Homework Help: Challenging problem about puck revolving on a table

1. Jul 16, 2009

go_ducks

1. The problem statement, all variables and given/known data

An air puck is rotating on a table top, attached to a hanging mass by a massless cord which passes through a hole in the centre of the table. Given variables are, radius R = 0.9 m. mass(puck) = 0.28 kg. mass(hanging mass) = 1.25 kg. Question is:

a) tension in cord
b) centripetal acceleration of puck
c) speed of puck
d) some of hanging mass is removed, so that its new mass is 1kg. Find the new radius and new speed. (hint : angular velocity is not constant)

(special note: not allowed to use conservation of angular momentum or energy concepts to arrive at solution - this material is not yet covered in class! i believe that linear momentum is also not allowed)

2. Relevant equations

F = mv2/R
v = 2 pi *r/t (perhaps)
think thats it.. newtons law and the kinematic equations are also ok

3. The attempt at a solution

First dont bother solve a,b,c except for your own sake, I can do those parts easily and I have found the question about 30 times on google up to part c.

Its part d which is blowing my mind, which is when you reduce the tension in the string and the puck moves out into a new radius of orbit with a new velocity and also a new angular velocity. Since angular velocity is non-constant then neither is the period. So basically I can't find any constants. I know you want to say "just use linear momentum, what are you doing" but the professor requires that they can't be used. Momentum and energy concepts are both out of the question.

Well I have been trying to tackle it three ways.

First way. Say T = mv^2/R ..which it is.. then if i multiply by mr^2/mr^2 i get T = (mvr)^2 / mr^3 which looks promising because mvr is conserved. Unfortunately I am not allowed to use this fact! So I have been trying to finish this way by showing d(mvr)/dt = 0 from the equations of motion in a circle to prove it . I have not been unable to do this successfully even though it seems like it should be doable since it is a fundamental fact.

Second way. Try to find T(r). i.e. find T in terms of only the value r. However I am unable to ever get rid of a v without introducing an w, another non-constant. I've also investigated dT/dR but this hasnt been fruitful at all - v depends on r somehow. I choose T = mv^2/r and get dT/dr = -mv^2)/r^2 * 2v *dv/dr It doesnt' look like it'll improve that way. Am I doing this derivative correctly?

Third way. (Actually this is the first thing I tried) Try starting from the hint which is angular velocity is not constant, and perhaps try to solve dw/dr and see what happens. w = v/r so dw/dr = -v/r^2 * dv/dr ... which is to say dw/dr = - a dv/dr. ... or dw/dr = - aw .. but an integral from 0.9 to R2 wont yield the answer since w2 is also unknown and contains v/r. I considered maybe some clever introduction of quantities leads from this to the conservation of angular momentum conditon dmvr/dt = 0, but I couldnt find a path. I apologise if this explanation was confusing to read.

This problem is to be graded, so please, gentle guidance only. Its also a very good problem so please dont spoil it with something kids will end up googling for. As you can see I have been really trying hard on this problem and at this point , I've started to wonder if I can actually solve it. The question was originally given on a mid-term so I feel I may be missing some vital clue (but that should just be the hint, right?).

Last edited: Jul 16, 2009
2. Jul 16, 2009

queenofbabes

Well, the mass moves to a new radius, but won't it stay constant then?

3. Jul 16, 2009

go_ducks

Yes after it moves. the question is find the new radius and new speed. I looked back over what I wrote here for method 3 and i put aw which isnt right but no worries, it still didnt go anywhere..

4. Jul 16, 2009

turin

Well, not exactly. But, as per your request, I will not tell you what equation to replace this with. Anyway, from the rest of your post, I will assume that this is only a typo (or brain fart).

I suspect that this is the root of your problem. Imagine that you remove the .25 kg, then hold the hanging 1 kg at its eq position, and then let go and allow the system to do its thing. (I mention the second step because, in general, the process of removing the .25 kg leads to oscillations, but I think you are supposed to assume that the 1 kg does not oscillate for this problem.) If the 1 kg is released from its eq position, then what can you say about the motion of the 1 kg? Then, what can you say about the length of string connecting the puck to the hole? Then, what can you say about the radius?

This is true if v and R are constant, but not true in general. However, see my previous comments.

Hint: if you can do parts a,b,c, and you can figure out what I'm talking about, then you can do part d.

Last edited: Jul 16, 2009
5. Jul 16, 2009

go_ducks

I think you were referring to v = rw which I have used subsequently.

Other than that, definitely still not getting it. It already seemed fairly self-evident that the radius of orbit should increase by the same amount that the hanging mass is raised by, but the professor said we weren't allowed to involve mg*dh... or any other energy concept in the calculation. I will try to figure out why you brought that thing about T up.

6. Jul 17, 2009

turin

Actually, now I'm scratching my head, too. I'll see if I can enlist some big guns to help us figure this out.

7. Jul 17, 2009

tiny-tim

Hi go_ducks!

(have a pi: π and an omega: ω and try using the X2 tag just above the Reply box )
Hint: r = rê, where ê is the radial unit vector,

so r'' = r''ê + 2r'ê' + rê''