Angular Momentum Problem: Determining Angular Acceleration

[JoeyBob]

Homework Statement

See attached

Relevant Equations

F=mg, L=I*rotation rate

So I first tried to find L using torque,

Torque=d/dt*L, and took the integral of this.

Ended up with 23.28484t

Now I square the equation L=rotation rate*I to get L^2=rotation acceleration *I^2

Angular acceleration=L^2/I^2

I feel like I am doing something wrong though, this doesn't give the right answer.

The answer is suppose to be 29.13

 

[kuruman]

You need to draw two free body diagrams, one for the cylinder and one for the hanging mass and write Newton's second law for each. Do that and show us your work.

 

[JoeyBob]

You need to draw two free body diagrams, one for the cylinder and one for the hanging mass and write Newton's second law for each. Do that and show us your work.

So a=g and a=v^2/r

Then v=1.498 and p=14.8302

L=r x p = 0.24*14.83=3.56

angular acceleration= L^2/I^2 = 3.56^2/0.229^2=242

Whats wrong now? Answer should be 29.13.

 

[JoeyBob]

You need to draw two free body diagrams, one for the cylinder and one for the hanging mass and write Newton's second law for each. Do that and show us your work.

Wait I forgot about tension. So

a=T/m-g and a=v^2/r

But this still doesn't help because I know neither the tension nor the velocity, nor the acceleration.

 

[kuruman]

First of all a = v²/r is centripetal acceleration that is irrelevant here.

Secondly, I asked you to draw two free body diagrams. Writing Newton's second law on the basis of these will give you two equations and you have two unknowns, the acceleration and the tension.

angular acceleration= L^2/I^2 = 3.56^2/0.229^2=242

Whats wrong now? Answer should be 29.13.

What's wrong now and always is your expression that "angular acceleration= L^2/I^2". I don't know where you got this, but it is not correct. Stop using it.

 

[JoeyBob]

First of all a = v²/r is centripetal acceleration that is irrelevant here.

Secondly, I asked you to draw two free body diagrams. Writing Newton's second law on the basis of these will give you two equations and you have two unknowns, the acceleration and the tension.What's wrong now and always is your expression that "angular acceleration= L^2/I^2". I don't know where you got this, but it is not correct. Stop using it.

So I need to find 3 separate equations that contain the unknowns to find the answer?

 

[kuruman]

You need two equations and you have two unknowns, the tension and the acceleration. You can only draw two free body diagrams, one for the disk and one for the hanging mass.

 

[JoeyBob]

You need two equations and you have two unknowns, the tension and the acceleration. You can only draw two free body diagrams, one for the disk and one for the hanging mass.

I don't know the acceleration of the mass. I don't know the Tension. I don't know the angular acceleration of the thing that spins.

Isnt that 3? I am pretty sure angular acceleration is different from acceleration.

 

[kuruman]

Yes, angular acceleration is a different beast from linear acceleration, but they are related. That relation is a third equation. The linear acceleration all points on the string is the same and matches the linear or tangential acceleration of all points on the rim of the disk if there is no slipping. The tangential acceleration is related to the angular acceleration. Read about it here. That gives you the third equation.

 

[JoeyBob]

Yes, angular acceleration is a different beast from linear acceleration, but they are related. That relation is a third equation. The linear acceleration all points on the string is the same and matches the linear or tangential acceleration of all points on the rim of the disk if there is no slipping. The tangential acceleration is related to the angular acceleration. Read about it here. That gives you the third equation.

So angular accel=R*normal acceleration

Normal acceleration=T/m-g

and Torque=r*T=angular acceleration*moment of Inertia

Is this correct?

 

[kuruman]

Normal acceleration=T/m-g

This is correct if the mass is accelerating up. Here it is accelerating down.

 

[JoeyBob]

This is correct if the mass is accelerating up. Here it is accelerating down.

It'll be negative if it accelerates down. If up is positive and down is negative I don't see the issue.

 

[kuruman]

So which is correct?
a = T/m - g
or
-a = T/m - g

The issue is that the correct sign affects the answer when you combine it with the other equation.

 

[JoeyBob]

So which is correct?
a = T/m - g
or
-a = T/m - g

The issue is that the correct sign affects the answer when you combine it with the other equation.

Ill explain it this way.

If T/m > g, then acceleration is in the positive k hat direction.

If T/m<g, then acceleration is in the negative k hat direction.

The first equation accounts for both. I suppose you could make down positive and up negative, but I am more used to making down negative.