Rotation, mass and velocity

Imagine a round turn table or one of those spinning tables at Las Vegas that they throw dice into. Now imagine this table has a groove or channel, I will call it a slot, deep enough to hold a marble. The slot runs in a straight line starting half way out from the center of the table to the edge circumference allowing the marble to roll away leaving the table, but first the marble is held in place in that slot while the table rotates at an edge circumference velocity of 1 meter a second. I press a button and the marble is released.

Is there a formula or method to figure out what direction the marble will end up leaving the table? Will it leave the table before 90 degrees of rotation or will it take 270 degrees? The velocity of the rotating table and the mass of the marble can vary.

Below is a drawing I made and some relevant information. Thanks.
M-long.jpg
kadinAsked:
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phoffricCommented:
Was this problem in your current text? Does it have an answer (or a solution)? If not, where did you find this problem? I am hoping that others will provide an answer so that we can compare results. If there is no corroboration, then unless you have the answer in your book, we have no way of knowing whether any of the posts are correct.

To do this problem with the approach I took, you need H.S. algebra and physics, college pre-algebra (including trigonometry), college calculus (for derivatives and vectors- I didn't use integrals), ordinary differential equations of the second order (i.e., two constants to be resolved using boundary conditions), and college Freshman physics.

You can learn all about these college subjects from Academic Earth, the MIT online website (where you can take the MIT famous 1999 physics class). For the H.S. math subjects, there are a number of online presentation classes that should prove invaluable presenting things in a logical order.
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TommySzalapskiCommented:
At the point the marble is released, it will have two independent (since they are orthogonal) vectors of velocity. It will have a velocity tangent to the disk at exactly 1 meter per second. It will also have a velocity going straight out from the disk at whatever speed the marble is moving away from the center.

The two velocities are completely independent of one another (except for the friction caused by the disk pushing on the marble) and can be computed separately.
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TommySzalapskiCommented:
The velocity going straight out from the center will be the same no matter how fast the disk is spinning (again, assuming that friction is negligable). You just add the two vectors together to get the final valocity.
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kadinAuthor Commented:
The velocity going straight out from the center will be the same no matter how fast the disk is spinning

If the velocity of the marble going straight out from the center is not effected by the rotation speed of the disk wouldn't the marble just sit in the slot without leaving the table? I am now thinking about those air force pilots experiencing g loads in those big rotation machines.
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TommySzalapskiCommented:
Right, sorry. It will actually be accelarating in the outward direction based on the cetripital force. What I meant was that you can pretend it's not moving and just use the cetripital force as if it were just a pushing force.
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kadinAuthor Commented:
Any idea how I can convert this into an equation to determine or choose at what point I want the marble to leave the table?

Maybe this equation is helpful. a=v^2/r  (a) is acceleration, (v) velocity and (r) radius. It looks like the radius and velocity of the marble changes as it nears the edge of the table.
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aburrCommented:
That is an excellent question. It is not simple. As TS pointed out, the velocity as it leaves the turntable is made up of two independent components,Tthe tangential component is easy. It is the velocity of the rim of the turntable. The radial velocity is another matter. The acceleration of the marble which leads to the radial velocity is not constant. Hence you will have to integrate. Unfortunately I do not have the time to do the integration just now but I hope to return to the problem in 24 hours.
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kadinAuthor Commented:
Thanks. Any help I can get to understand how to do this would be greatly appreciated.

