# Vertical Circular Motion

1. Sep 18, 2006

### Delzac

Hi all,

A roller-coaster train is upside down when it is at the top of a circular vertical structure which is 20 m high. In order to get to the top without falling out, the velocity of the roller-coaster train at the top of the circle has to be more than
a 25 m s-1
b 14 m s-1
c 9.9 m s-1
d zero

Ans : C

The question didn't give me any mass of anything at all to work with except for height. Is mass not required for this question? i couldn't get C, in fact i chose D, zero instead.

Any help will be greatly appreciated. ( probably need clarification of concepts.)

2. Sep 18, 2006

### Hootenanny

Staff Emeritus
Think about centripetal force. For any body undergoing circular motion the sum of all the forces acting must be equal to the centripetal force. In this case there are only two forces acting (ignoring friction); the normal reaction force R and gravity mg, therefore we can write;

$$\frac{mv^2}{r} = R + mg$$

Do you follow? Now, what will the reaction force if the coaster only just makes the loop?

3. Sep 18, 2006

### Delzac

R = 0 i guess?

For R to be non-zero the roller coaster must have some velocity so as to be able to "press" on to the track is that right?

Last edited: Sep 18, 2006
4. Sep 18, 2006

### Hootenanny

Staff Emeritus
This is correct. So can you solve the problem from here?

5. Sep 19, 2006

### Delzac

so,

$$V = \sqrt{rg} =\sqrt{(10)(9.81)}= 9.9$$

But i have a qns, if there is velocity by the time i reached the top, wouldn't there be normal reaction force already? then R will be non-zero.

Last edited: Sep 19, 2006
6. Sep 22, 2006

### Hootenanny

Staff Emeritus
Sorry, for the delay in replying but I have been increasingly busy of late. Why does velocity imply a normal reaction force? Do not forget that the reaction force (if any) is not the only force that is acting, HINT: Gravity.

7. Oct 29, 2006

### BSCS

I have a question: If R were to drop to zero, the coaster would fall, right? So, for v=9.9 m/s, it would actually lose contact with the track and fall. But for any value even minutely larger than R=0 (and thus v=9.9 m/s) the coaster would just barely "stick" to the track? Even for a super small value of R?

8. Apr 1, 2008

### dre028

(mv2/r) = T top + mg

The mass from the (mv2)/r and mg cancel each other out thus giving

v2/r = T top + g

where g = 10 or 9.81 whichever you are using.
For the coaster to remain in contact with the tracks, the force T top has to be greater then zero, so that the resultant force (mv2/r) is greater than the weight of the coaster pulling it downwards.

From there on it is relatively simple. Your problem, I think, was more of mathematics than physics.

r is the radius, and since it is at a height of 20m this is the diameter.

v2/10 = 0 + 9.81
v2 = 98.1
v = square root of 98.1
v = 9.9m/s

Sorry for the format, but I don't really know how to use this thing well. I hope that you understand that v2 means v squared etc

9. Apr 1, 2008

### dre028

Exactly yes, in theory that is what should happen, so that the centrepetal force forcing the coaster outwards are greater than the forces pulling it down. Remember though that in reality there are other external factors which may change this.

I think that even if R = 0, the coaster should not fall, because this would just mean that the forces pushing it up are equal to the forces pulling it down. I am not sure of this though

10. Apr 1, 2008

### dre028

Hi,
How

11. Apr 1, 2008

### dre028

Hi,
How con I find the velocity of an object at the top of the loop, if it has a radius of 0.4m, a mass of 0.025kg, and an energy value of 9J? This was an exam question and I couldn't work it out. The lecturer has now given it to us for homework and I still cannot work it out. I hace become an expert in these problems trying to solve this, but perhaps there is something I am not seeing. Thanks

12. Dec 14, 2011