A 2.50 kg ball is attached to a 3.00 m bar and swung in a vertical circle. If the ball does not leave the circular loop, what minimum speed must it have at the top of the arc?

1 Answer
Jun 5, 2017

v=sqrt(29.4)~~5.42217668469m/sv=29.45.42217668469ms

Explanation:

Use Newton's second law:

sumF=maF=ma

Since the ball is moving in a circle, we can say that a=v^2/ra=v2r

There are two forces acting on the ball. We have the force of gravity and the normal force and they both point downwards at the top of the arc:

mg+N=m(v^2/r)mg+N=m(v2r)

We are finding the minimum speed that the ball must have. That means that there is no normal force:

mg+0=m(v^2/r)mg+0=m(v2r)

Now plug in the values we know and solve for vv. m=2.5kgm=2.5kg, g=9.8m/s^2g=9.8ms2, r=3mr=3m:

2.5(9.8)=2.5(v^2/3)2.5(9.8)=2.5(v23)

Solve for vv:

9.8=v^2/39.8=v23

3(9.8)=v^23(9.8)=v2

v=sqrt(29.4)~~5.42217668469m/sv=29.45.42217668469ms