How Do Two Connected Discs with Masses Affect Angular Acceleration?

AI Thread Summary
The discussion revolves around a physics problem involving two connected discs with equal masses attached to strings. Participants express confusion about neglecting the discs' rotational inertia while solving for angular acceleration and the direction of rotation. The key equations derived include the relationship between tensions and accelerations, leading to the conclusion that the system accelerates to the right. Clarifications are made regarding the treatment of torque and the assumptions made in the problem statement. The conversation highlights the importance of understanding the implications of simplifying assumptions in physics problems.
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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

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


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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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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.
 
phinds said:
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.
 
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??
 
Bump, I uploaded the full problem in attachemtn
 

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lmao2plates said:
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.
 
phinds said:
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.
 
lmao2plates said:
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
haruspex said:
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
lmao2plates said:
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
haruspex said:
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
lmao2plates said:
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
lmao2plates said:
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
haruspex said:
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
lmao2plates said:
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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