Physics circular motion lab help?

Question

please look at my lab and tellme what i did wrong because in theory you should have weight of washers= to centripetal Force. Circular Motion Lab

Purpose: The purpose of this lab is to examine the relationship between a centripetal force acting on an object experiencing uniform circular motion, velocity, and its centripetal acceleration.

Data table 1: Circulation Motion of the stopper, washer and time

Mass of the Stopper (Kg) Mass of the Washer (Kg)
.0138.00546


Length (m) Revolutions Number of Washers Time
.50 10204.89
.5010205.50
.5010185.0
.5010184.77
.5010165.91
.5010165.78
.5010145.72
.5010145.69
.5010126.62
.5010126.47











Washer (m) Tangential Vel Centripetal a Centripetal F

20 (.109) Kg

.048 m/s.11.66 m/s .
18 (.098) Kg.0511 m/s.824 m/s.0107 N
16 (.087) Kg.042 m/s.823 m/s.0010 N
14 (.076) Kg.043 m/s.611 m/s .0079 N
12 (.066) Kg .038 m/s.461 m/s.0059 N

Sample Calc: m= .109 Kg
a.Tangetial Velocity V=d/t = .50/5.2=.096 r/s
Vt= r( angular speed)
.50* .096= .048 m/s
Centripetal Accel: Vt^2/R
VT= 2pieR/t= 2*3.14*.50/5.2 s= .606
Centripetal Accel = (.606)^2/.50= .729 m/s
Centripetal Force= FC= mstopper* ac= .729 * .013 Kg


Weight of washers (Force) Percent Difference

Answer 1

This is too confused a question to go over in detail. (For instance:
"a.Tangetial Velocity V=d/t = .50/5.2=.096 r/s
Vt= r( angular speed)
.50* .096= .048 m/s"
is not essential to a description of the problem. Fortunately, you didn't use it anywhere else so it's harmless.)
You need to do further study on rotational and translational velocity and acceleration and their units. I gather you think the expected result is .729*.013 kg = 0.0095 kg but you are getting 0.109 kg. A few points:
From web descriptions of circular motion labs I'm going to assume that the rotation is either produced by a turntable with a vertical rotation axis or by swinging the stopper from a tube held over your head, that the radius is the horizontal distance from the rotation center to the stopper mass, and that there is negligible friction of the stopper mass against the turntable or tube.
Next, forces are in Newtons, not kg. If the stopper and washer masses are in fact in kg as stated, then you are getting a force of G * washer mass = 9.8 * .109 = 1.068 N and (more or less correctly) expecting 0.0095 N. This is a big (factor of >100) disagreement. First check that the 5.2 seconds is the time for 1 revolution, not 10 or however many you were advised to use to improve accuracy. If it was for 10 revs, then the rotation rate is 10 * your assumption and the expected centripetal force is 100 * your calculation or 0.95 N, not too far from the observed 1.068 N. If it was really that slow, I think you need to check on the weights of the stopper and the 20 washers. Basically your experiment shows that G * 20-washers mass = 0.729 * stopper mass, which indicates that the stopper mass is G/.729 or about 13.5 times the mass of the 20 washers. If that's the case, recompute and read on; the mass data you cited is in error. If not, I'm out of suggestions...
If you find that your answer is close but still has a minor error, be sure you have used all the available significant figures, or at least have rounded off correctly. 0.0138 shouldn't be rounded to 0.013. Using 0.0138 gives you a 6% improvement in accuracy. Also consider possible error sources. Are all the washers known to be the same weight? Is there friction in the setup (string and pulley or whatever it is) that transfers the centripetal force to the washers? Does the string move during the experiment, changing the radius of rotation? Was the tube also rotating with a significant radius? This would add to the total radius of rotation. I imagine you can think of similar realities that can move the results away from the ideal expectation.
P.S. If you have similar questions to ask, do yourself and those who read your question a favor and try to be clear. Add spaces to separate data items. Use consistent variable names. For instance:
Centripetal Accel CA=Vt^2/R, where Vt is tangential velocity and R is radius
Vt = 2piR/t = 2*3.14*.50/5.2 = .606 m/s
CA = (.606)^2/.50 = .729 m/s^2
Centripetal Force = mstopper*CA = .729 * .0138 = .01006 N (not even near the experimental result of ----- N)

Answer 2

Circular Motion Labs

Answer 3

Your answer is not wrong it is absolutely right. Donot worry you are correct.