Physics · Simulation and virtual lab Free
Circular motion: centripetal force with a whirling bung
Whirl a bung in a circle, time its turns, check that F = mv² ÷ r, and cut the string to see where it goes.
- You will learn to
- Measure the period and speed of an object moving in a circle.
- Verify that the centripetal force is F = mv² ÷ r.
- Explain why an object leaves a circle along the tangent when the force is removed.
- You will measure
- Time for 10 revolutions (s), period (s), speed (m/s), Mg and mv² ÷ r (N).
- Time needed
- About 30–40 min
- Equipment
- Nothing extra. Real version: rubber bung, nylon string, glass or plastic tube, slotted masses, paper clip marker, stopwatch, metre rule.
- Price
- Free: no payment, no account.
Virtual lab
Follow the steps with the simulation above (go back to the simulation ↑). Each time you tap Record, a row is added to the table and a point to the graph.
Aim
To investigate the relationship between centripetal force, mass, speed and radius for an object in uniform circular motion, and to verify that F = mv² ÷ r.
What to do
- Keep r = 0.5 m, bung 50 g, hanging mass 100 g. Tap Time 10 revolutions. Calculate T and v = 2πr ÷ T. Record.
- Look at the arrows: the force points to the centre, the velocity along the tangent. Tap Cut the string. Which way does the bung go?
- Increase the hanging mass to 150, 200 and 300 g and repeat. What happens to the speed?
- Keep the force fixed and change the radius. Record each result.
- Calculate mv² ÷ r for each row and compare it with Mg. Look at the graph.
- Use slow motion for the small, fast circles to see the turns clearly.
Results
Questions
What is centripetal force?
The resultant force on an object moving in a circle; it points towards the centre and keeps changing the direction of the velocity.
Why is an object moving at constant speed in a circle accelerating?
Its direction, and therefore its velocity, is always changing. The acceleration v² ÷ r points to the centre.
A bung takes 6.0 s for 10 turns on a circle of radius 0.40 m. Find its speed.
T = 0.60 s; v = 2π × 0.40 ÷ 0.60 = 4.2 m/s.
Find the centripetal force on a 50 g bung moving at 4.2 m/s in a circle of radius 0.40 m.
F = mv² ÷ r = 0.050 × 4.2² ÷ 0.40 = 2.2 N.
Why does the bung fly off along the tangent when the string is cut?
No force acts towards the centre any more, so by Newton’s first law it moves in a straight line in the direction of its velocity at that moment.
What provides the centripetal force for a car going round a roundabout?
Friction between the tyres and the road.
For teachers: expected results and how the model works
Ideal model (horizontal circle, no friction in the tube): the tension in the string equals the weight of the hanging mass, Mg with g = 9.78 m/s², and this is the centripetal force, so v = √(Mg r ÷ m). Example: r = 0.5 m, bung 50 g, hanging mass 100 g: F = 0.978 N, v = 3.13 m/s, T = 1.00 s, 10 turns in 10.0 s. Doubling the hanging mass multiplies v by √2. The timing button stops the clock automatically after exactly 10 turns. In real experiments the string is not quite horizontal and the radius is hard to keep fixed, so results scatter about the line.
Exam practice Exam set
National-exam style questions on this simulation (16 marks). Try each one on paper first, then open the marking scheme.
A 0.80 kg stone is whirled in a horizontal circle of radius 1.2 m and makes 2 revolutions per second. Calculate (a) the period, (b) the speed, (c) the centripetal force. (6 marks)
Marking scheme: (a) T = 1 ÷ 2 = 0.50 s (1). (b) v = 2πr ÷ T = 2π × 1.2 ÷ 0.50 (1) = 15.1 m/s (1). (c) F = mv² ÷ r = 0.80 × 15.1² ÷ 1.2 (2) = 152 N (1).
In the whirling bung experiment the hanging mass is doubled. By what factor does the speed change if the radius stays the same? Explain. (3 marks)
Marking scheme: F = mv² ÷ r, so v ∝ √F (1); doubling F multiplies v by √2 (1) ≈ 1.41 (1).
Explain why a moto rider leans inwards when going round a bend. (3 marks)
Marking scheme: A centripetal force towards the centre of the bend is needed (1); it is provided by friction at the tyres (1); leaning makes the resultant of the friction and the normal reaction pass through the centre of gravity, so the moto does not topple (1).
Give two sources of error in the whirling bung experiment and how to reduce them. (4 marks)
Marking scheme: Radius changes while whirling: use a marker on the string and keep it just below the tube (2); timing errors: time 10 or 20 revolutions and repeat (2).
Donat Sciences and MathsPhysics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Circular motion: centripetal force with a whirling bung
Aim: To investigate the relationship between centripetal force, mass, speed and radius for an object in uniform circular motion, and to verify that F = mv² ÷ r.
Apparatus (real laboratory): rubber bung, nylon string, glass or plastic tube, slotted masses, paper clip marker, stopwatch, metre rule.
