# Curved motion

1. Dec 26, 2004

### Clari

A particle m is moving in a vertical circle of radius R inside a track. There is no friction. When m is at its lowest position, its speed is v0.
Suppose v0 is 0.775 vmin [ vmin = root(5gR) ]. The particle will move up the track to some point at P at which it will lose contact with the track and travel along a path roughly like a line connected to the centre of the circle. Find the angular position theta of point P.

I solved the question by conservation of energy. energy at the bottom = 1/2 mv^2............1/2m (0.775)(5gR)
energy at the top, when the particle is momentarily at rest.....mgh.......

1/2m(0.775^2)(5gR) = mgh
h= 1/2(0.775^2)(5R)
h is less than 2R, so h-R will give the height of a triangle form , where theta is the angle in the triangle....

(h-R) = R sin(theta).........and I get theta = 30 degrees......but the answer is 19.5 degrees.....what's wrong with my steps? please tell me....

2. Dec 26, 2004

### Staff: Mentor

You are assuming that the mass loses contact with the track when its speed is zero. Not so. There is a minimum speed required to maintain contact; find it by considering the forces on the particle and Newton's 2nd law.

3. Dec 26, 2004

### apchemstudent

I'm in the middle of solving this problem. I'm just wondering, the minimum centripetal acceleration at the top of the track is 6g. Shouldn't it be 5g as indicated by v^2(min)/R? It would make more sense considering it's
v^2(min)/R... Which is why, i don't think the answer should be 19.5 degrees.

Last edited: Dec 26, 2004
4. Dec 26, 2004

### Clari

Hello Doc AI.......I get the minimum speed to maintain contact as v^2 = gr.....Now I set the equation as 1/2 m(0.775^2)(5gR) = 1/2m v_min^2 + mgh.........but h = 1.0015625 R.....so I don't get a sensible theta.....or am I misunderstanding something?

Hello apchemstudent.............yes, you're right, the minimum centripetal acceleration at the top is v^2 = 5gR.....but now, the particle is not at the top of the circle, because it has a smaller speed, that's why it can only moves to some point P....

5. Dec 26, 2004

### apchemstudent

However though, you can use the ratio of the min Centripetal acceleration of the track and the min centripetal acceleration of the particle (if it were travelling in its own circle) to figure out theta...

Last edited: Dec 26, 2004
6. Dec 26, 2004

### Parth Dave

You're so close Clari. Keep going with that idea you and Doc Al have been toying with. Your problem lies with the minimum velocity it must have to remain on the track. You said it was: v^2 = gr. I'm going to tell you that this is not quite right. Remember how the two forces you are comparing are oriented. You need to make a slight modification before this is right. Can you tell me what you have to do?

7. Dec 26, 2004

### apchemstudent

I'll show you what a i meant by the 19.5 degrees not making any sense.

Ao = Vo^2/R = 29.43...
However, this is at the bottom of the track. To figure out the min (I'll just called centripetal acceleration now as A(c)) A(c), we'll need to subtract 29.43 - 9.8 = 19.63...

We know the min A(c) for the track which is 5g so:

sin(theta) = 19.63/5g

theta = 23.6 degrees

However for the 19.5 degrees the min A(c) for the track will have to be 6g, which is incorrect, for the A(c) min is stated as 5g. You can try that out yourself. Adding on the extra g does not make sense for the min A(c) for the track. Can anyone explain why it is done this way?

8. Dec 27, 2004

### Clari

umm....Thank you for all of ur help....I give up the use of conservation of energy in this problem...because I really don't know what I need to do.....>_<......instead, I consider the forces acting onthis particle.....mg/sin theta - N = mv^2 / r......for N is the normal reaction force acting on the particle....when it is 0, the particle will falls off the track.....so mg/sin theta = mv^2 /r........sin theta = gr / 0.775^2 * 5gr....theta = 19.5 degrees....^-^

9. Dec 27, 2004

### Parth Dave

Good job! The thing you forgot when you were finding the minimum velocity was the sin(theta), it should have been: v^2*sin(theta) = gr. Well, you found a more direct approach, but either way would have worked.

Actually, I'm abit confused about the answer as well. The way I see it, the centripetal force is always changing (decreasing) because of gravity. But your solution is assuming that is remains constant. Because you are saying, at some instant, the force of gravity will completely counter the centripetal force which will cause the particle to fall off. But you don't know what the centripetal force at that point is because it is always changing. I think I sort of see apchemstudents point. Can anyone reconcile this for us?

10. Dec 27, 2004

### Staff: Mentor

Sorry, but this answer makes no sense to me. For one thing, the radial force on the particle at point P is $mg cos\theta$ (or $mg sin\theta$, depending on how the angle is defined). It's not $mg/ sin\theta$. Secondly, the radial acceleration ($v^2/R$) depends on the speed at point P (call it $v_{min}$), not the speed at the bottom ($v_0$).

To solve this you'll need to use both conservation of energy and Newton's 2nd law. Let's call the angle that the particle leaves the track $\theta$ (measured from vertical). The radial forces on the particle at that moment are $mg cos\theta$, so $mg cos\theta = mv_{min}^2/R$.

Now apply conservation of energy. At the bottom, there is KE = $mv_0^2/2$. At the point P, there is PE ($mgR(1+cos\theta)$) and KE ($mv_{min}^2/2 = mgR cos\theta /2$). Write down the equation for conservation of energy and solve for $\theta$.

11. Dec 27, 2004

### Parth Dave

Using that idea you get the same answer of 19.5 degrees. That is where my confusion is. Why does her idea work?