In the hints below, we intersperse discussion with actual sample data , following sig. fig rules every step of the way. The most important thing is to NOT round off until the last final step; keep the extra digits, but remember to label them as significant or not, usually with an underline. For example 45.98567132  suggests  only 4 sig. figs. But we keep the digits to the right of the 8  to avoid  round off errors in intermediate computations.

 

DATA SHEET:

(Q-1) Here we assumed an error of plus or minus 1 g or  +/- 1 g.  Suppose you have a data set like  below. The thing to remember is the number of significant digits increases from 3 to 4 when you compute the average =
(sum of  numbers) /5 = 3150/5 = 630.0 since 3150 has 4 sig. fig. You choose the maximum of R/N and
Δminst, which in this case is the latter.  If you were reporting a  final result you would write,

 m =
mBEST + Δm = 630.0 g  +/- 1g. 

Note we would  consider  mBEST   to have only 3 sig. fig in this case. On the other hand, if we were to have used 
0.8 g as the uncertainty,  we would write 630.0 g  +/- 0.8 g; clearly we are not  doing  that since the instrument error is larger, but it's good to see this possibility. More hints later . 

 

MASS   m (g)

 

 630

 

 631

 

 629

 

 632

 

 628

mBEST

 630.0

R/N

 (632-628)g/5 = 0.8 g

Δminst

 1 g

Δm

 1 g

(Q-1) 

The radius r is a continuation of the process of determining the dynamic centripetal force magnitude F. Measured in m, r takes on the following values using a meter stick  to  4 sig. figs. Note that the last digit is either a 5 or a 0 since we are rounding to the nearest 0.10 cm/4 = 0.025 cm = 0.0025 m, which we will round to 0.03 cm = 0.0003 m in the final step. Now, when I compute the average, I still get 4 sig figs since the  sum of the numbers is also 4 sig fig: (0.8220 m )/5 = 0.1644 m.  

 

 

RADIUS  r   (m)

 

 0.1640

 

 0.1645

 

 0.1645

 

 0.1650

 

 0.1640

rBEST

 0.1644

R/N

 (0.1645 - 0.1640)/5 = 0.00020 m = 0.020 cm.

Δrinst

 0.03525 cm = 0. 0003525 m = 0.0004 m = 
0.04 cm.

Δr

 0.04 cm

 

 (Q-2)  Derive equation  (3)

(Q-3) Bob mass is a continuation of the process of determining the dynamic centripetal force magnitude F. Measured in g, M takes on the following values using a mass scale to  4 sig. figs. Note that the last digit is either a 5 or a 0 since we are rounding to the nearest 0.10 g/4 = 0.025 g,  which we will round to 0.03 g m in the final step. Now, when I compute the average, I get  5 sig figs since the  sum of the numbers is 5 sig fig: (2255.90/5 = 451.180 g. 

 

MASS   M (g)

 

 451.15

 

 451.15

 

 451.20

 

 451.25

 

 451.15

MBEST

 451.180

R/N

 (451.25 - 451.15)/5 = 0.10/5 = 0.020 g = 0.02 if rounded to one sig. fig.

ΔMinst

 0.03

ΔM

 0.03

 More hints later. Check back soon .  However, you should be able to 
complete the data sheet based on the above discussion. Follow
the suggested rules.

(Q-4) 

time

PERIOD  T  (s)

 17.1921

0.687684

 17.2252

0.689008

 17.0672

0.682688

 17.1541

0.686164

 17.2015

0.688060

 17.1458

0.685832

 17.0612

0.682448

 17.1234

0.684936

 17.3215

0.692860

 17.0982

0.683928

TBEST

 0.6863608000 (note 7 sig. figs.)

R/N

 (0.692860 - 0.682688)/10 = 0.0010171 = 0.001 if rounded to one sig. fig.

ΔTinst

 0.000004

ΔT

 0.001

 

(Q-5)  and (Q-6)
Below, I will show you how to compute F = FBEST +  ΔF. First off, we  compute FBEST   (formula (3) previous page) 
then compute  uncertainty under (Q-6), ΔF= 






#  Formula (3): FBEST    = 4π2 *MBEST*rBEST/TBEST2   = 2 *(0.451180)*(0.1644)/(0.6863608000 )2  = 6.215865913 (N).

# (Q-6)'s uncertainty formula:  
  ΔF = (FBEST)*(  ΔM/MBEST + Δr/rBEST   + 2*ΔT/TBEST ) = 

(6.2158659)*(0.025/451.180    +     0.03525/16.44    +      2*0.0010171/0.6863608000) 

Now, let's look at each term within the parentheses. The first term is the  smallest to be included since 

first term = 0.025/451.180 = 0.000554103. Compare this with the second and third terms: 

second term = 0.03525/16.44 = 0.00214416

third term =   2*0.0010171/0.6863608000  = 0.002963747.

I have underlined the least sig. fig. in all three cases and will now add them, realizing the least sig. fig.  is in the 0.000001 place: 
first term + second term + third term = 0.00566201 = 0.00566 when rounded to the correct mathematical place or 0.006 when rounded to one sig. fig.  Thus we can now get ΔF = 
(6.2158659)*(0.025/451.180    +     0.03525/16.44    +      2*0.0010171/0.6863608000) =
(6.2158659)*(0.00566201) = 0.0351943.  Now,  if we round this to one sig. fig. and round FBEST to the same place we get:

F = 6.215865913 +/- 0.04 = 6.21 (N) +/- 0.04 (N)

F   =  6.21 (N)  ±  0.04 (N)

 (Q-7) Is the theory that the static and dynamic values of F are the same supported? Explain.

 Now check whether or not the discrepancy test is true: |FBEST – mBEST··g|  < 2(ΔF + Δmg). Do not round off when you first substitute 
values. Let 's evaluate each side,  then compare:  

Left side: | 6.215865913  -  (0.6300 )*(9.8) | = |   6.215865913 - 6.174   | =  0.041865913 = 0. 
Right side: 2*(
0.0351943 + 0.001*9.8)  = 2*(0.0351943 + 0.0098) = 2*(0.0449943)= 0.0899886.

Since the left hand is less than the right hand side, the discrepancy test is true. We can say  the centripetal force equals the spring force keeping the mass moving in a circle.