Physics 11–12 · Year 12
Forces in circular motion: conical pendulum and a car on a banked bend
Module 5: Advanced Mechanics (Circular Motion)
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The idea
On a banked surface or a slanted string, the horizontal component of the normal force or tension supplies the centripetal force, which fixes the speed for a given angle.
What you need
- Conical pendulum: 1.000 m string, 0.100 kg bob, retort stand, protractor, stopwatch, phone video from below
- Banked track: a marble on the inside of a large bowl or a toy-car loop track with a marked bank angle, and video for speed
How to do it
- Set the bob circling steadily so the string makes 30 degrees with the vertical (check on video); time 20 revolutions.
- Compute the predicted period T = 2 pi sqrt(L cos(theta) / g) and compare.
- Repeat at 20 and 45 degrees and plot T^2 against cos(theta).
- Banked track: roll the marble at the speed that keeps it on a marked line of the bowl; measure the bank angle there and the radius, and compare v with sqrt((5/7) r g tan(theta)). A marble rolling without slipping needs a friction force to keep turning its spin as it goes round, so it circles about 15 per cent slower than the sqrt(r g tan(theta)) that suits a frictionless car; a toy car, whose wheels hold little of its energy, is closer to that frictionless value.
What you should see
A 1.000 m conical pendulum at 30 degrees has a period of 1.867 s and the bob moves at 1.68 m/s around a circle of radius 0.500 m. T^2 against cos(theta) is a straight line of gradient 4 pi^2 L / g. A car on a bend of radius 100 m needs a bank of 22.2 degrees to round it at 20 m/s with no friction. A marble rolling on a 30 degree bank at 0.100 m radius should circle at sqrt((5/7) x 0.100 x 9.806 65 x tan 30 degrees) = 0.636 m/s, against 0.752 m/s for a frictionless slider.
What changes
- What you change
- angle of the string to the vertical
- What you measure
- period (time for one circuit of the bob)
- What you keep the same
- string length
- bob mass
Common misconceptions
Each of these ideas is wrong, and the activity is a chance to test it.
- The bob is in equilibrium because it moves steadily; the net force is not zero, it points to the centre.
- Banking helps only because of friction; the bank itself provides an inward force component.
Safety card
Hazards
- swinging bob
Controls
- clear area, light bob
Note
Record the activity in RiskAssess (https://www.riskassess.com.au/) and follow the Science ASSIST risk management information sheet (https://asta.edu.au/resource/ais-risk-management-and-risk-assessment/).
Curriculum references
The NSW syllabus outcomes and Australian Curriculum v9 codes this activity supports. They are references, not a verified or complete curriculum alignment.
- Physics Stage 6 Syllabus (2017), current: Year 11 until the end of 2026, Year 12 until Term 3 2027PH12-12PH11/12-6
- Physics 11-12 Syllabus (2025), not yet taught: Year 11 from Term 1 2027, Year 12 from Term 4 2027, first HSC examination 2028PY-12-01PY-12WS-06
- Australian Curriculum v9No Australian Curriculum v9 code is listed.
Sources
The pages the author read to write this activity.