The method and the results are on the circular motion lab. Do the analysis yourself first, then check the answers below.
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. These answers go with the Physics 11-12 lab on circular motion: the graph, the calculations, the uncertainties and the conclusion explained, for all 8 analysis questions.
The answers use the sample results in the table, so your own values will differ slightly. Check that you timed 20 revolutions, divided by 20 for the period, and squared the period before plotting T² against 1/M.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: A Level 3.6
Question 1 Why the weight equals the tension
The hanging mass is in equilibrium, so the tension in the string equals Mg. The same string tension acts on the bung, and it is the only horizontal force on it, so it is the centripetal force.T = M g.
Question 2 Period
T = 20.1 ÷ 20 = 1.005 s (to 3 d.p. 1.005 s). ω = 2π ÷ T = 6.283 ÷ 1.005 = 6.25 rad/s.T = 1.005 s; ω = 6.25 rad/s.
Question 3 Predicted tension
F = m ω² r = 4π² m r ÷ T² = 39.48 × 0.0205 × 0.60 ÷ 0.203 = 2.40 N. Mg = 0.250 × 9.81 = 2.45 N. They agree to within 2 %.predicted 2.40 N against 2.45 N.
Question 4 Straight-line graph
From M g = 4π² m r ÷ T², T² = (4π² m r ÷ g) × (1 ÷ M). So T² is proportional to 1/M and the gradient is 4π² m r ÷ g = 39.48 × 0.0205 × 0.60 ÷ 9.81 = 0.0495 s² kg.gradient = 4π² m r ÷ g.
Question 5 Gradient from the data
m = gradient × g ÷ (4π² r) = 0.050 × 9.81 ÷ (39.48 × 0.60) = 0.0207 kg, in agreement (within about 1 %) with the measured 0.0205 kg.m = 20.7 g.
Question 6 Percentage uncertainty
Uncertainty in time = 0.3 ÷ 9.0 × 100 % = 3.3 %. T² doubles this, so the percentage uncertainty in T² is about 6.7 %. Timing 20 revolutions already reduces the effect of reaction time.±6.7 %.
Question 7 Sources of error
The bung also has weight, so the string makes a small angle below the horizontal, which makes the real tension a little larger than the horizontal component. Friction between the string and the tube also reduces the tension at the bung. Both effects make the measured period slightly longer than expected.measured T is a little larger than predicted.
Question 8 Conclusion
At constant radius, the centripetal force is inversely proportional to T²: doubling the tension reduces T² by a factor of about two. The results agree with F = 4π² m r ÷ T².F ∝ 1/T².
