physics - BOB

PLEASE SEE ADDITIONAL QUESTION AT BOTTOM.

A pendulum was set up and measurements were made to enable the mechanical energy to be calculated at the start position S and the lowest point of the pendulums swing L.

The mass of the pendulum bob was determined on an electronic scale and its diamter was measured using calipers. The initial height was measured with a meter stick. At the lowest point of its swing, the pendulum bob broke a photogate light beam. The time interval that the light was interrupted was recorded on an electronic timer attached to the photogate.
Use the following data to complete a report.
MAss of pendulum bob = 240.3 g
Diameter of pendulum bob = 3.50 cm
Initial height of pendulum bob = 48.0 cm
Length of pendulum string = 2.14 m
Time interval of photogate light interruption = 11.8 ms

Your report should include the following:
(a) conclusion as to whether or not the pendulum demonstrated the law of conservation of energy
(b) calculations of the efficiency of the pendulum as a mechanical machine

I am really confused how to even start this question. I think that I should figure out the Emechanical = Ek + eg

=1/2(240.3)(0) + 240.3(9.8)(.48)
=1130.37J

but have no idea how to calculate the speed of when it hit. I don't know where to go from here at all, please help!

PHYSICS - Damon, Saturday, January 30, 2010 at 4:10pm
The clue is the diameter of the bob
it broke the beam for 11.8 *10^-3 seconds
it is .035 meters in diameter
so it went
.035 meters in 11.8^10^-3 seconds
which is about 2.97 meters/s

typo, left m out - Damon, Saturday, January 30, 2010 at 4:16pm
(1/2) 240.3 v^2 = mgh = 1130.37
v^2 = 9.407
v = 3.067 ideally but we only measured 2.97 m/s so some energy got lost along the way like air friction and stuff

DAMON: I understand how you got the 1130.37, but don't know how you got v^2. Can you please explain how you got there?...and what are you using for height in the second part of the question...

physics - Damon - bobpursley, Saturday, January 30, 2010 at 5:10pm
1/2 240.3 v^2=1130.37

120.1 v^2=1130.37

v^2= 1130.37/120.1= 9.407

DAMON; thanks for helping so much but I still need some clarification. So the speed when it hit the beam was 2.97 m/s, what does the 3.07 m/s speed represent?

When i subbed in the new speed of 3.07 into the Ek + Eg formula, i got this:
=1/2mv^2 + mgh
=1/2(240.3)(3.07)^2
=1132.4

i am not sure what the 1132.4 actually represents though..please help me more!


physics - Damon - bobpursley, Saturday, January 30, 2010 at 5:36pm
It represents the mechanical energy in joules at the speed gate, at the bottom.

Bob - but if this 1132.4 J is the mechanical energy at the bottom, and the mechanical energy at the top is 1130.37 J, then wouldn't that make is seem like energy was GAINED, rather than lost like DAMON first said?

also,what would percent efficiency equal?

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  1. The 3.07 is what it should have been if it had all the energy at the bottom that it started with at the top.
    It ended up going slower, 2.97, with less energy. Some was lost.

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  2. (1/2) 240.3 v^2 = mgh = 1130.37
    v^2 = 9.407
    v = 3.067
    this is simply saying that if there is no loss of energy due to friction, the kinetic energy at the bottom (1/2) m v^2 will be the same as the potenital energy at the top m g h

    the actual speed measured was less so the actual energy at the bottom is
    (1/2) 240.3 (2.97)^2 = 1059.83
    so you lost
    1130.37 - 1059.83 Joules to friction
    and your efficiency is
    100 (1130.37 - 1059.83)/1130.37

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