Mr. Maloney's pendulum is swing back and forth. He has a 1 kg mass on the end, the string length is 2 meters, and he started it by pulling it back 35 degrees, determine:

a. The initial potential energy

b. Speed at the bottom

c. Speed half way to the bottom

d. How high would it go if a student have it a 3 m/s push

Thank you for the opportunity to help you with your question!

The height y can be found by subtracting Lcos(θ) from L. Therefore, the potential energy is given by:
U=mgy
U=mg(L-Lcos(θ))
U=mgL(1-cos(θ))

b.)In order to sketch a potential energy function of the angle θ, make
the y-axis potential energy, U, and make the x-axis the angle θ. The
pendulum has its greatest potential energy when its farthest away from
the origin. So start with a large potential energy at -210° and slowly
decrease until there is zero potential energy at 0°. The potential
energy will again increase as displacement from the origin increases so
draw an increasing potential energy to 210°. Remember that the
potential energy of a pendulum cannot be negative, so your graph will
only appear in the first and second quadrants. The sketch should look
periodical.

c.)The tangential force is equal to the sine of the angle time the mass
and gravity of the pendulum. In this case: Ftangential=sinθ(mg)

d.)At the initial state of the pendulum we have zero potential energy
and nonzero kinetic energy. Thus, by the Conservation of Energy, we
have:
Ei=Ef
KEi+Ui=KEf+Uf
1/2mvi^2+0=0+mgL(1-cos(180°))
vi^2=2gL(2)
vi=2sqrt(gL)

Please let me know if you need any clarification. I'm always happy to answer your questions.

Jun 3rd, 2015

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