Energy conservation and circular motion

[w3390]

Homework Statement

A ball with a mass m swings from a string of length L from a pivot. A distance D directly below the pivot there is a nail. The ball is released making an angle theta=pi/2 with the vertical. For this particular problem, there is no friction.

Assuming the ball swings completely around the nail in a circle, what is the velocity of the ball at the top of the circle?

Homework Equations

K=(1/2)mv^2
U=mgh
F=m(v^2/r) for circular motion

The Attempt at a Solution

So as soon as the string hits the nail, the radius of circular motion will be r=(L-D). Therefore, the centripetal force will be m(v^2/(L-D)). When the ball is at the top of the circle, if the ball is to just make it over the centripetal force should be equal to the gravitational force, right?

So:

m*(v^2/(L-D))=mg

v^2=(L-D)*g

v= sqrt((L-D)*g)

I think this is correct because if the velocity at the top of the circle were equal to zero, then it would not be able to complete the rest of the circle which is a necessary condition.

However, it seems unsettling to me that I did not calculate the velocity of the ball right before the string hits the nail.

Does my answer look correct?

 

[PhanthomJay]

Homework Statement

A ball with a mass m swings from a string of length L from a pivot. A distance D directly below the pivot there is a nail. The ball is released making an angle theta=pi/2 with the vertical. For this particular problem, there is no friction.

Assuming the ball swings completely around the nail in a circle, what is the velocity of the ball at the top of the circle?

Homework Equations

K=(1/2)mv^2
U=mgh
F=m(v^2/r) for circular motion

The Attempt at a Solution

So as soon as the string hits the nail, the radius of circular motion will be r=(L-D).

Yes, correct.

Therefore, the centripetal force will be m(v^2/(L-D)).

only after it hits the nail.

When the ball is at the top of the circle, if the ball is to just make it over the centripetal force should be equal to the gravitational force, right?

not necessarily, there still could be a tension in the string contributing to an increase in the centripetal force (centripetal force = T+ mg).

So:

m*(v^2/(L-D))=mg

v^2=(L-D)*g

v= sqrt((L-D)*g)

I think this is correct because if the velocity at the top of the circle were equal to zero, then it would not be able to complete the rest of the circle which is a necessary condition.

this is not correct

However, it seems unsettling to me that I did not calculate the velocity of the ball right before the string hits the nail.

That should be unsettling, so why not settle yourself and calculate the velocity right before the string hits the nail? Then use conservation of energy to find V at the top.

 

[w3390]

The velocity right before hitting the nail would be (1/2)m*v^2=mgh.

Therefore, v=sqrt(2gh) which in this case is:

v=sqrt(2gL)So:

Ei=Ef

(1/2)m*(2gL) = (1/2)m*v'^2 + mg*2(L-D)
g*L=(1/2)v'^2 + 2g(L-D)
v'^2 = 2[g*L - 2g(L-D)]

So:

v' = sqrt[2[g*L - 2g(L-D)]]

v'=sqrt[2gL - 4gL + 4gD]

v'=sqrt[4gD - 2gL] (velocity at the top of the circle)

 

[PhanthomJay]

Excellent!

 

[w3390]

Cool. Thanks a lot PhanthomJay