Uniform circular motion and angular momentum?

In summary, a sphere is attached to a rope and is moved in a circular motion with a radius that is changed. The water added to the bucket causes the radius to change and the mass and force needed to cause the circular motion is known.
  • #1
PhyIsOhSoHard
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Homework Statement


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A sphere on top of a table is attached to a rope which goes through a hole in the table and is attached to a bucket at the other end. The sphere moves in a uniform circular motion with radius R.

Water is then added to the bucket and the radius for the sphere's circular motion is changed. Find an expression for the new radius.

The mass of the sphere (m) and bucket (M) is known. The mass of the water added to the bucket is known as well. The initial radius of the circular motion is known and the initial speed of the sphere is known. Angular momentum is conserved.

Friction can be neglected.

Homework Equations


Angular momentum?

The Attempt at a Solution


I have no idea how to start on this?
 
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  • #2
The force needed to cause a circular motion is ?
How big is it?
And it is provided by ?
So if that changes, when is there equilibrium again ?

If 2. and 3. are basically blank, providing assistance at a suitable level is difficult. Show something more than just '?'
 
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  • #3
Well, you stated that angular momentum is conserved, which is true, so try that. Write out expressions for the angular momentum before and after the water is added, and see if that can get you to the radius. A few more equations that should be helpful:

[itex]a = \frac{v^{2}}{r}[/itex] (uniform circular motion)

[itex]l = |\stackrel{\rightarrow}{r} \times \stackrel{\rightarrow}{p}| = I\omega[/itex] (angular momentum)
 
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  • #4
BvU said:
The force needed to cause a circular motion is ?
How big is it?
And it is provided by ?
So if that changes, when is there equilibrium again ?

If 2. and 3. are basically blank, providing assistance at a suitable level is difficult. Show something more than just '?'

The force needed to cause a circular motion is ?

It is based on Newton's second law, am I right?
The radial acceleration combined gives you:
[itex]\sum F=m \frac{v^2}{R}[/itex]

How big is it?

By drawing a FBD on the bucket, the weight of the bucket (M*g) is equal to the tension in the rope for when [itex]\sum F_y=0[/itex]

Thus:
[itex]M\cdot g=m \frac{v^2}{R}[/itex]

And it is provided by ?

The weight of the bucket?

So if that changes, when is there equilibrium again ?

I'm not sure.
 
  • #5
jackarms said:
Well, you stated that angular momentum is conserved, which is true, so try that. Write out expressions for the angular momentum before and after the water is added, and see if that can get you to the radius. A few more equations that should be helpful:

[itex]a = \frac{v^{2}}{r}[/itex] (uniform circular motion)

[itex]l = |\stackrel{\rightarrow}{r} \times \stackrel{\rightarrow}{p}| = I\omega[/itex] (angular momentum)

I'm a little bit confused by the equation for angular momentum that you wrote.

In my textbook, it writes two different formulas.

Angular momentum for a particle:
[itex]L=r \times p = r \times mv[/itex]

Angular momentum for a rigid body rotating about axis of symmetry:
[itex]L=I\omega[/itex]

I'm assuming I have to use the formula for a particle?
 
  • #6
What about filling in the new weight of the bucket + water in the "thus" equation ?
 
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  • #7
BvU said:
What about filling in the new weight of the bucket + water in the "thus" equation ?

Do you mean like this:
[itex](M+M_{water})\cdot g=m \frac{v^2}{R}[/itex]
 
  • #8
What happened to the angular momentum ? (particle)
 
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  • #9
BvU said:
What happened to the angular momentum ? (particle)

My guess is that it got bigger?

For a particle, we have that
[itex]L=mvR[/itex]

Since the angular momentum is conserved, that means that the angular moment before and after is adequate:
[itex]mvR=mvr[/itex]

Where R is the radius before water was added and r is the radius after water is added (the radius that we want to find an expression for).

The problem here is that the mass used here is the mass for the sphere, not the bucket + water?
 
  • #10
Don't forget that v also may change...

Originally you had [itex]M\cdot g=m \frac{v^2}{R}[/itex] where all, except R are known. So R is known.
From your expression for angular momentum (particle), L is known too. All you have to do is combine this mvR = L = L' = ##mv'R'## with [itex](M+M_{water})\cdot g=m \frac{v'^2}{R'}[/itex]
 
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  • #11
BvU said:
Don't forget that v also may change...

Originally you had [itex]M\cdot g=m \frac{v^2}{R}[/itex] where all, except R are known. So R is known.
From your expression for angular momentum (particle), L is known too. All you have to do is combine this mvR = L = L' = ##mv'R'## with [itex](M+M_{water})\cdot g=m \frac{v'^2}{R'}[/itex]

Originally I had my speed:
[itex]M\cdot g=m \frac{v^2}{R}[/itex]

After the water was added, I have:
[itex](M+M_{water})\cdot g=m \frac{v'^2}{R'}[/itex]

But I'm not sure how to combine these equations with L = mv'R'? Can you explain that?
 
  • #12
If you take your original conservation equation:
$$mvR = mv'R'$$
You have three unknowns: [itex]v[/itex], [itex]v'[/itex] and [itex]R'[/itex]. You can solve for [itex]v[/itex] from the first equation in your last post. Then, in your second equation (the one where you account for the mass of the water) you have [itex]v'[/itex] and [itex]R'[/itex] again, so if you now have a system of equations, which you can use to solve for [itex]R'[/itex].
 
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  • #13
jackarms said:
If you take your original conservation equation:
$$mvR = mv'R'$$
You have three unknowns: [itex]v[/itex], [itex]v'[/itex] and [itex]R'[/itex]. You can solve for [itex]v[/itex] from the first equation in your last post. Then, in your second equation (the one where you account for the mass of the water) you have [itex]v'[/itex] and [itex]R'[/itex] again, so if you now have a system of equations, which you can use to solve for [itex]R'[/itex].

Oh yes, now I see it. I have 3 equations and 3 unknowns. I don't know why I didn't see that before.

Thank you both for your help!
 
  • #14
It was a pleasure. I learned too.
 

Related to Uniform circular motion and angular momentum?

1. What is uniform circular motion?

Uniform circular motion is the motion of an object along a circular path at a constant speed. This means that the object travels the same distance in the same amount of time, resulting in a constant velocity.

2. How is angular velocity different from linear velocity in uniform circular motion?

Angular velocity is the rate of change of angular displacement, while linear velocity is the rate of change of linear displacement. In uniform circular motion, angular velocity remains constant while linear velocity changes as the object moves along the circular path.

3. What is the relationship between angular velocity and angular momentum?

Angular velocity and angular momentum are directly proportional to each other. This means that as angular velocity increases, angular momentum also increases. Therefore, in uniform circular motion, as the object's speed increases, its angular momentum also increases.

4. How does centripetal force affect uniform circular motion?

In uniform circular motion, the object is constantly changing direction, meaning it is accelerating towards the center of the circle. This acceleration is caused by the centripetal force, which is directed towards the center of the circle.

5. Can an object in uniform circular motion have a constant speed but changing velocity?

Yes, an object in uniform circular motion can have a constant speed but changing velocity. This is because velocity is a vector quantity that takes into account both speed and direction. In uniform circular motion, the direction of the object's velocity is constantly changing, even though its speed remains constant.

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