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TommySzalapskiCommented:
Okay. I have some time to sit and look at the problem a bit more closely. Centripital force is a lot like a normal force in that it's opposing the natural tendency of the object to move. The forces in both directions obviously cancel one another out. So you actually see centrifugal force if you let your frame of reference be the rotation. (http://xkcd.com/123/) I will assume we are using such a system, and the edge of the disk where the marble comes off is at the origin. So x is the tangential axis and y goes to the center of the disk.
We don't want the force though, we want the acceleration. But you already have that anyway. It's v^2/r where v is the tangential velocity (the x velocity in our system).
Now, if the marble is released from rest at the edge of the circle it will travel only in the x direction because it had no time to accelerate so t is 0 and Vt = V0 which is of course 0 (in the y direction)..
Now the tangential velocity will not be constant in our problem since the marble is moving outward in the circle.
We really need v in terms of r.
v = 2*pi*r/T. You get get T from the radius of your disk. So your acceleration is
4pi^2/(rT^2). All you need to do is integrate that over r from the initial r to the final r to get the velocity along the y axis. Since pi and T are constants, the integral is easy.
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TommySzalapskiCommented:
Uh, oops 4*r*pi^2/T^2.
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thehagmanCommented:
In the rotating frame, the radial acceleration is a = omega^2*r, the force is F = m*a.
By moving outward from r to r+dr, the marbel gains energy F*dr = m*a*dr = m*omega^2*r*dr.
Hence at radius r the energy is (by integration) 1/2*m*omega^2*r^2 +E0 where E0 is chosen such that at the inner radius R1 the energy is 0, i.e. we have E(r) = 1/2*m*omega^2*(r^2 -R1^2).
From E=1/2 m v^2 we conclude that the speed is v(r) = omega * sqrt(r^2 -R1^2)
To solve dr/dt = omega * sqrt(r^2-R1^2) transform to
integral  dr/sqrt(r^2-R1^2) = integral omega dt
and solve for r(t)
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phoffricCommented:
Hello kadin,

  I've been recently trying to study rotational dynamics, so I'm taking a stab at this problem. To simplify matters, I assume that the marble is in a frictionless groove, and therefore does not roll, since rolling gives the marble some angular momentum; and that would make the equations quite a bit more complicated.

  I took a look at some previous questions that you asked; and I didn't see any questions with calculus. Some of the previous posts refer to integration and Work/Force/Energy discussions, and my approach was solving a differential equation. So, I'm not sure which of these methods you are most comfortable with.

  I admit that I'm a bit confused about the post that says "All you need to do is integrate that over r from the initial r to the final r to get the velocity along the y axis" since the usual form to compute v_final from v_initial is to integrate a(t) with respect to t, rather than r.

If acceleration were constant (and it is not), then we know that this integral would result in
   v_final = v_initial + a * delta_t
which is one of the formulas that you posted.

I admit that I'm also a bit confused about the post that says "By moving outward from r to r+dr, the marbel gains energy F*dr = m*a*dr". Usually the F*dr is in terms of dot products between the force vector, F, and the displacement vector dr. Reason for confusion is that the only force on the marble is a normal force of the groove; but dr has both a tangential and radial component; so I would have expected that the dot product would produce a different result.

Had I spent more time delving into those two posts, I might be able to unravel the confusion. But I didn't, because I was working the problem in yet another approach. (There are even more approaches that others may offer.) I used differential equations.

It would be interesting if others would confirm whether or not they get the same answers as below when using their approaches.

===========

Letting r(t) be the radial displacement, I came up with:
    r(t) = 0.25 * [ e^t + e^(-t) ]
Notice that
    r(0) = 0.25 * [ 1 + 1 ] = 1/2 meter (satisfies drawing)

We want to solve for the exit time, t_exit (i.e., when r = 1):
       1 = 0.25 * [ e^t_exit + e^(-t_exit) ]
  ==>  t_exit = ln [ 2 + sqrt(3) ] = 1.317 seconds

The radial speed, v_r, is just d/dt [ r(t) ] = r'(t) = 0.25 * [ e^t - e^(-t) ]
So, v_r_exit = 0.25 * [ e^(t_exit) - e^(-t_exit) ] meters/second

   The differential equation derivation is a bit involved. I didn't assume any centrifugal forces existed; and there certainly are no centripetal forces present (again, assuming frictionless groove).

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>> what direction the marble will end up leaving the table?

The velocity has the radial and tangential components.
You have v_tangent = 1 m/s (this is consistent with omega^2 R, since omega and R are both 1).

The speed at which the marble exits is then sqrt( v_tangent^2 + v_r_exit^2 )
The marble's angle relative to the tangent to the circle at exit is just
    angle = arctan[ magnitude(v_tangent) / magnitude(v_r_exit) ]

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>> Will it leave the table before 90 degrees of rotation or will it take 270 degrees?
v_tangent = 1 m/sec for R = 1 m

==> T = C = (2 pi) sec
and t_exit = 1.317 sec

The fraction of the circumference traversed when marble started its journey = t_exit/T = 1.317/(2pi) = 0.21
This corresponds to an angle of 0.21 * 360 = 75.6 degrees (i.e., less than 90 degrees of rotation).

