What is the velocity of a ball on a frictionless curved track?

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The discussion centers on calculating the velocity, time to hit the floor, distance from the base of the table, and total energy of a ball on a frictionless curved track. The correct velocity of the ball as it leaves the track is determined to be sqrt(2gR). The time to hit the floor is confirmed as correct, while the distance the ball lands from the base of the table is calculated to be sqrt(4RH). The total energy upon striking the floor is mg(H + R), with potential energy referenced from floor level. Overall, the calculations provided clarify the dynamics of the ball's motion on the track.
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I am stuck on this problem:

A ball with a mass of M is on a frictionless curved track with a radius of R. The track sits atop a table that has height of H. Solve the following in terms of R,g,H, and M.
a) the velocity of the ball
b) the time it takes to hit the floor
c) the distance D the ball lands from the base of the table
d)the total amount of energy the ball has when it strikes the floor

Here are my answers, are these correct.
a) v= sqrt(2gH) ?
b) t=sqrt(2gh)/g ?
c)?
d) would i add 1/2mv^2 + MgR+MgH ?
 
Last edited:
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You've told us the track is on the table, and the ball is on the track. What's this about the ball hitting the floor? And the velocity could be anything, from the information you have given.
 
I have attached a diagram to show what I am trying to solve for, and the velocity is as the ball leaves the table.
 

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Ive tried to work out the problem , can anyone see if these answers are correct.
a)sqrt(2gR)=velocity
b)sqrt(2H/g)=time
c)H+ R =distance
d)mgh=total energy

Thanks
 
Last edited:
a) the velocity of the ball

Assuming it is released from rest at the top of the track, and the track is frictionless, the speed as it leaves the track will be sqrt(2gR).

b) the time it takes to hit the floor

Your answer is correct.

c) the distance D the ball lands from the base of the table

Answer is:

sqrt(4RH)

d)the total amount of energy the ball has when it strikes the floor

Answer is mg(H + R), taking the zero of potential energy to be floor level.
 
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