Centripetal force lab

  • Thread starter Elbobo
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  • #1
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Homework Statement


Since I ran out of time in class, I have to make up my own times for this lab using calculations.

The lab consists of our providing a little force by shaking this tube. Inside this tube is a string, and at one end of the string is attached a hanging mass, the other end a rubber stopper. Essentially, the hanging mass's weight is what provides the centripetal force of the rubber stopper.

However, my calculations for the time in seconds are waaay too high. I tried doing the same method for my other trials done in class and they were also far too high. I don't know what I'm doing wrong, because human error couldn't have possibly been that high all 10 trials in class.

m1 = 0.011 kg
m2 = 0.030 kg
r = 0.545 m
g = 9.81 m/s2

Homework Equations



d = r * 2 pi * 20 (we measured the time it took for 20 revolutions)
t = d / v
W = mg
Fc = (mv2) / r

The Attempt at a Solution


Since the force we provide with our arms shouldn't factor in too much, I did:
(m2v2) / r = m1g
v = 1.400 m/s

t = d / v
t = (2pi * 0.545 * 20) / 1.4
t = 48.915 s,
which varies WAY too much with the other trials (Each trial, a variable was altered a bit, such as the hanging mass or the radius, but this change only affected the time by 3 seconds or less).

The other trials were in the 12.0 - 15.0 s range.
What am I doing wrong?
 
Last edited:

Answers and Replies

  • #2
PhanthomJay
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Check to see if you may have your m1 and m2 reversed in your formula. The stopper mass at the end of the string is what you are calling 'm2'. Is it?
 
  • #3
98
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Homework Statement


Since I ran out of time in class, I have to make up my own times for this lab using calculations.

The lab consists of our providing a little force by shaking this tube. Inside this tube is a string, and at one end of the string is attached a hanging mass, the other end a rubber stopper. Essentially, the hanging mass's weight is what provides the centripetal force of the rubber stopper.

However, my calculations for the time in seconds are waaay too high. I tried doing the same method for my other trials done in class and they were also far too high. I don't know what I'm doing wrong, because human error couldn't have possibly been that high all 10 trials in class.

m1 = 0.011 kg
m2 = 0.030 kg
r = 0.545 m
g = 9.81 m/s2

Homework Equations



d = r * 2 pi * 20 (we measured the time it took for 20 revolutions)
t = d / v
W = mg
Fc = (mv2) / r

The Attempt at a Solution


Since the force we provide with our arms shouldn't factor in too much, I did:
(m2v2) / r = m1g
v = 1.400 m/s

t = d / v
t = (2pi * 0.545 * 20) / 1.4
t = 48.915 s,
which varies WAY too much with the other trials (Each trial, a variable was altered a bit, such as the hanging mass or the radius, but this change only affected the time by 3 seconds or less).

The other trials were in the 12.0 - 15.0 s range.
What am I doing wrong?

For clarification, I read this as
m1 = hanging mass
and
m2 = "swinging mass" (rubber stopper)

Is that correct? If not and you reversed these, you will have a very big difference in your calculation of v.

The other potential sources of error that I see (if I am envisioning this correctly) may include friction of the string on the tube, the mass of the string and its effect on the moment of inertia/radius of gyration, and potential changes in the radius due to motion of your arm. I would assume all of these would be fairly small and shouldn't have too big an effect on the outcome.
 
  • #4
145
0
I see where you guys might think I have reversed the masses, but no, Stovebolt is right with the labels.

EDIT:

Well since that didn't work for some reason, I resolved to simply find the factor by which it should change if I doubled the mass. I already had 2 trials that had the same conditions except the swinging mass was halved.

Doubling the swinging mass should change the time by a factor of (1/2)-(1/2), right?
 
Last edited:

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