Quick answer: Circular motion at constant speed involves a constant angular velocity ω = 2π/T and a centripetal acceleration a = v²/r = ω²r directed towards the centre. This Physics 11-12 lab on circular motion gives the aim, the equipment, the safety points, a step-by-step method, sample results and 8 analysis questions; the answers are on a separate page and the lab sheet is a free printable PDF.
How does the period of a whirled bung change as the centripetal force increases? In this lab you keep the radius fixed, vary the tension with hanging masses, time 20 revolutions and test the relationship F = 4π² m r ÷ T² against your data.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: A Level 3.6
Aim
To test how the period of a horizontal circular motion depends on the centripetal force, and to use the results to find the mass of the bung.
Equipment
- Rubber bung (mass 0.0205 kg) on 1.5 m of nylon thread
- Glass or plastic tube, smooth at both ends, about 12 cm long
- Hanger and slotted masses, 0.050 kg to 0.250 kg
- Paperclip or tape marker, metre rule, stopwatch, balance
- Safety glasses
Safety
Wear safety glasses and swing the bung in a clear area, away from people. Tie the thread securely to the bung and the hanger. Do not let the bung hit anyone if the thread breaks, and keep the heavy masses over a soft box, not your feet.
Method
- Thread the string through the tube. Tie the bung to the top end and the hanger to the bottom end. Place the marker on the string so that the distance from the bung to the top of the tube is 0.60 m.
- Add 0.050 kg to the hanger. Hold the tube vertically and whirl the bung in a horizontal circle so that the marker stays just below the tube.
- Time 20 revolutions with the stopwatch while the marker stays at the same place.
- Calculate the period T from the time for 20 revolutions.
- Repeat with 0.100, 0.150, 0.200 and 0.250 kg, keeping r = 0.60 m.
Results
Sample results by Bilal in Lahore (m = 0.0205 kg, r = 0.60 m, g = 9.81 N/kg). The weight Mg of the hanging masses equals the tension.
| M (kg) | time for 20 revolutions (s) | T (s) | T² (s²) | 1/M (kg⁻¹) | tension Mg (N) | predicted tension 4π²mr/T² (N) |
|---|---|---|---|---|---|---|
| 0.050 | 20.1 | 1.005 | 1.010 | 20.0 | 0.49 | 0.48 |
| 0.100 | 14.2 | 0.710 | 0.504 | 10.0 | 0.98 | 0.96 |
| 0.150 | 11.6 | 0.580 | 0.336 | 6.7 | 1.47 | 1.44 |
| 0.200 | 10.0 | 0.500 | 0.250 | 5.0 | 1.96 | 1.94 |
| 0.250 | 9.0 | 0.450 | 0.203 | 4.0 | 2.45 | 2.40 |
Analysis questions
Question 1 Why the weight equals the tension
Explain why the weight of the hanging masses equals the tension that supplies the centripetal force.Question 2 Period
Calculate the period and the angular velocity for the 0.050 kg load.Question 3 Predicted tension
Show how the last column was obtained for the 0.250 kg load and compare it with Mg.Question 4 Straight-line graph
Show that a graph of T² against 1/M should be a straight line through the origin and state its gradient.Question 5 Gradient from the data
Use the gradient of the best-fit line through the origin, about 0.050 s² kg, to calculate the mass of the bung.Question 6 Percentage uncertainty
The time for 20 revolutions is read to ±0.3 s. Find the percentage uncertainty in T² for the 0.250 kg load.Question 7 Sources of error
Give two reasons why the string is not exactly horizontal and say how this affects the results.Question 8 Conclusion
State the relationship between centripetal force and period that your results support.
