# Centripetal Force and velocity

1. Jan 23, 2015

### terryds

1. The problem statement, all variables and given/known data

http://www.sumoware.com/images/temp/xzfrbardcbliotkt.png [Broken]

A box travels in a circular track like the picture above.
How much velocity it must have in point A to get out of the track when reaching point B ?

2. Relevant equations
Fcentripetal = m V^2/R
E = E'

3. The attempt at a solution

First, I draw the free body diagram of the forces in point B
http://www.sumoware.com/images/temp/xzstkkpbgobpmjmi.png [Broken]

Actually, I want to draw the normal force, but I don't know where the normal force direction is.
I know that the centripetal force is the force that's accelerating to the center.
So, I think that Wb sin Θ is the same as the centripetal force.
I use
Wb sin Θ= F centripetal
mg sin Θ= m v^2/r
g sin Θ= v^2/r
v = √(g r sin Θ)

Then, I use the mechanical energy conservation formula. (I assume that the point A is zero in y axis)
E = E'
1/2 mva^2 = mgh + 1/2 m vb^2
1/2 va^2 = gh + 1/2 vb^2
va^2 = 2gh + vb^2
va^2 = 2g(R+R sin Θ) + vb^2
va^2 = 2g(R+R sin Θ) + (√(g r sin Θ))^2
va^2 = 2g(R+R sin Θ) + g r sin Θ
va^2 = 2gR+2gRsinΘ+gr sin Θ
va^2 = 2gR+3gRsinΘ
va^2 = gR (2+3sinΘ)
va = √(gR (2+3sinΘ))

But, I'm not sure my answer is right.
What I doubt is when I think the W sinΘ equals centripetal force and when I don't know where the normal force direction is pointing. Actually, I want to use ∑F = ma , but I just know the weight force to draw.

Last edited by a moderator: May 7, 2017
2. Jan 23, 2015

### haruspex

At the point where the box is just falling out of the track, what do you think the magnitude of the normal force will be?

3. Jan 23, 2015

### terryds

Hmm... I think it is the same as W but with opposite direction ? Or maybe W sin theta ?

4. Jan 23, 2015

### haruspex

If there's still all that force exerted between the box and the track, why is it about to fall of the track?

5. Jan 23, 2015

### terryds

Maybe because the weight is bigger than the normal force, or there is no normal force..
But, i don't know how to determine the normal force in this problem

6. Jan 23, 2015

### haruspex

Those forces both act sort-of downwards, so that isn't going to explain it.
Exactly. You lose contact with a surface when there's no force holding you to it, and so no normal force reacting.
Hence the direction of the normal force does not matter for this question, but, for future reference, if either body has a tangent plane at the point of contact then the normal force is normal to that plane. Note that if they both have tangent planes they will necessarily be the same. If neither has a tangent plane it becomes impossible to say.

7. Jan 23, 2015

### terryds

Thanks for your explanation.
But, what about its centripetal force ? Does it equal W sin theta ?
And, is my answer to the problem correct ?

8. Jan 23, 2015

### haruspex

Yes. To keep on the track, the acceleration must be v2/r, no more, no less. The net force in the radial direction will never be less than mv2/r because the normal force will increase as necessary to prevent the block breaking into the track. But if the net force exceeds the centripetal force, even when the normal force has fallen to zero, then it will accelerate faster than v2/r radially and leave the track. The only source of such a force is g, and its radial component will be W sin(theta).