved for angular speed (radians/sec). L is usual, but I made the mistake of using that for the bar's length. How about λ?
3. I'm not sure if I'm overthinking or underthinking this one... Does this have to do with L = r x p (in this case, L = d x vc) and L = Iω?
No, as I said, it's just geometry, so masses won't come into it. But it is very similar to the equations you quoted. The speed of the clay after impact will be the same as the speed of the centre of the bar, plus a bit for the bar's rotation. If the rotation rate is ω and the distance is d, how much extra speed?
4. psystem = mbvb + mcvc
5. Conservation of momentum, so 1 and 4 need to be equal
Yes and yes.
6. If it's the same O, would it not be different from 2?
It will be different because the clay's speed has changed. Use the same equation but with the new speed.
7. ω = mb*u*d
No. The bar doesn't know anything about u, the initial speed of the clay. The bar is rotating at rate ω about O. What is its moment of inertia about O? So what is its angular momentum about O?
8. These also need to be equal to 2 from conservation of angular momentum
Individually equal to 2? Or... what?
 
  • #14
3. I think this is where I'm really not getting the problem. I'm having a tough time visualizing it. Would it be vc = vb + ω/d?
6. Then λ = mc*vc*d ?
7. The moment of inertia of the bar would be (I think) I = 1/12 mb*L2. And its angular momentum is Iω.
8. 6+7 need to be equal to 2.
 
  • #15
baubletop said:
3. I think this is where I'm really not getting the problem. I'm having a tough time visualizing it. Would it be vc = vb + ω/d?
Close, but think about dimensions (always a good sanity check). ω is a rate (like, 1/time) and d is a distance. How would you expect to combine 1/time with distance to get a speed?
6. Then λ = mc*vc*d ?
7. The moment of inertia of the bar would be (I think) I = 1/12 mb*L2. And its angular momentum is Iω.
8. 6+7 need to be equal to 2.
Yes, yes and yes!
 
  • #16
3. Then vc = vb + ω*d?
 
  • #17
baubletop said:
3. Then vc = vb + ω*d?

Bingo.
 
  • #18
Great! So this gives me:
1. mcu = mbvb + mcvc
and
2. mc*u*d = (mc*vc*d) + 1/12 mb*L2
and
3. vc = vb + ω*d

Using 2 I can get
vc = (12*d*mc*u - mb*L2*ω)/12*d*mc
This is getting messy so I'll try to plug in some numbers...
vc = (12*.38m*0.5kg*6.6m/s - 1.58kg*1.69m2*ω)/12*.38m*0.5kg
vc = -1.17*(ω-5.636)

If I plug this into 1 I get:
mcu = mbvb + mc(-1.17*(ω-5.636))
0.5kg*6.6m/s = 1.58kg*vb + 0.5kg*(-1.17*(ω-5.636))
Solving for vb:
vb = 0.37ω + 0.0019

Then plugging these both into 3 I get:
-1.17*(ω-5.636) = (0.37ω + 0.0019) + 0.38m*ω
Solving for ω:
ω = 3.433 rad/s

Which is...drum roll...CORRECT!
Thank you SO SO MUCH for helping me! I've been stuck on this problem all day!
 
  • #19
Good work!

Just for the record, here's how I would have done it.

You found the center of mass of the system. The clay travels along a line that is dc = 0.289 above the system center of mass and the bar's center is at distance of db = 0.091 below the system center of mass.

Since the center of mass of the system travels at 1.587 m/s, you know that the bar travels at that speed relative to the center of mass. The angular momentum of the bar about the system center of mass (before the collision) is thus Lb = dbmbvb.

Similarly, the speed of the clay relative to the center of mass is vc = 6.60 m/s - 1.587 m/s. The angular momentum of the clay about the system center of mass (before the collision) is thus Lc = dcmcvc.

Add them up to get the total angular momentum, which is conserved during the collision.

Find the rotational inertia of the system about its center of mass. (It's just the sum of the individual Is about the system center of mass.) Then set Ltotal = Isystemω, to find ω after the collision.

There are many ways to skin a cat. Take your pick! (I like to use the center of mass frame, but it's all good.)
 

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