# Normal force of a ball at the top and bottom of a circular path

## Homework Statement

A ball whirls around in a vertical circle at the end of a string. The other end of the string is fixed at the center of the circle. Assuming that the total energy of the ball-Earth system remains constant, show that the tension in the string at the bottom is greater than the tension at the top by six times the weight of the ball.

See below...

## The Attempt at a Solution

I tried to use W = ∆K = mgy and W₁ + U₁ + W = W₂ + U₂ to solve the problem,
but it did not work out.

I'm pretty lost on this one.

HELP!!!

Last edited:

rl.bhat
Homework Helper
Minimum velocity of the ball at the top of the circular motion is given by
m*vi^2/R = mg = T Or vi^2 = Rg.
When it falls to the lowest point, let its velocity be vf and displacement be 2R.
Now apply the conservation of energy to find vf.
What is the tension in the string at the bottom of the circular motion of the ball?

Thanks.. but I still don't quite see how to obtain the solution.

Last edited:
rl.bhat
Homework Helper
vf^2 = vi^2 + 2*g*2R
Bur vi^2 = gR. So
vf^2 = gR + 4gR = 5gR
So the centripetal acceleration is 5g and gravitational acceleration is g.
So the total acceleration at the bottom is = ......
Compare it with that of at the top.

vf^2 = vi^2 + 2*g*2R
Bur vi^2 = gR. So
vf^2 = gR + 4gR = 5gR
So the centripetal acceleration is 5g and gravitational acceleration is g.
So the total acceleration at the bottom is = ......
Compare it with that of at the top.
When I use the conservation of energy equation, do I account for T on both sides?