ES190 Lab Report – Inertia part
Name
Student ID
Date
H&S Considerations:
Refer to the ‘Risk Assessment’ and fill the table below
(2 marks)
Hazards Counter measures
Falling weights / discs Wear appropriate protective shoes and remain
stood up while conducting the experiment.
Finger burn or cut from moving spindle / string. Ensure low rotational speeds and mount the rig on
a level bench.
Q1a. Error Analysis Data for Disc 1 (small) - include formulae used and units
(8 marks)
Formula Unit 20 g 50 g 70 g 100 g
Weight Weight Weight Weight
Sample mean of velocity, 𝑣̅ 0.454 0.496 0.531 0.572
Standard error of velocity, 0.00176 0.00393 0.00306 0.00338
Δ𝑣
Estimate of 𝑣̅ ! 0.206 0.246 0.282 0.328
Estimate of the error in 𝑣̅ ! 0.00160 0.00390 0.00324 0.00387
Q1b. Error Analysis Data for Disc 2 (large)
(4 marks)
units 20 g 50 g 70 g 100 g
Weight Weight Weight Weight
Sample mean of 𝑣̅ 0.317 0.355 0.377 0.408
Standard error Δ𝑣 0.000882 0.00100 0.00173 0.000577
Estimate of 𝑣̅ ! 0.100 0.126 0.142 0.166
Estimate of the error in 𝑣̅ ! 0.000559 0.000710 0.00131 0.000471
, Q2. For each disc, produce a graph in Excel/Matlab of 𝑣 ! against 𝑚 (using the estimated
values of 𝑣 ! from Q1) and show the equation of the fitted line for each disc. (8 marks)
Figure 1 – shows the velocity-mass graph of the small disc. Figure 2 – shows the velocity-mass graph of the large disc.
Q3.
a. Explain why the curves do not pass through the origin and what the values of the 𝑦
intercepts represent. (5 marks)
The curve does not pass through the origin because although there is no mass attached to the holder,
the brass holder weight still pulls on the nylon cord. Therefore, when there is no mass, there is still an
acceleration. This would then mean that the y-intercept gives the value of 𝑣 ! of the brass holder with
no attached mass.
b. Estimate the mass of the brass holder for experiments with small and large discs.
(3 marks)
".$%&'
Small disc: 𝑚 = = 115.7𝑔
".""$'
"."()!
Large disc: 𝑚 = "."""( = 105.3𝑔
Derivation of the equation:
Figure 3 – shows the derivation of the equation for the mass of the brass holder.
The mass is shown to be equal to the absolute value of the x-intercept, since this is the point at which
the velocity of the object is zero.
Q4. Write down the estimated values of 𝐼 for the two discs from the experimental results and
compare theoretical values. The result for which disk will have the higher relative error?
(rationalize your statements) (6 marks)
Small Disc Large Disc
Experimental Values 0.001086 𝑘𝑔 𝑚! 0.002243 𝑘𝑔 𝑚!
Theoretical Values 0.0009418 𝑘𝑔 𝑚! 0.002053 𝑘𝑔 𝑚!
Relative Error 13.3% 9.25%
2/6
Name
Student ID
Date
H&S Considerations:
Refer to the ‘Risk Assessment’ and fill the table below
(2 marks)
Hazards Counter measures
Falling weights / discs Wear appropriate protective shoes and remain
stood up while conducting the experiment.
Finger burn or cut from moving spindle / string. Ensure low rotational speeds and mount the rig on
a level bench.
Q1a. Error Analysis Data for Disc 1 (small) - include formulae used and units
(8 marks)
Formula Unit 20 g 50 g 70 g 100 g
Weight Weight Weight Weight
Sample mean of velocity, 𝑣̅ 0.454 0.496 0.531 0.572
Standard error of velocity, 0.00176 0.00393 0.00306 0.00338
Δ𝑣
Estimate of 𝑣̅ ! 0.206 0.246 0.282 0.328
Estimate of the error in 𝑣̅ ! 0.00160 0.00390 0.00324 0.00387
Q1b. Error Analysis Data for Disc 2 (large)
(4 marks)
units 20 g 50 g 70 g 100 g
Weight Weight Weight Weight
Sample mean of 𝑣̅ 0.317 0.355 0.377 0.408
Standard error Δ𝑣 0.000882 0.00100 0.00173 0.000577
Estimate of 𝑣̅ ! 0.100 0.126 0.142 0.166
Estimate of the error in 𝑣̅ ! 0.000559 0.000710 0.00131 0.000471
, Q2. For each disc, produce a graph in Excel/Matlab of 𝑣 ! against 𝑚 (using the estimated
values of 𝑣 ! from Q1) and show the equation of the fitted line for each disc. (8 marks)
Figure 1 – shows the velocity-mass graph of the small disc. Figure 2 – shows the velocity-mass graph of the large disc.
Q3.
a. Explain why the curves do not pass through the origin and what the values of the 𝑦
intercepts represent. (5 marks)
The curve does not pass through the origin because although there is no mass attached to the holder,
the brass holder weight still pulls on the nylon cord. Therefore, when there is no mass, there is still an
acceleration. This would then mean that the y-intercept gives the value of 𝑣 ! of the brass holder with
no attached mass.
b. Estimate the mass of the brass holder for experiments with small and large discs.
(3 marks)
".$%&'
Small disc: 𝑚 = = 115.7𝑔
".""$'
"."()!
Large disc: 𝑚 = "."""( = 105.3𝑔
Derivation of the equation:
Figure 3 – shows the derivation of the equation for the mass of the brass holder.
The mass is shown to be equal to the absolute value of the x-intercept, since this is the point at which
the velocity of the object is zero.
Q4. Write down the estimated values of 𝐼 for the two discs from the experimental results and
compare theoretical values. The result for which disk will have the higher relative error?
(rationalize your statements) (6 marks)
Small Disc Large Disc
Experimental Values 0.001086 𝑘𝑔 𝑚! 0.002243 𝑘𝑔 𝑚!
Theoretical Values 0.0009418 𝑘𝑔 𝑚! 0.002053 𝑘𝑔 𝑚!
Relative Error 13.3% 9.25%
2/6
