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## John Palcewski's Journal

### Works In Progress

Physics Vittoria swings on a 23.8 m long rope initially inclined at an angle of 36° from the vertical. Questions:

(a) What is her speed (m/s) at the bottom of the swing if she starts from rest?

(b) What is her speed at the bottom of the swing if she starts with an initial speed of 2.22 m/s? This question is all about Vittoria's energy in the two separate situations, that is, her potential and kinetic energy at the starting point, and then her kinetic energy at the bottom of the swing.

To determine these answers, you first need to find how far above the bottom of the swing she starts. Draw a diagram of the rope inclined at an angle of 36° and use trigonometry to figure out the starting height above the bottom. You should get that h=23.8*(1-cos(36°)). For (a), Vittoria only has potential energy at the beginning, and only kinetic energy at the bottom of the swing. So all the potential energy is converted to kinetic, which gives the equation: m*g*h = 1/2*m*v^2. Solve for v, and you get v = sqrt(2*g*h). For (b), Vittoria has both kinetic and potential energy at the beginning, and again, only kinetic at the bottom of the swing. This situation gives the equation m*g*h + 1/2*m*(v-initial)^2= 1/2*m*(v-final)^2. You want to solve for v-final, and you should get that v-final = sqrt(2*g*h + v-initial^2). Plug in the value you found for h, and the value you are given for v-initial (her initial speed), and remember that g is the acceleration due to gravity (about 9.8 m/s^2).

Have I made myself clear?    My eyes glazed over with the math bits there, but I marveled at Victoria's wonderfully fit body. I wish mine looked like that!

And a question. is her watch waterproof? Big issues on my mind! ;-D To be perfectly honest, I never noticed her watch as we both swung at the river, and I still never noticed the watch in the many years since then whenever I looked at the photos. You might say that I was, and remained, distracted! 