# Rotational Dynamics Question

1. Nov 30, 2015

### Dusty912

1. The problem statement, all variables and given/known data
I attached the stated questions. Mainly need help with question 2. See below for my answer to question 1.

2. Relevant equations
ΣT=Iα
ΣF=ma
moment of inertia of sphere: I=(2mR^2)/5
moment of inertia of disk: I=(mr^2)/2
v=rω
V(center of mas)=V(tangential) -because of smooth rolling
ΔΘ=ωt+(1/2)αt^2
ω(final)=ω(initial)+αt
ω^2(final)=ω^2(initial) +2αΔΘ
3. The attempt at a solution
1. C) 4F1 -I figure the lever arm is 1/4 as big for the smaller disk so it would require a force 4 times as big to counteract the force applied to the larger disk.

2. a) I know how to draw the FBD for this.
B) what I need the most help with. It has been a while since we have done this section and I had trouble with it the first time. But I am assuming we use the analogy for Newton's second law ΣT=Iα and the sum the torques. But do I do this for the sphere and the the disk together? not really sure. My guess is that it would look something like this. T1=the tension in the rope tangential to the sphere and disk.
T2=the tension in the rope tangential to the disk and connected to the block.
So summing the torques for the sphere: T1R=(2/5)mR^2α -the radius would be of the sphere of course

Summing of the torques for the disk: T2r=(1/2)mr^2α

and summing of the forces for the block would be T2-mg=ma (the m's would be the mass of the block)
from here I'm stuck. Sorry I couldn't show more work but like I said I really need help with this one.

Part (c) should be pretty easy once I know the acceleration of the black from part B

#### Attached Files:

• ###### Quiz Rotational Dynamics TAKE HOME FA15.pdf
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2. Nov 30, 2015

### J Hann

Basically, there are two tensions in the string.
One tension accelerates the sphere.
The other tension is between the disk and the mass.
The difference of these tensions accelerates the disk.

3. Nov 30, 2015

### Dusty912

Okay that somewhat helped. But I'm not really sure what I should do with that information. Could you elaborate? or anyone else care to add on?

4. Nov 30, 2015

### Dusty912

okay so this is what I have so far: I summed the forces for the block resulting in T2=m2g-m2a
then I summed the torques for the sphere resulting in: T1R=(2/5)MR^2α then I used the relationship between tangential acceleration and alpha at=Rα and subbed in α resulting in T1=(2/5)Mat then I summed the torques for the disk resulting in : T1r-T2r=Iα then subbing in α and the moment of inertia for a disk and canceling all of the radii yields T1-T2=(1/2)m1at subbing in T1 and T2 and solving for at yields
at=(m2g)/((2/5)Mat+m2-(1/2)m1)
which gave me 2.58m/s

Am I in the ballpark?

oh and by the way I'm now using the variables from the attached document, sorry for the confusion

5. Nov 30, 2015

### haruspex

Just to clarify, those two uses of alpha are for different angular accelerations, yes?

6. Nov 30, 2015

### Dusty912

yes they are, sorry knew I should have clarified that

7. Nov 30, 2015

### haruspex

Ok.
Your equation for at in post #4 doesn't look right. Does it make sense that increasing the mass of the disk would increase the acceleration?
Even with the equation as you have it, your final numerical answer looks about ten times too large.

8. Dec 1, 2015

### Dusty912

okay so are you saying I did something wrong in my algebra?

9. Dec 1, 2015

### Dusty912

wouldn't increasing the mass of the disk make the acceleration smaller? just for clarification m2 is the block and m1 is the black

10. Dec 1, 2015

### haruspex

Quite so, but your equation says otherwise. Check the signs. If you cannot spot the error, post all your steps.
(The black?)

11. Dec 1, 2015

### Dusty912

sorry m1 is the disk, and im looking at it now. So I'm on the right track then?

12. Dec 1, 2015

### haruspex

Yes, there is just the one error in the equation, and it looks like you had an arithmetic error later too.

13. Dec 1, 2015

### Dusty912

Okay so I'm still getting the same answer here are my steps:
1.I summed the forces for the block and solved for T2 T2=m2g-m2a
2.then I summed the torques for the sphere and solved for T1 and subbed in a=Rα for α. The R's cancel and yield T1=(2/5)Ma
3.then I sum the torques for the disk and sub in a=rα for α. That yields: T1-T2=(1/2)m1a
4. then I substitute T1 and T2 into the disk torque summation which yields: (2/5) Mat-m2g +m2at=(1/2)m1at
5. I move the terms with at onto one side and factor out at yielding: at((2/5)M +m2-(1/2)m1)=m2g
6. then I solve for at yielding at=(m2g)/((2/5)M+m2-(1/2)m1)

14. Dec 1, 2015

### haruspex

Which ways do T1 and T2 act? Which way does the disk accelerate?

15. Dec 1, 2015

### Dusty912

oh I see so:
at=(m2g)/((1/2)m1 +m2 +(2/5)M)

?

16. Dec 1, 2015

### haruspex

Yes.

17. Dec 1, 2015

### Dusty912

okay thank you and as far as question #1 goes would C) 4F1 be correct?

18. Dec 1, 2015

### Dusty912

okay thank you and as far as question #1 goes would C) 4F1 be correct?

19. Dec 1, 2015

### haruspex

Yes.

20. Dec 1, 2015

### Dusty912

So I already turned in my quiz, but I was a little curious about the radii canceling out. Since that was information given to me, I figured Id have to use them.

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