Swing 360: Solve for Velocity of Swingset

[Mathg09]

Homework Statement

Someone is going to push another person sitting on a swingset with the goal to make a whole 360 degree loop. What should the velocity be that the person pushes the swing to make this happen? You can make your own data about the height of the swingset but it's a swingset with chainlinks if that matter.

Homework Equations

(mv^2)/r = F - mg
mgh= (mv^2)/2
v = ω*r

The Attempt at a Solution

mg=m(v^2)/r in the top which gives v = sqrt (gr)
(m(initial v)^2)/2 = m((sqrt gr)^2)/2 + mg2r
Intial velocity should be sqrt 5gr I'm not sure I'm doing it right. would really appreciate some help, thanks in advance!

 

[gneill]

Just carry on with your algebra. So far you've got (clearing out the common m's):

\frac{v_o^2}{2} = \frac{(\sqrt{g r})^2}{2} + 2gr

 

[Mathg09]

Ok thanks, then it gives intial velocity should be = \sqrt{5gr} which is what the person pushing should aim for, right?
I was just wondering if this is the only correct answer? I found an example in my textbook where it says that a ball of mass m is rotated at constant speed v in a vertical cirle with radius r. Then (v^2) > gr which prevents the object from falling when its on the top while swinging. Is it the same with my swing set, I'm hesitating since the swing set doesn't have constant velocity I guess?
Thanks in advance !

 

[gneill]

Note that the example you refer to indicates that by some means not explained, the ball is being kept at a constant speed around its course despite the acceleration due to gravity. This is not the same situation for your swing. What is the same between the two is the minimum velocity that must obtain at the top of their paths.

 

[Mathg09]

Okay thank you verry verry much! :)