Question

2 Experiment: In the experiment conducted, the mass of the counter weight M was chosen to be 50.0 +/- 0.1 g The mass of the rubber stopper was measured to be 12.0 +/- 0.1 g The stopper was spun at five different values of the radius (measured with a meter stick). For each value of r, the time to complete 10 full revolutions was measured with stopwatch. The following data was collected: 40.0 60.0 80.0 100.0 (cm) 1) 120.0 10.74 6.20 7.65 8.77 9.85 4) Notice that as the radius of the circle is increased, the time to complete 10 revolutions also increases. Explain why this should be the case in this experiment Note: The fact that the stopper has to travel a longer distance when the radius is greater does not explain the time increase because the speed of the stopper increases as well when the radius increases (see below). The explanation is a little more subtle. In order to calculate the speed v for each value of the radius, calculate the circumference (210) for each value ofr. The speed v can be calculated as v -10 (207). 32 Calculate v for each of the five values of r in the above table. Notice that the speed increases as ricases. Calculate the speed square (v) for each of the five values of Since the magnitude of centripetal force is constant in this experiment, the magnitude of centripetal acceleration of the stopper should be constant as well. Because av Ir and hence, vr, the graph of vvs. should be a straight line with the slope equal to a, and y-intercept close to zero. 6) Plot vs. using a computer or a graphing calculator (v on the yaxis and on the x-axis). Add the regression line to your graph. Display the regression results (the parameters of the fit) on your graph. 7) Does the data support the war equation? (Is the value of correlation close to one? Is y-intercept close to zero) 8) Based on your graph, what was the centripetal acceleration of the stopper in the experiment? 9) Now, calculate Mgm from the measured values of Mand. Calculate its uncertainty. Which quantity introduces the greater uncertainty into the result, Mor? (Treat g = 9.80 m/s as exact in the uncertainty calculation.) 10) Report the value of Me (mean +/- uncertainty, units) 11) Compare the results in questions 8 and 10. Do the values agree? Defend your answer 12) What value of a would you expect to measure if the hanging counter weight mass were doubled? 13) Would it be possible to swing the stopper without any counter weight in this experiment? Why or why not?

Answer 1

Given,
mass of counter weight M = 50.0 +- 0.1 g
mass of ruber stopper, m = 12.0 +- 0.1 g

given data is for time taken for 10 revolutions

r(cm)	t (n = 10)	t ( n = 1)	v = 2*pi*r/t	v^2
40	6.2	0.62	4.053668	16.43222
60	7.65	0.765	4.927988	24.28507
80	8.77	0.877	5.731526	32.85039
100	9.85	0.985	6.378868	40.68996
120	10.74	1.074	7.020319	49.28488

4. Now, we can see as r increases, t increases as well
now, from force balance
Mg = mv^2/r
where v = 2*pi*r/t
hence
Mg = m*4*pi^2*r/t^2
so we can see
r/t^2 = Mg/4*pi^2*m = constant
hence, as r increases, t increases too

5. spped is calculated as v = 2*pi*r/t in the table, v^2 as well
6. the plot of v^2 vs r is as under

7. from the relation we can see
v^2 = 41.06r - 0.1356
value of corelation is 0.9998 ( close to 1)
intercept is close to 0
hence it obeys v^2 = ar

8. from the graph, slope of graph is v^2/r
now, centripital acceleration is v^2/r
hence slope is centripital acceleration = 41.06 m/s^2

9. Mg/m from measured values is   
50*9.81/12 = 40.875 m/s/s
uncertianity is d(u)/u = dm/m + dM/M
du = 0.422375
hence
Mg/m = 40.875 +- 0.422 m/s/s

greater uncertianity is introduced by mass of rubber stopper

10. Mg/m = 40.875 +- 0.422 m/s/s
11. % error is |(40.875 - 41.06 )|*100/40.875 = 0.45259938 %
hence experimental value is close to theoretical value