Torque on a Pulley Homework: Find Speed & Average Torque

[veronicak5678]

Homework Statement

Two paint buckets, initially at rest, are connected by a lightwieght rope which is wrapped over a pulley which has a mass 2 kg and a radius of .5 m. Only conservative forces do work.
Mass A is 4 kg and starts on the ground. Mass B is 6 kg and starts 5 m above the ground.
a- Find the speed at which the top bucket hits the ground.
b- What was the average torque on the pulley?

Homework Equations

The Attempt at a Solution

a - 4.22 m/s
b- Using torque = (tension 1 - tension 2) r, I get 9.8 N for net torque. Is this correct?

 

[LowlyPion]

Homework Statement

Two paint buckets, initially at rest, are connected by a lightwieght rope which is wrapped over a pulley which has a mass 2 kg and a radius of .5 m. Only conservative forces do work.
Mass A is 4 kg and starts on the ground. Mass B is 6 kg and starts 5 m above the ground.
a- Find the speed at which the top bucket hits the ground.
b- What was the average torque on the pulley?

Homework Equations

The Attempt at a Solution

a - 4.22 m/s
b- Using torque = (tension 1 - tension 2) r, I get 9.8 N for net torque. Is this correct?

Because the pulley has mass and hence moment of inertia, your acceleration won't be entirely just the difference in the paint buckets accelerating just the paint buckets. There is also the increase in angular kinetic energy that goes into the pulley.

 

[veronicak5678]

So how do I account for that in the equation? Should I add 1/2 Iω^2?

 

[LowlyPion]

So how do I account for that in the equation? Should I add 1/2 Iω^2?

Yes. You could approach it from the point of view of kinetic energy of the buckets and changes in potential energy and the change in kinetic energy of the pulley and noting along the way that ω is equal to v/r.

 

[aniketp]

or the acceleration = (driving force)/(Inertia) will work quite well, giving you the tension in both strings (which will be unequal, since the pulley has mass) :)

 

[veronicak5678]

I don't know how to use these equation in a system like this. For example, do I need to focus on 1 mass? In the term for initial potential, only one has initial potential. How do I find info about the pulley using this?

Kf + Uf + RKf = Ki + Ui + Rki

 

[aniketp]

i think that applying force dynamics and rotational dynamics will make this problem easier.

 

[LowlyPion]

I don't know how to use these equation in a system like this. For example, do I need to focus on 1 mass? In the term for initial potential, only one has initial potential. How do I find info about the pulley using this?

Kf + Uf + RKf = Ki + Ui + Rki

Well that equation certainly would work.

What is the initial KE then? 0 right? And same for Rotational KE. For PE the initial is the m*g*h of the 6kg bucket.

For final KE you have the masses of the two buckets in motion at some velocity and your rotational KE related by v/r and you have the 4 kg bucket with its m*g*h when the heavier bucket touches down.

 

[veronicak5678]

So I don't consider the mass of the pulley in the Kf term? Should it just be .5 ( 6 + 4 kg) vf^2? And the info about the pulley is only used in the RKf term?

 

[LowlyPion]

So I don't consider the mass of the pulley in the Kf term? Should it just be .5 ( 6 + 4 kg) vf^2? And the info about the pulley is only used in the RKf term?

The pulley is the one rotating.

The buckets are the ones with mv²/2.

You just have to account for the energy wherever it is. That after all is what's conserved isn't it?

 

[veronicak5678]

OK, I plugged that al in and found ω = 8.49 m/s. Using I ω= τ , I end up with τ = 18.02. Does this look right?

 

[LowlyPion]

OK, I plugged that al in and found ω = 8.49 m/s. How do I use this to find average torque?

They asked you for v in part a) didn't they? Not ω.

As for Torque isn't that merely Net force * radius?

 

[veronicak5678]

So the answer is incorrect? I would rather use I ω= τ and not find all the separate forces.

 

[LowlyPion]

OK, I plugged that al in and found ω = 8.49 m/s. Using I ω= τ , I end up with τ = 18.02. Does this look right?

I ω= is not torque. That's Angular momentum sometimes written as L.

I * α = τ

 

[LowlyPion]

http://hyperphysics.phy-astr.gsu.edu/hbase/rotv.html#rvec2

 

[veronicak5678]

Argh! Thanks for being so patient. So I can't find α without time. I can't find time without rotations. So I do need to find all the forces. Would that be -pulley weight - mass1 weight - mass 2 weight + tension 1 + tension 2?

 

[LowlyPion]

Argh! Thanks for being so patient. So I can't find α without time. I can't find time without rotations. So I do need to find all the forces. Would that be -pulley weight - mass1 weight - mass 2 weight + tension 1 + tension 2?

If you wanted to find α that's a/r and a can be found by V² = 2*a*x = 2*5*a

So take your answer from part a) square it and divide by 10 and that will be a.

 

[veronicak5678]

Oh! So I alpha = I (a/r) = torque
3.56*.25 = .89

torque = .89 Nm