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Rotational motion problem

  • #1

Homework Statement


Two discs of radius R and r are fixed to each other, i.e., they rotate
together. Strings are wound around both discs and two equal masses M
are connected to the ends of the strings (see Figure 1). Find the angular
accelerations of the discs, the accelerations of the masses and indicate
the direction of rotation of the discs (neglect the rotational inertia of the
discs).

http://www.physics.louisville.edu/wkomp/teaching/summer2005/p298/quizzes/quiz4.pdf [Broken]

The diagram looks like question 3 from above webpage except with the masses being equal.

Homework Equations


at=αr
F=ma


The Attempt at a Solution


[/B]
The problem I have while attempting a solution is that the question tells me to neglect the rotational inertia (moment of inertia). However, if we consider the moment of inertia, it is obvious that the system will rotate to the right since the smaller wheel has a smaller moment of inertia.
 
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Answers and Replies

  • #2
phinds
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Why do you have a problem with that? They are just saying to assume the disks have no mass. That kind of assumption if very common in beginning instruction problems such as this.
 
  • #3
Why do you have a problem with that? They are just saying to assume the disks have no mass. That kind of assumption if very common in beginning instruction problems such as this.
So if I assume that the discs have no mass, we are also neglecting the gravity? I think I am missing something here.
I just find it difficult when a question does not give me any quantitative value to work with.

I appreciate your help.
 
  • #4
bump

Here's my working so far:

Angular acceleration is the same for both discs = α
=> acceleration of mass 1 = αR and mass 2 = αr

I then drew free body diagrams of the 2 masses with mass 1 on the left and mass 2 on the right.

Fnet=ma

T1 - Mg = M(a1) - eq.1

T2 - Mg = M(a2) - eq.2

eq1-eq2

T1 - T2 = M(a1-a2) = M(αR-αr)

since R>r, (αR-αr) is > 0 => T1-T2 > 0

therefore T2 < T1 so the system is accelerating to the right.

I am not really sure on this solution.

Could a helper possibly review this for me and suggest the correct approach??
 
  • #5
Bump, I uploaded the full problem in attachemtn
 

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Bumping is not allowed on PF.
 
  • #7
phinds
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So if I assume that the discs have no mass, we are also neglecting the gravity?
No, I don't understand why you would think that. Look, if it makes you happy, then DO consider the mass of the disks and DO take rotational inertia into account. Ignore the fact that the problem statement doesn't tell you the mass of the disks and specifically tells you not to take it into account. I was just trying to help you understand why the problem was stated as it was stated.
 
  • #8
No, I don't understand why you would think that. Look, if it makes you happy, then DO consider the mass of the disks and DO take rotational inertia into account. Ignore the fact that the problem statement doesn't tell you the mass of the disks and specifically tells you not to take it into account. I was just trying to help you understand why the problem was stated as it was stated.
Yea I understand what you mean now. Sorry about that.

The problem I am really facing with this problem as evident by my attempt at the solution in post number 4. The question asks to find the angular acceleration, do I just give it as α? or do I derive that from other variables stated in the problems? There aren't many to begin with either.

Thank you.
 
  • #9
haruspex
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do I derive that from other variables stated in the problems?
Yes.
Consider the FBD of the pair of discs. What torque v. acceleration equation can you write there?
 
  • #10
Yes.
Consider the FBD of the pair of discs. What torque v. acceleration equation can you write there?
Okay. Torque = Iα. However since we are neglecting the moment of inertia of the discs, torque would be 0?
 
  • #11
haruspex
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Okay. Torque = Iα. However since we are neglecting the moment of inertia of the discs, torque would be 0?
Yes. So what is the net torque in terms of the tensions?
 
  • #12
Yes. So what is the net torque in terms of the tensions?
net torque = radius * tension

=> 0 = T1R - T2r
=> T1R = T2r
=> T1 = T2(r/R)

so T2(r/R) -T2 = M(αR-αr)
T2(r/R-1) = M(αR-αr)

It seems like I am going in circles haha.
 
  • #13
net torque = radius * tension

=> 0 = T1R - T2r
=> T1R = T2r
=> T1 = T2(r/R)

so T2(r/R) -T2 = M(αR-αr)
T2(r/R-1) = M(αR-αr)

It seems like I am going in circles haha.
EDIT: so from there, assuming that the above equation is valid,

α = T2(r/R-1)/M
 
  • #14
haruspex
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net torque = radius * tension

=> 0 = T1R - T2r
=> T1R = T2r
Yes. You have your original eq 1 and eq 2, plus a relationship between a1 and a2 that you used without stating. You now have four equations and four unknowns.
 
  • #15
Yes. You have your original eq 1 and eq 2, plus a relationship between a1 and a2 that you used without stating. You now have four equations and four unknowns.
Okay so now I have
T1 - Mg = M(a1)

T2 - Mg = M(a2)

a1/R = a2/r

T1 = T2(r/R)
 
  • #16
haruspex
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Okay so now I have
T1 - Mg = M(a1)

T2 - Mg = M(a2)

a1/R = a2/r

T1 = T2(r/R)
Not quite. Your eqn 1 and eqn 2 are only correct if you are taking up as positive everywhere, including the accelerations. The third equation above therefore has both masses accelerating the same way.
 

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