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=29.45.42217668469ms

Explanation:

Use Newton's second law:

F=ma

Since the ball is moving in a circle, we can say that a=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(v2r)

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

mg+0=m(v2r)

Now plug in the values we know and solve for v. m=2.5kg, g=9.8ms2, r=3m:

2.5(9.8)=2.5(v23)

Solve for v:

9.8=v23

3(9.8)=v2

v=29.45.42217668469ms