# Homework Help: Acceleration of mass moving along a radial track.

1. Feb 9, 2014

### peripatein

Hi,
1. The problem statement, all variables and given/known data
I am trying to determine the magnitude of the acceleration, registered by an observer in the lab, of mass m forced to move along a radial track, immediately as it begins moving along the track. Please see attachment. ω=αt and the entire set up is horizontal. The initial position of the mass is marked as r0 and it is stated that only static friction is present along the track, with coefficient μ, whereas kinetic friction is null.

2. Relevant equations

3. The attempt at a solution
As there is no friction once the mass begins moving, there are no forces in action (coriolis will not have an effect as the mass couldn't move in any direction but the radial). Hence, its acceleration, as registered by an observer in the lab, will be zero. Is this correct?

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2. Feb 9, 2014

### Staff: Mentor

Let me understand the problem you are posing. You have a man on a rotating table moving radially outward on the table. Correct?

Chet

3. Feb 9, 2014

### peripatein

Not at all :-). On a rotating horizontal platform (whose angular velocity changes in time) there is a mass forced to move radially along a track. I, the observer, am trying to determine the magnitude of its acceleration as it begins moving from its initial position r0 on the track. There is no kinetic friction and static friction's coefficient is given as mu.

4. Feb 9, 2014

### Staff: Mentor

Gotcha. OK. Is the equation you gave for the angular velocity correct, so that the initial angular velocity is zero, or is there an initial angular velocity? I'm trying to understand the kinematics of the motion, since this should be the first step in determining the acceleration.

chet

5. Feb 9, 2014

### peripatein

At t=0 the mass doesn't move and I am asked to determine the acceleration immediately after it does begin to move along the track. I am quite sure the magnitude of the acceleration as noted by the observer in the lab should be zero. Is this incorrect?

6. Feb 9, 2014

### BvU

That there are no forces in action is a misconception. The constraint forces are very much in action !

7. Feb 9, 2014

### peripatein

But the mass is forced to move along the track so the forces in action could not affect its acceleration.

8. Feb 9, 2014

### Staff: Mentor

It depends on whether it is rotating initially. Let's try this a little differently. I'm going to write down a couple of equations for the kinematics of the motion as reckoned from a non-rotating frame, and you correct them if they are wrong.

r = r0+ vrt

θ = θ00t + αt2/2

These are the polar coordinates of the mass at time t.

Corrections?

9. Feb 10, 2014

### peripatein

Let's examine the forces rather. What forces are in action here? Once the mass begins moving there is no friction (kinetic friction is zero, so it is stated); the normal from the platform and the weight cancel themselves out; coriolis would have to be perpendicular to r in this case and hence would not carry any effect on the mass' acceleration which is limited to the radial track. Ergo, there are no forces acting on the mass as it begins moving along the track.

10. Feb 10, 2014

### voko

If the were no forces, the mass would be moving in a straight line. But that is clearly not the case here.

11. Feb 10, 2014

### peripatein

The mass is moving in a straight line along the designated track. It couldn't derail from the track.

12. Feb 10, 2014

### voko

That is not an inertial frame. In an inertial frame, the mass does not move in a straight line.

13. Feb 10, 2014

### BvU

I'm getting intrigued by this one: It is one of the simple examples in Goldstein's classical mechanics chapter 1, with a presentation that is aimed at making things as easy as possible. So it could just as well be part of an introduction to rotational motion. Here's my two cents worth input: shoot at it if you disagree:

Can be interpreted as "There are no forces along the track as the mass begins moving", which is correct.
Can also be interpreted as "There are no forces on the mass at the moment it begins moving, with the motion being along the track", which is wrong. Was already pointed out a few times, but perhaps hasn't penetrated yet.

Earlier in post #5:
Looks as if you are jumping to conclusions and don't want to question them, nor the considerations that bring you to these conclusions. On two counts: dismissing coriolis (which I do think is correct at t=0, but only because the bead is not moving in the rotating frame of reference. It definitely can't be dismissed once the bead is moving) and thinking the acceleration in the lab should have magnitude zero.
Especially the latter is at least questionable to me: the bead is going to fly outwards, real fast. If the kinetic energy increases, something has to deliver the work. I will require energy to maintain $\omega$. A lot, I would expect.

Perhaps if you work out things, some of the premature conclusions may appear as a result, but you can't be certain until you sniffed at it from all sides.

From the description of the OP in a few installments:
• Bead describes circular trajectory in the lab up to t=0: r(t) = r0 $t\le 0$.
• Centripetal force/acceleration is provided by friction $F = \mu N$ so $\ddot{\vec r} = -\omega^2(t) r_0\ \hat r$
• At t=0 friction goes to 0 instantaneously. Clearly the opposite of the centripetal force now governs the motion in the rotating frame of reference and we have $\ddot{\vec r} = \omega_0^2 r_0\ \hat r$.
Next thing to do: transform this back to the laboratory frame of reference (if that is really desired -- the rendering of the OP in post #1 would suggest a yes).

Last edited: Feb 10, 2014
14. Feb 10, 2014

### Staff: Mentor

You are asking us for help. Three people with lots of experience (Voko, BuV, and myself) are pleading with you to look at the kinematics first. I know you would rather look at the forces first, but please reconsider. In our judgement, your understanding will solidify much more rapidly once you look at the kinematics (from the perspective of the laboratory inertial frame of reference). I hope you will reconsider.

Chet