- #1
nchin
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The string in the Figure is L = 120 cm long, has a ball attached to one end, and is fixed at its other end. The distance d to the fixed peg at point P is 75.0 cm. When the initially stationary ball is released with the string horizontal as shown, it will swing along the dashed arc. What is its speed when it reaches (a) its lowest point and (b) its highest point after the string catches on the peg?
Picture of figure:
http://panda.unm.edu/Courses/Price/Phys160/F17-1.jpeg
a) I understand part a
i just use v = sqrt (2gh) = sqrt (2(9.8)(1.20m)) = 4.85 m/s.
b) I found the solution for part b online but i don't understand it at all like why is y = 2r?
Solution for part b:
http://panda.unm.edu/Courses/Price/Phys160/p17-1.pdf
mgL = 1/2mv^2 + mgy
does that formula equals E = KE + PE? So at the highest point the total energy is always KE + PE? Also why is y = 2r?
Help please this is really bothering me! Thanks! :)
Picture of figure:
http://panda.unm.edu/Courses/Price/Phys160/F17-1.jpeg
a) I understand part a
i just use v = sqrt (2gh) = sqrt (2(9.8)(1.20m)) = 4.85 m/s.
b) I found the solution for part b online but i don't understand it at all like why is y = 2r?
Solution for part b:
http://panda.unm.edu/Courses/Price/Phys160/p17-1.pdf
mgL = 1/2mv^2 + mgy
does that formula equals E = KE + PE? So at the highest point the total energy is always KE + PE? Also why is y = 2r?
Help please this is really bothering me! Thanks! :)