How Can I Determine the Mass of a Rubber Stopper in a Circular Motion Lab?

In summary, the conversation discusses a lab experiment involving circular motion and the determination of the mass of a stopper. The experiment involves swinging a rubber stopper attached to a glass tube with a hanging mass, and plotting graphs to determine the centripetal force and acceleration. However, the results were not making sense and there were difficulties in measuring the time and accounting for friction and the angle of droop. The speaker is seeking help to figure out any mistakes in their calculations and acknowledges that the experiment may not yield perfect results.
  • #1
rebekahje
1
0

Homework Statement


So i have to do a lab about circular motion
we had to swing a rubber stopper around while holding a glass tube and when there was a mass hanging at the end of it
"plot a graph so the slope of the straight line graph may be used to determine the mass of the stopper"
we changed the mass at the bottom and varied the length of the string, but i don't know if this is a problem or not
we were going to plot a centripetal force as a function of centripetal acceleration graph, and find the slope to find the mass of the stopper
we measured time (t) in seconds, the length of the string (r) in meters, the mass hanging mass (m1) in kilograms, and we were supposed to calculate centripetal acceleration (ac) in meters per second squared, centripetal force, (Fc) in Newtons, and velocity (v) in meters per second, and we had to find the mass of the stopper (m2) in kilograms, but we measured it for percent error
(this is one of the trials
m1 = .05 kg, t =.486 s, r =.295 m, Fc = .49N, ac = ?, v = ? m2 = ?

Homework Equations


Fc=m1g
Fc=m2ac = m2v2/r
C=2∏r
v= C/t


The Attempt at a Solution


So, i tried to find the velocity and the acceleration and they just did not make any sense at all, so I'm wondering if anyone can help me figure out what i did wrong.
C=2∏r
C=(2)(∏)(.295m)
C=1.854m
v=C/t
v=1.854m/.45s
v=4.12m/s
ac=v2/r
ac[/SUP]=(4.12m/s)2/.295m
ac=57.5m/s2
which does not make any physical sense at all, HELP.
 
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  • #2
Welcome to PF, Rebeka.
This is a difficult experiment I have done many times with grade 11 classes. For starters it is difficult to measure the time for one turn. Typically, you measure the time for 20 turns and divide by 20. But it is hard to keep turning at the same speed that long. Your "t" is the time for one turn, isn't it?
Secondly, there is a good deal of friction where the string turns at 90 degrees on the top of the glass tube. You must not expect near perfect results.
Thirdly, the string "droops" at an angle below horizontal so there is a vertical component of force on the string that balances gravity in addition to the horizontal component that is the centripetal force. The force of gravity on the mass hanging below the glass tube is the combination of these two forces. It turns out that the results are very sensitive to the angle of droop and you must work very hard to measure that angle in order to successfully show that F = mv²/r to within experimental error. Of course the angle is different for every run. Oh - and you need to estimate the error in your measurements, especially of the angle because it has the greatest effect.

It looks like you did not measure the angle of droop, so you must make note of it in your writeup and proceed anyway - not expecting good results. It would be impressive to your marker if you showed how the angle matters and come up with a good way to measure it. Any chance of repeating it? This is one experiment you really can do at home.

The usual approach is to make a table of values with your measurements and calculations of F and v²/r. I always tried to keep the answer from my students, so they were supposed to try combinations other than v²/r, too. Anyway, you graph F vs v²/r, put error bars on if you have any error estimates, and then try to draw a straight line through the error bars. If you can, you have F = k*v²/r and perhaps you can show that k is somewhere near the mass. At least it has the right units.
 

FAQ: How Can I Determine the Mass of a Rubber Stopper in a Circular Motion Lab?

1. What is circular motion in physics?

Circular motion in physics is the movement of an object along a circular path at a constant speed. It involves the concept of centripetal force, which is the force that keeps an object moving in a circular path.

2. How is circular motion related to Newton's laws of motion?

Circular motion is related to Newton's laws of motion in that it follows the first law of motion, also known as the law of inertia, which states that an object in motion will stay in motion with a constant velocity unless acted upon by an external force. In circular motion, the centripetal force acts as the external force that keeps the object moving along the circular path.

3. What is the purpose of a circular motion lab in AP Physics?

The purpose of a circular motion lab in AP Physics is to demonstrate the principles of circular motion and to allow students to apply their knowledge of Newton's laws of motion and centripetal force. It also helps students understand the relationship between an object's speed, radius, and centripetal force in circular motion.

4. What equipment is needed for a circular motion lab?

The equipment needed for a circular motion lab typically includes a rotating platform, a string, a weight, a stopwatch, and a protractor. Optional equipment may include a motion sensor and a computer for data collection and analysis.

5. What are some real-life applications of circular motion?

Circular motion has many real-life applications, such as the movement of planets around the sun, the rotation of tires on a car, and the motion of a roller coaster. It is also used in industries such as aerospace, where satellites and spacecraft use circular motion to maintain their orbit around Earth.

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