Tangential movement in a waterslide

[Fibo112]

Homework Statement

A person enters a curve of a waterslide with velocity v. The curve is fully horizontal and has the shape of a circle segment of radius R, there is no friction and width of the curve can be neglected for calculating the centripedal force. The waterslide itself can be described as a semi circle with radius r. The question is to describe the tangential equation of motion for the angle a the person in the slide. We must than linearise this equation of motion with the small angle approx for sin and cos.

Homework Equations

F(centripedal)=mv^2/r

The Attempt at a Solution

The only forces acting are g and the normal force. Since the person is traveling in a circle of radius R with speed v there must be a centripedal force of mv^2/R (problem: the motion is not really circular since the person also has tangential movement in the slide). This centripedal force is horizontal so it cannot stem from gm -->it must come from the normal force. The horizontal component of the normal force is sin(a) N= aN=mv^2/R. The remaining force acting on the person is the vertical componenet of N and g =( mv^2/(Ra))-gm. The tangential component of this force is (mv^2/Ra)-g)sin a=(mv^2/Ra)-gm)a). This component is equal to a''mr. This gives a''=v^2/Rr-ga/r.

 

[Hiero]

Homework Equations

F(centripedal)=mv^2/r

It is not quite as simple as circular motion. (This equation isn't immediately useful if we don't know the radius of curvature of the motion.)

I would like to guide you based off your attempt, but I don't think I understand it. You seem to be taking the tangential component of the vertical component of the net force and equating it with mra'' ... Did you mean to take the tangential-component of the entire force instead of just a component of a component?

Assuming that is what you meant, it is still wrong to equate this with mra'' ... That would be the tangent-acceleration if the mass were oscillating in the slide (r) but not moving around the curve (R) ... but when it is also moving around the curve, things are not the same.A safe approach would be to write the position of the mass in terms of some (inertial) coordinates and differentiate twice in time to find the acceleration.
An example of what I mean;
Position during circular motion can be described by the coordinates as R=<Rcos(θ), Rsin(θ)> so when we differentiate twice we get,
R''=<-θ'²Rcos(θ), -θ'²Rsin(θ)> = -θ'²R = -ω²R, the familiar centripetal-acceleration result.

 

[Fibo112]

It is not quite as simple as circular motion. (This equation isn't immediately useful if we don't know the radius of curvature of the motion.)

I would like to guide you based off your attempt, but I don't think I understand it. You seem to be taking the tangential component of the vertical component of the net force and equating it with mra'' ... Did you mean to take the tangential-component of the entire force instead of just a component of a component?

Assuming that is what you meant, it is still wrong to equate this with mra'' ... That would be the tangent-acceleration if the mass were oscillating in the slide (r) but not moving around the curve (R) ... but when it is also moving around the curve, things are not the same.A safe approach would be to write the position of the mass in terms of some (inertial) coordinates and differentiate twice in time to find the acceleration.
An example of what I mean;
Position during circular motion can be described by the coordinates as R=<Rcos(θ), Rsin(θ)> so when we differentiate twice we get,
R''=<-θ'²Rcos(θ), -θ'²Rsin(θ)> = -θ'²R = -ω²R, the familiar centripetal-acceleration result.

Thanks for the reply, I'm still stuck though :/

 

[Hiero]

Thanks for the reply, I'm still stuck though :/

You want the general equations of motion for when the angle a can change in time right?

The particle is restricted to move on (part of) the surface of a torus (donut shape). I am suggesting you describe the cartesian coordinates of a point on this surface in terms of two angles (a and θ) analogous to my example:

An example of what I mean;
Position during circular motion can be described by the coordinates as R=<Rcos(θ), Rsin(θ)> so when we differentiate twice we get,
R''=<-θ'²Rcos(θ), -θ'²Rsin(θ)> = -θ'²R = -ω²R, the familiar centripetal-acceleration result.

If you paramerterize it correctly then finding equations of motion should just be a matter of carefully differentiating (and recognizing how v relates to θ').

If you don't understand what I am trying to say still then I apologize, but you'll have to explain your attempts more clearly if I am to help you more.

 

[haruspex]

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Thanks for the reply, I'm still stuck though :/

You need another relationship between v and a. What's conserved?

 

[haruspex]

The horizontal component of the normal force is sin(a) N

You seem to be taking a as angle to the vertical (consistently). To use the small angle approximations you need it to be a deviation from an average angle.

 

[Fibo112]

.