What to do (with the simulation on www.donatstem.co.rw)
- Keep r = 0.5 m, bung 50 g, hanging mass 100 g. Tap Time 10 revolutions. Calculate T and v = 2πr ÷ T. Record.
- Look at the arrows: the force points to the centre, the velocity along the tangent. Tap Cut the string. Which way does the bung go?
- Increase the hanging mass to 150, 200 and 300 g and repeat. What happens to the speed?
- Keep the force fixed and change the radius. Record each result.
- Calculate mv² ÷ r for each row and compare it with Mg. Look at the graph.
- Use slow motion for the small, fast circles to see the turns clearly.
Results
| # | Hanging mass (g) | Radius (m) | Time for 10 turns (s) | Period T (s) | Speed v (m/s) | Mg (N) | mv² ÷ r (N) |
|---|
| 1 | | | | | | | |
| 2 | | | | | | | |
| 3 | | | | | | | |
| 4 | | | | | | | |
| 5 | | | | | | | |
| 6 | | | | | | | |
| 7 | | | | | | | |
| 8 | | | | | | | |
Graph
Plot Mg (N) on the vertical axis against mv² ÷ r (N). Draw the best line and find its gradient.
Questions
- What is centripetal force? [2]
- Why is an object moving at constant speed in a circle accelerating? [2]
- A bung takes 6.0 s for 10 turns on a circle of radius 0.40 m. Find its speed. [2]
- Find the centripetal force on a 50 g bung moving at 4.2 m/s in a circle of radius 0.40 m. [2]
- Why does the bung fly off along the tangent when the string is cut? [2]
- What provides the centripetal force for a car going round a roundabout? [2]
- A 0.80 kg stone is whirled in a horizontal circle of radius 1.2 m and makes 2 revolutions per second. Calculate (a) the period, (b) the speed, (c) the centripetal force. [6]
- In the whirling bung experiment the hanging mass is doubled. By what factor does the speed change if the radius stays the same? Explain. [3]
- Explain why a moto rider leans inwards when going round a bend. [3]
- Give two sources of error in the whirling bung experiment and how to reduce them. [4]
Donat Sciences and MathsPhysics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Marking scheme: Circular motion: centripetal force with a whirling bung
Expected results
Ideal model (horizontal circle, no friction in the tube): the tension in the string equals the weight of the hanging mass, Mg with g = 9.78 m/s², and this is the centripetal force, so v = √(Mg r ÷ m). Example: r = 0.5 m, bung 50 g, hanging mass 100 g: F = 0.978 N, v = 3.13 m/s, T = 1.00 s, 10 turns in 10.0 s. Doubling the hanging mass multiplies v by √2. The timing button stops the clock automatically after exactly 10 turns. In real experiments the string is not quite horizontal and the radius is hard to keep fixed, so results scatter about the line.
Answers
- What is centripetal force? [2]
The resultant force on an object moving in a circle; it points towards the centre and keeps changing the direction of the velocity.
- Why is an object moving at constant speed in a circle accelerating? [2]
Its direction, and therefore its velocity, is always changing. The acceleration v² ÷ r points to the centre.
- A bung takes 6.0 s for 10 turns on a circle of radius 0.40 m. Find its speed. [2]
T = 0.60 s; v = 2π × 0.40 ÷ 0.60 = 4.2 m/s.
- Find the centripetal force on a 50 g bung moving at 4.2 m/s in a circle of radius 0.40 m. [2]
F = mv² ÷ r = 0.050 × 4.2² ÷ 0.40 = 2.2 N.
- Why does the bung fly off along the tangent when the string is cut? [2]
No force acts towards the centre any more, so by Newton’s first law it moves in a straight line in the direction of its velocity at that moment.
- What provides the centripetal force for a car going round a roundabout? [2]
Friction between the tyres and the road.
- A 0.80 kg stone is whirled in a horizontal circle of radius 1.2 m and makes 2 revolutions per second. Calculate (a) the period, (b) the speed, (c) the centripetal force. [6]
(a) T = 1 ÷ 2 = 0.50 s (1). (b) v = 2πr ÷ T = 2π × 1.2 ÷ 0.50 (1) = 15.1 m/s (1). (c) F = mv² ÷ r = 0.80 × 15.1² ÷ 1.2 (2) = 152 N (1).
- In the whirling bung experiment the hanging mass is doubled. By what factor does the speed change if the radius stays the same? Explain. [3]
F = mv² ÷ r, so v ∝ √F (1); doubling F multiplies v by √2 (1) ≈ 1.41 (1).
- Explain why a moto rider leans inwards when going round a bend. [3]
A centripetal force towards the centre of the bend is needed (1); it is provided by friction at the tyres (1); leaning makes the resultant of the friction and the normal reaction pass through the centre of gravity, so the moto does not topple (1).
- Give two sources of error in the whirling bung experiment and how to reduce them. [4]
Radius changes while whirling: use a marker on the string and keep it just below the tube (2); timing errors: time 10 or 20 revolutions and repeat (2).