Proving the Law of Conservation of Energy Using a Pendulum

AI Thread Summary
The discussion revolves around using a pendulum to demonstrate the law of conservation of energy through calculations. Amy provides initial measurements and calculations for potential and kinetic energy at two positions, A and B, but realizes her speed calculation may be incorrect due to a misunderstanding of units. After clarifying that 11.8 ms refers to milliseconds, she recalculates energy values and considers the impact of friction and air resistance on energy loss. There is also a discussion about the importance of measuring the pendulum's height at its center of gravity and whether the weight of the pendulum string affects the overall calculations. The conversation highlights the need for accurate measurements and understanding of energy principles in the experiment.
physicsgal
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im suppose to do some calculations to prove whether or not the pendulum demonstrates the law of conservation of energy.

mass of pendulum = 240.3g = 0.2403 kg
diamter of pendulum bob = 3.50
initial height of pendulum bob = 48 cm = 0.48 m
length of pendulum string = 2.14 m
time interval of photogate light interruption = 11.8 m/s

so there's position 'A' and position 'B'
for 'A"
Emechanical = Ek + Eg
= (0.5)mv^2 + mgh
= 0 + (0.2403kg)(9.8)(0.48m)
= 1.13J

for 'B'
Emechanical = Ek + Eg
= (0.5)mv^2 + mgh
= (.5)(0.2403kg)(11.8^2) + 0
= 16.73J

i must be missing something (my calculations aint proving anything).. any help will be appreciated!

~Amy
 
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Did you calculate the speed correctly? You list it as time but give units of speed, suggesting an error there. Furthermore, 12 m/s is pretty fast for a 2m long pendulum.
 
k, i think i see the mistake! it says "11.8 ms" and i thought they meant meters per second.

so 11ms = 0.011s?

so for B i'd go:
Emechanical = Ek + Eg
= (0.5)mv^2 + mgh
= (.5)(0.2403kg)((d/0.011)^2) + 0 = 1.13

im short on time but will figure this out tomorrow and post my results.


~Amy
 
just thought of something else. to find the velocity i just take the diameter of the bob divided by 11ms?

~Amy
 
physicsgal said:
just thought of something else. to find the velocity i just take the diameter of the bob divided by 11ms?

so 11ms = 0.011s?

~Amy
Yes, but use the 11.8 ms as it was in your data. And yes, ms is milliseconds so 11.8 ms = .0118 s
 
i changed all the units to regular units (meters, kg, s, etc.) just to play thing safe.

for Ek (A)
i got: mgh
= (0.2403kg)(9.8 m/s^2)(0.48m)
= 1.13J

for Eg (B)
i got: 0.5mv^2
=(0.5)(0.2403kg)(2.966^2)
=1.057J

(to get the 2.966 velocity i took 0.035/0.01185s).

and then i just explain that mechanical energy becomes heat energy so the total mechanical energy gradually decreases. ?

~Amy
 
Yes, a certain amount of potential energy is lost to friction and air resistance, although 6% may be a little high. All measurements have uncertainty or inaccuracy, and that will also be a factor. One final question: when you measured the height of the pendulum, did you measure at the center of the weight at both positions A and B? It's important to measure the heights at the center of gravity.
 
i didnt do any of the measurements myself. I am taking an independent learning course, and the measurements are listed in the book, I am just suppose to do some qualculations and write a lab report explaining why and whether or the pendulum lab demonstrates the law of conservation of energy.

slight ot: does the weight of the pendulum string make a difference? like a light weight vs. a heavy one?

~Amy
 
physicsgal said:
i didnt do any of the measurements myself. I am taking an independent learning course, and the measurements are listed in the book, I am just suppose to do some qualculations and write a lab report explaining why and whether or the pendulum lab demonstrates the law of conservation of energy.
Got it. Thanks!

physicsgal said:
slight ot: does the weight of the pendulum string make a difference? like a light weight vs. a heavy one?

~Amy
Good question. What do think?
 
  • #10
shouldnt the string weight be added to the bob weight?

~Amy
 
  • #11
Not exactly, because it isn't in the same place. Think about where its center of gravity (COG) is and where the bob's is. Any ideas on how you would add the string's mass to your problem?
 

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