Uniform circular motion and angular momentum?

[PhyIsOhSoHard]

Homework Statement

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?

 

[BvU]

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 '?'

 

[jackarms]

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:

$a = \frac{v^{2}}{r}$ (uniform circular motion)

$l = |\stackrel{\rightarrow}{r} \times \stackrel{\rightarrow}{p}| = I\omega$ (angular momentum)

 

[PhyIsOhSoHard]

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:
$\sum F=m \frac{v^2}{R}$

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 $\sum F_y=0$

Thus:
$M\cdot g=m \frac{v^2}{R}$

And it is provided by ?

The weight of the bucket?

So if that changes, when is there equilibrium again ?

I'm not sure.

 

[PhyIsOhSoHard]

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:

$a = \frac{v^{2}}{r}$ (uniform circular motion)

$l = |\stackrel{\rightarrow}{r} \times \stackrel{\rightarrow}{p}| = I\omega$ (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:
$L=r \times p = r \times mv$

Angular momentum for a rigid body rotating about axis of symmetry:
$L=I\omega$

I'm assuming I have to use the formula for a particle?

 

[BvU]

What about filling in the new weight of the bucket + water in the "thus" equation ?

 

[PhyIsOhSoHard]

What about filling in the new weight of the bucket + water in the "thus" equation ?

Do you mean like this:
$(M+M_{water})\cdot g=m \frac{v^2}{R}$

 

[BvU]

What happened to the angular momentum ? (particle)

 

[PhyIsOhSoHard]

What happened to the angular momentum ? (particle)

My guess is that it got bigger?

For a particle, we have that
$L=mvR$

Since the angular momentum is conserved, that means that the angular moment before and after is adequate:
$mvR=mvr$

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?

 

[BvU]

Don't forget that v also may change...

Originally you had $M\cdot g=m \frac{v^2}{R}$ 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 $(M+M_{water})\cdot g=m \frac{v'^2}{R'}$

 

[PhyIsOhSoHard]

Don't forget that v also may change...

Originally you had $M\cdot g=m \frac{v^2}{R}$ 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 $(M+M_{water})\cdot g=m \frac{v'^2}{R'}$

Originally I had my speed:
$M\cdot g=m \frac{v^2}{R}$

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

But I'm not sure how to combine these equations with L = mv'R'? Can you explain that?

 

[jackarms]

If you take your original conservation equation:
$$mvR = mv'R'$$
You have three unknowns: $v$, $v'$ and $R'$. You can solve for $v$ 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 $v'$ and $R'$ again, so if you now have a system of equations, which you can use to solve for $R'$.

 

[PhyIsOhSoHard]

If you take your original conservation equation:
$$mvR = mv'R'$$
You have three unknowns: $v$, $v'$ and $R'$. You can solve for $v$ 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 $v'$ and $R'$ again, so if you now have a system of equations, which you can use to solve for $R'$.

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!

 

[BvU]

It was a pleasure. I learned too.