Circular Motion: Find an Expression for Mass M

[GeneralOJB]

Homework Statement

The diagram shows a smooth thin tube through which passes a string with masses m and M attached to its ends. The tube is moved so that the mass m travels in a horizontal circle of radius r at constant speed v

http://quickpic.info/z/yb.jpg

Find an expression for M.

Homework Equations

F = \dfrac{mv^2} r

The Attempt at a Solution

The string will be slanting down slightly to provide a vertical component of tension to keep the mass m in a horizontal circle. Let \theta be the angle that the string makes to the vertical.

Then T \sin{\theta} = mg and T \cos{\theta} = \dfrac {mv^2} r

So T = \sqrt{(mg)^2 + \left(\dfrac {mv^2} r \right)^2 }

We are told the radius is constant, which happens if the bottom mass M is in equilibrium, so T = Mg.

So M = m \sqrt{\left(\dfrac{v^2} {r g}\right)^2 + 1}

 

[Nathanael]

We are told the radius is constant, which happens if the bottom mass M is in equilibrium,

Mass M only needs to be in vertical equilibrium, right? Isn't it possible that mass M is moving in a horizontal circle?

(I'm actually not sure what would happen, I've never encountered a problem like this.)

If not though, then everything you did seems correct.

 

[GeneralOJB]

Mass M only needs to be in vertical equilibrium, right? Isn't it possible that mass M is moving in a horizontal circle?

(I'm actually not sure what would happen, I've never encountered a problem like this.)

If not though, then everything you did seems correct.

Yes, only in vertical equilibrium. Maybe M could be moving in a horizontal circle, though we aren't told that it is.

 

[Nathanael]

I just brought that possibility up because I assumed your answer was wrong (since you posted this question) but I suppose you could have posted this to get confirmation on your answer (if you don't have a solutions manual).


The problem is tricky to me because of this part:

The tube is moved so that the mass m travels in a horizontal circle of radius r at constant speed v

To me, that seems to imply there is an outside force (I imagine someone's hand) moving the tube in small circles. How this effects the tension is a little mysterious to me. (Is the outside force causing the centripetal force, or does the tension cause it? How does this outside force create more tension?)


But from all the information given (and the diagram in which mass M is straight down) I would say that your answer is likely correct.

 

[GeneralOJB]

What's interesting is the "correct" answer is \dfrac{mv^2}{rg}, according to the exam board AQA.

 

[AlephZero]

If you igmore air resistance, there is no need to move the tube in small circles to keep the motion going. Of course there must be some forces acting on the tube to keep it in place but they are not relevant to the question.

I think you can assume the mass M is not rotating from the way the diagram is drawn. If it was rotating, there is no reason why it has to rotate at the same angular velocity as mass m, and there are too many unknowns to get a unique solution.

 

[Nathanael]

What's interesting is the "correct" answer is \dfrac{mv^2}{rg}, according to the exam board AQA.

I'm thinking they made a mistake... Because that's the answer you would get if you neglected to consider that the string would be slanted down a bit.
Edit:

If you igmore air resistance, there is no need to move the tube in small circles to keep the motion going. Of course there must be some forces acting on the tube to keep it in place but they are not relevant to the question.

Good point

 

[GeneralOJB]

I'm thinking they made a mistake... Because that's the answer you would get if you neglected to consider that the string would be slanted down a bit.

Indeed. If it is a mistake, then it's shocking that it made its way to an A level exam.

Source: http://filestore.aqa.org.uk/subjects/AQA-PHYA4-1-QP-JUN12.PDF (question 6)

 

[AlephZero]

What's interesting is the "correct" answer is \dfrac{mv^2}{rg}, according to the exam board AQA.

They assumed that $\theta = 0$, which is how the diagram is drawn.

In real life engineering situations with rotating machines, the centripetal accelerations are often hundreds or thousands of times bigger than g, so that would be a good approximation. But IMO your solution is "better" than the one given by AQA, and if this was in a real exam I hope you would have got full marks for it!

Edit: I didn't know it was a multiple choice question when I wrote that. Still, you should be able to see which answer to pick.

Actually, you can find the right answer just by looking at the dimensions of the four choices, without doing any algebra at all. Only one of them has the dimension of "mass".

 

[GeneralOJB]

They assumed that $\theta = 0$, which is how the diagram is drawn.

In real life engineering situations with rotating machines, the centripetal accelerations are often hundreds or thousands of times bigger than g, so that would be a good approximation. But IMO your solution is "better" than the one given by AQA, and if this was in a real exam I hope you would have got full marks for it!

Edit: I didn't know it was a multiple choice question when I wrote that. Still, you should be able to see which answer to pick.

Actually, you can find the right answer just by looking at the dimensions of the four choices, without doing any algebra at all. Only one of them has the dimension of "mass".

Yes, I would have picked that in the exam, but the question is still wrong. It can't be exactly horizontal if there is no vertical force to balance gravity.

 

[mafagafo]

AlephZero
Same here, they could have at least put 2 "mass" alternatives.

GeneralOJB
I agree with you.