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 ?
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
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.
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.
Please help me.
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