A tennis ball is rolling without slipping at 4.03m/s on a horizontal section of a track. It then rolls into a vertical circular loop that is 0.45m in radius and the ball leaves the loop at a point 0.20m below the horizontal section. Find the speed of the ball at the top of the loop.

I am unsure of where I should start with this question, I have calculated the circumference of the circle and found that the top of the circle is about 1.41 m from the start of the loop. I have also used the equation " v = r w " to find that w is equal to 8.96 m/s. I am not sure if anything I have found is relevent but I have gotten very stuck on this. Any help or ideas would be very much appreciated! Thanks so much for your time.

Could you attach a figure please? I cant picture the ball leaving the loop below the horizontal.

This is the best picture I can get, I hope it works.

Attached Images

15753_332873340042_683485042_9919986_7037535_n.jpg (44.7 KB, 7 views)

I still cant see it leaving the loop below the horizontal.

This should happen at some point at a height greater than the radius, above the horz.

When the ball leaves the loop, the normal reaction = zero. The centripetal force is a combination of the normal reaction and the radial component of the weight.

If you have the answer, we can check if they meant something else.

I do not have any answers to this question, but do you have any idea how I would go about finding the speed of the ball at the top of the loop? I need to prove that the ball will not fall off of the track.

As explained, the centripetal force is determined by the normal reaction and the weight of the ball.

Let the vel at the top be v and the mass of the ball be m

The centripetal force = m v^2 / r , and the weight of the ball = mg

At the top of the loop, the normal reaction N points vertically downwards and so does the weight. The centripetal force is thus given by

m v^2/r = N + m g or, N = m [ (v^2 / r) - g ].

For N to be zero v^2/r - g = 0, or v^2 = r g

This is the min vel required for it to stay on the loop is N = 0.

For lower values, N would be -ve.

For larger values of v, it will clearly remain in the loop.

When it enters the loop, at the bottom it has a k.e. = (K.E.)b

Let its k.e at the top of the loop be (K.E.)t .

From conservation of energy, (K.E.)b - m g ( 2 r) = (K.E.)t.

From this find the vel a the top and find how it compares with the reqd min velocity.

That makes sense, but I was not given any mass for the tennis ball. So I am not sure I can use those equation. Unless the masses cancel out and I am missing that, but if so I am not sure how they might cancel out.

Consider

From conservation of energy, (K.E.)b - m g ( 2 r) = (K.E.)t.

0.5 m Vb ^2 - m g (2r) = 0.5 m Vt ^2.

m will cancel out .

And in the condition for minimum vel, the quantity in [ ] which = 0 does not contain m.