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This result is independent of the mass of the marble. The heavier the mass, the more inertia it has; however, since you constrain the spin of the table to have a constant angular velocity of 1 radian per second (since v_tangent = 1, and R = 1), then the heavier the mass, the more energy it requires to move that mass. And so the equations balance out removing mass from the radial acceleration and exit time equations. But the heavier the mass, the larger will be the normal (i.e., perpendicular) force of the groove on the marble.

I hope I did this correctly.
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kadinAuthor Commented:
I did not realize this would be such a difficult problem to solve. It sounds like all three of you are discussing the use of calculus because I see the word integral in some of your statements. I just know some algebra so I don't fully understand your explanations. Does anyone know any books that could teach me how to solve such a problem?

These are the closest two I could find on Amazon but I'm not 100% sure what to look for.
700 Solved Problems In Vector Mechanics for Engineers Volume II: Dynamics
Joseph Shelley

Engineering Mechanics: Statics & Dynamics, 10th Edition
Russell C. Hibbeler
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TommySzalapskiCommented:
I just know some algebra
That's going to be a problem. Since the acceleration is not constant, there is no way to solve this without advanced forms of math (like calculus and differential equations which is taught after calc 3 at my university). So you are at least two semesters behind where you need to be to understand the math needed. I'm not sure what book would address this exact type of problem. It's not standard. Physics books will give you all the force equations, but you'll need integrals (or DEs) to calculate the final velocity.
All you need to do is integrate that over r from the initial r to the final r to get the velocity along the y axis
By that I meant integrate dr/dt. Saying "integrate over r" was a very bad way of putting it. Rereading it, it looks like I was saying to use something/dr which is silly. Since r is the only non constant in the equation, it's the important one in the integration but you're actually integrating r over time.

I didn't do any problems exactly like this in any of my math classes (I have a B.S. in math) so I'm just extrapolation from what I did learn.
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TommySzalapskiCommented:
extrapolating*
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phoffricCommented:
For a college physics text, here's one that was very popular
   http://cgi.ebay.com/University-physics-/310241727234?pt=US_Texbook_Education&hash=item483bd91302

I see that there are solution manuals for later editions.
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aburrCommented:
I do not have the solution yet. but some comments.
You will sooner of later have to integrate over r, that is dr, because there is no time in the problem.
You can use one of two frames of reference, fixed or rotating with the turntable. The later is simplest but it is an accelerating frame and you have to do with ficticious forces (rather easy)
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phoffricCommented:
I didn't do any integration. Just derived a differential equation and solved it. Constants determined from boundary conditions for r(t) and r'(t) at t=0.
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phoffricCommented:
I didn't use any ficticious forces (but I think I proved that you can use them).
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phoffricCommented:
I used thehagman approach, but instead of getting
     r(t) = 0.25 * [ e^t + e^(-t) ]
I got
    r(t) = 0.50 * [ e^t + e^(-t) ]

Maybe I got some constant wrong.

(Note that omega = 1)

@Tommy,
If you do not have time, could you just jot down the definite integral you are refering to. I got a little lost in your description. If you do this, I will try to compute the answer using your approach.

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kadinAuthor Commented:
Thanks to all of you for helping me on this.

The book I have Introductory Physics with Algebra Mastering Problem-Solving Stuart E. Loucks has a lot of good learning problems I have been studying. The books focus is on converting a word problem into an equation. This problem however I just thought up so I have no answer for it. With your help I have learned that because there is some variation over time I must learn some calculus first before approaching this kind of problem. I guess algebra can only go so far.

I ordered one book "The Calculus Direct: An intuitively Obvious Approach to a Basic Understanding of the Calculus for the Casual Observer" and over time I will likely be ordering two or three others that cover engineering math through calculus and physics.

Thanks Phoffric and TS for the info I need to know and where to get it. I will investigate your sources.
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phoffricCommented:
I think you should reconsider which books you order. There are excellent online college video courses from universities all over the world. If you choose a book that goes with the course, then you have the video lectures and the corresponding book that goes with the course. I have used this approach, and it works. But, when choosing a book, make sure that you are able to obtain a solutions book (not just answers) for a good number of the problems.
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kadinAuthor Commented:
Thanks. I will investigate the video courses and make sure I get a book that has solutions.
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