Non Calc - Potential Energy/Conserv. & Centrepital Acc?

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The discussion revolves around two physics problems involving potential energy and centripetal acceleration. The first problem requires calculating the tension in a vine when Jeff swings from an angle of 32 degrees, starting from rest, with a mass of 73 kg. The second problem involves determining the change in gravitational potential energy for a pendulum bob of 0.43 kg swinging from point A to B at a 35-degree angle. Key points include using the formula U=mgh for gravitational potential energy and understanding the relationship between potential energy and kinetic energy to find velocity and centripetal force. Additionally, it is noted that a swing released from a horizontal position will experience a force of 3g at the lowest point, regardless of the swing's length.
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OMG these 2 problems are driving me insane! If you are able to help, please show steps so I can gain a good understanding...

1.) Jeff of the Jungle swings on a 7.6- vine that initially makes an angle of 32 with the vertical. If Jeff starts at rest and has a mass of 73 , what is the tension in the vine at the lowest point of the swing?

* I do know that the tension on the vine at rest is m(g) at rest...how am I supposed to incorporate theta?


2.) A 0.43 pendulum bob is attached to a string 1.2 long, making an angle of 35 with the vertical. What is the change in the gravitational potential energy of the system as the bob swings from point A to point B (from right to left)?
 
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on the second one you could find the change in height and use. U=mgh
U= gravitational potential energy
And also on the first one you could find the change in height and use mgh and then equal this to kinetic energy to get a velocity and then use the centripetal force. This should work but I am also really tired right now.
 
As cragar says - just calculate the change in gpe and this will tell you the KE increase, which will give you the force involved.

It's interesting to note that, on a swing of any length, if you are released with the string horizontal, you will always pull 3g at the bottom. It's just that, on a long swing, the experience lasts longer.
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...

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