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?

Question

2027 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-05T00:42:28

#v=sqrt(29.4)~~5.42217668469m/s#

#sumF=ma#

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

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)#

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)#

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

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

Solve for #v#:

#9.8=v^2/3#

#3(9.8)=v^2#

#v=sqrt(29.4)~~5.42217668469m/s#