Learn the techniques to avoid forgery and phishing attacks and the types of attacks an application or network may face.

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

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

Experts Exchange Solution brought to you by

Enjoy your complimentary solution view.

Get every solution instantly with Premium.
Start your 7-day free trial.

I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

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.

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.

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.

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)

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

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

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).

---

>> 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) ]

-----

>> 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).

-----

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.

These are the closest two I could find on Amazon but I'm not 100% sure what to look for.

Joseph Shelley

Russell C. Hibbeler

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.

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.

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.

Experts Exchange Solution brought to you by

Your issues matter to us.

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Start your 7-day free trialhttp://cgi.ebay.com/University-physics-/310241727234?pt=US_Texbook_Education&hash=item483bd91302

I see that there are solution manuals for later editions.

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)

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.

The book I have

I ordered one book

Thanks Phoffric and TS for the info I need to know and where to get it. I will investigate your sources.

Math / Science

From novice to tech pro — start learning today.

Experts Exchange Solution brought to you by

Enjoy your complimentary solution view.

Get every solution instantly with Premium.
Start your 7-day free trial.

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.