Find the Maximum Mass for Riding a Swing

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The problem involves calculating the maximum mass of a person who can ride a swing without exceeding the rope's maximum tension of 800N. The swing is initially pulled back at a 60-degree angle and released, requiring an understanding of centripetal force and energy conservation. The maximum tension occurs at the lowest point of the swing, where the equation T - mg = mv^2/r is applied. After using energy conservation to find v^2, the calculations lead to a maximum mass of approximately 40.8 kg. The solution appears correct based on the derived equations and values.
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Homework Statement



A swing is made from a rope that will tolerate a maximum
tension of 800N without breaking. Initially, the swing
hangs vertically. The swing is then pulled back at an
angle of 60.0 degree with respect to the vertical and
released from rest. What is the mass of the heaviest
person who can ride the swing?



Homework Equations


Fc = mv^2/r
trig ratios


The Attempt at a Solution



I initially just thought of doing 800N / 9.81m/s^2 which give 81.5kg or so but i thought that was too easy and can't be right so I did some more complex stuff:

using unit circle:
r = sin60r = 0.5r


Fc = mv^2/r
I forgot how but i somehow ended up in the step below, where 0.5r cancels out and in the end and the end result goes back to m = 800/9.81
800 = (9.81m(0.5r))/0.5r

still pretty sure that is wrong and I am really lost, so help please o;
 
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The swing is basically, undergoing motion in a vertical circle. So for this to happen, at every instant the swing needs a radially inward component of the net force(centripetal) which is equal to mv^2/R.

Now it is crucial in this problem that you understand, the tension in the string will be maximum, when it is at, the lowest point in it's motion, I.e when weight and tension are anti parallel.

So at this instant,

T - mg= mv^2/RNow you want to know the maximum mass of the person, so put the maximum value in for the tension.

Use conservation of energy to find v^2 and solve the above eqn for m
 
At what point after release will the tension in the rope be the greatest? Once you decide on that, then draw a free body diagram of the weight on the swing noting all forces. Relate those forces to the tension in the rope. From the sound of the problem statement, I assume this is a one rope swing.
 
The problem can be worked from an energy standpoint or from the acceleration of a mass as it is affected by gravity. I don't know which subject you are studying.
 
Ok so I made some progress from all the help you guys give and now I am at :

T - mg = mv^2/r

mgh = 1/2mv^2
v^2 = 2gh

and since i figured h was 0.5r so:

v^2 = 9.81r

replacing that in initial yields:
800 - 9.81m = 9.81rm/r
800 = 19.62m
m = 40.8kg

DID I DO IT RIGHT?
 
Looks good to me.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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