You need another relationship between v and a. What's conserved?

If the mechanical energy is conserved I would have another relationship since I know the relationship between the potential energy and the angle a. Is the mechanical energy conserved though? It would be if the normal force was always perpendicular to v..

 

[haruspex]

It would be if the normal force was always perpendicular to v..

Does the normal force do work?

 

[Fibo112]

Does the normal force do work?

With circular motion in a plane I know it doesn't since it is perpendicular to v, in 3d I'm not sure

 

[haruspex]

With circular motion in a plane I know it doesn't since it is perpendicular to v, in 3d I'm not sure

A force only does work if there is motion parallel to it. I think you are getting confused with the constant speed condition.

 

[Fibo112]

A force only does work if there is motion parallel to it. I think you are getting confused with the constant speed condition.

Yeah that's why I said it doesn't do work since it is perpendicular?

 

[haruspex]

Yeah that's why I said it doesn't do work since it is perpendicular?

Yes, sorry. Anyway, work is conserved here.

 

[Fibo112]

Now I have this relationship.

 

[haruspex]

Now I have this relationship.

Sin(a) for PE?

 

[Fibo112]

Sin(a) for PE?

I thought sin(a)r is the height and then I just multiply that with mg?

 

[haruspex]

I thought sin(a)r is the height

Think again. E.g. consider 45 degrees.

 

[Fibo112]

Think again. E.g. consider 45 degrees.

Ups, PE= r(1-cos a)mg

 

[Fibo112]

I still don't know how to continue though...

 

[Fibo112]

At this point I would really appreciate it if someone could just give me a solution...

 

[haruspex]

I still don't know how to continue thoough...

Don't differentiate. Just use the energy expression to substitute for v in your acceleration equation.

At this point I would really appreciate it if someone could just give me a solution...

Not allowed on homework forums.
As I noted, you cannot use the small angle approximation for a. It will not generally be that small. Instead, find the equilibrium value for a and make the small angle approximation for deviations from there.

 

[Fibo112]

Don't differentiate. Just use the energy expression to substitute for v in your acceleration equation.

Not allowed on homework forums.
As I noted, you cannot use the small angle approximation for a. It will not generally be that small. Instead, find the equilibrium value for a and make the small angle approximation for deviations from there.

Which acceleration equation are you referring to?

 

[haruspex]

Which acceleration equation are you referring to?

The last part of your post #1, $\ddot a=$...

 

[Fibo112]

The last part of your post #1, $\ddot a=$...

But how is that equation valid? Didn't I make mistakes in it's derivation?

 

[Fibo112]

The solution given is the same as mine in post#1, but I know the way I got there is very wrong...can someone tell me how they would solve the problem?

 

[haruspex]

But how is that equation valid? Didn't I make mistakes in it's derivation?

Yes, but obviously I mean the corrected version of that equation.
Write the equations without making a small angle approximation. (You can still make the approximation that velocity away from the horizontal can be ignored in the energy equation.) Can you find the equilibrium value of a?

The solution given is the same as mine in post#1, but I know the way I got there is very wrong...can someone tell me how they would solve the problem?

Do you mean the method is the same or the answer? Please quote the given answer. I might disagree.

 

[Fibo112]

Yes, but obviously I mean the corrected version of that equation.
Write the equations without making a small angle approximation. (You can still make the approximation that velocity away from the horizontal can be ignored in the energy equation.) Can you find the equilibrium value of a?

Do you mean the method is the same or the answer? Please quote the given answer. I might disagree.

Only the answer.

 

[Fibo112]

Yes, but obviously I mean the corrected version of that equation.
Write the equations without making a small angle approximation. (You can still make the approximation that velocity away from the horizontal can be ignored in the energy equation.) Can you find the equilibrium value of a?

Do you mean the method is the same or the answer? Please quote the given answer. I might disagree.

Why can velocity away from the horizontal be ignored?

 

[Fibo112]

I think this is the equillibrium a

 

[haruspex]

Why can velocity away from the horizontal be ignored?

Snce you are told to make small angle approximations, we must only be interested in small peturbations. That means any velocity away from the horizontal will be small compared with velocity in the horizontal.
If you prefer to develop, first, a completely accurate equation, before making any approximations, I would endorse that. It is a very sound way to proceed.

 

[haruspex]

I think this is the equillibrium a

Yes. Can you write your equations in terms of that angle plus a perturbation, then make the small angle approximation?