# Centripetal Force and velocity

• terryds
In summary, a box is in a circular track and needs to exit at point B. If it has a centripetal force of W sin(theta) then it will stay on the track.

## Homework Statement

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

## Homework Equations

Fcentripetal = m V^2/R
E = E'

## 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:
terryds said:
Actually, I want to draw the normal force, but I don't know where the normal force direction is.
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?

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

terryds said:
Hmm... I think it is the same as W but with opposite direction ? Or maybe W sin theta ?
If there's still all that force exerted between the box and the track, why is it about to fall of the track?

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

terryds said:
Maybe because the weight is bigger than the normal force
Those forces both act sort-of downwards, so that isn't going to explain it.
terryds said:
or there is no normal force
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.

haruspex said:
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.

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

terryds said: