B. Dynamics · Dynamics: Forces and Motion in a Plane

Centripetal Force: Flat Curves, Conical Pendulums and Rotor Rides

B2.6 B2.1 B3.3About 75 minutes

Learning goals

I can explain that centripetal force is not a new kind of force but the net force toward the centre, supplied by real forces.
I can use Fnet = mv²/r = 4π²mr/T² to solve problems with strings, friction and normal forces in horizontal circles.
I can find the maximum safe speed on a flat curve from the coefficient of static friction.
I can analyse a conical pendulum and a rotor ride with free-body diagrams.

Words to know

centripetal force (Fc)
The net force toward the centre that keeps an object moving in a circle: Fc = mv²/r. It is always supplied by real forces such as tension, friction, gravity or a normal force.

conical pendulum
A mass on a string that moves in a horizontal circle, so the string sweeps out a cone.

rotor ride
An amusement ride in which riders stand against the wall of a spinning cylinder; the wall's normal force supplies the centripetal force, and friction holds riders up.

Why cars slide off icy curves

On a dry day a car rounds a flat highway ramp with no trouble. On a January morning with black ice, the same car at the same speed slides toward the outside of the curve. It is not pushed outward. Instead, nothing pulls it inward strongly enough, so it continues more nearly in a straight line, as the first law says it should.

In the last lesson you found that circular motion needs an acceleration v²/r toward the centre. Newton's second law says that takes a net force toward the centre. This lesson is about what supplies that force.

Centripetal force is a job, not a force

Never draw "Fc" as an extra arrow on a free-body diagram. Centripetal force is the name of a job: the net inward force. Real forces do the job:

a string's tension (a ball on a string, a hammer throw)
static friction (a car on a flat curve, a coin on a turntable)
a normal force (clothes against the drum of a washing machine, riders against a rotor wall)
gravity (the Moon around Earth, satellites; see Gravitational, Electric and Magnetic Fields)
a combination (a conical pendulum: tension and gravity)

Then apply Newton's second law along the radius:

ΣF(toward centre) = mv² ÷ r = 4π²mr ÷ T² = 4π²mrf²

Choose one axis pointing toward the centre (the direction of the acceleration) and, if needed, one perpendicular to the plane of the circle, where a = 0.

Worked example

Example 1: A puck on a string

A 0.50 kg puck on frictionless ice is tied to a 1.2 m string whose other end is fixed to a peg. The puck goes around once every 0.80 s. Find the tension in the string.

On ice, gravity and the normal force balance vertically. The only horizontal force is the tension, so it supplies the centripetal force.
FT = 4π²mr ÷ T² = 4π²(0.50)(1.2) ÷ (0.80)² = 23.7 ÷ 0.64 = 37 N.

About 37 N, directed along the string toward the peg.

Worked example

Example 2: The maximum speed on a flat curve

A highway ramp is a flat curve of radius 60 m. Find the maximum speed without skidding (a) on dry pavement with μs = 0.80 and (b) on ice with μs = 0.10.

Vertical: FN = mg. Horizontal: static friction supplies the centripetal force, and it can be at most μs FN = μs mg.
mv² ÷ r ≤ μs mg, so v(max) = √(μs g r). The mass cancels.
(a) v = √(0.80 × 9.8 × 60) = √470 = 21.7 m/s (78 km/h).
(b) v = √(0.10 × 9.8 × 60) = √58.8 = 7.7 m/s (28 km/h).

(a) 22 m/s (about 78 km/h) (b) 7.7 m/s (about 28 km/h). On ice, the safe speed drops to about a third. Why static friction? The tires roll, so their contact patches are not sliding sideways on the road.

A car on a flat curve, seen from above. Static friction from the road points toward the centre. If friction is too small (ice), the car carries on more nearly along the tangent: it is not thrown outward.

Worked example

Example 3: The conical pendulum

A 0.20 kg ball on a 1.0 m string swings in a horizontal circle, with the string at 30° to the vertical. Find (a) the tension, (b) the speed and (c) the period.

Radius: r = L sin θ = 1.0 × sin 30° = 0.50 m.
Vertical (a = 0): FT cos 30° = mg, so FT = 0.20 × 9.8 ÷ 0.866 = 2.26 N.
Horizontal (toward the centre): FT sin 30° = mv² ÷ r.
Divide the two equations: tan θ = v² ÷ (rg), so v = √(rg tan θ) = √(0.50 × 9.8 × 0.577) = 1.68 m/s.
T = 2πr ÷ v = 2π(0.50) ÷ 1.68 = 1.87 s.

(a) 2.3 N (b) 1.7 m/s (c) 1.9 s. The mass cancels in the speed and the period: a heavier ball at the same angle goes around at the same rate.

A conical pendulum. The tension has two jobs: its vertical component holds the ball up (FT cos θ = mg), and its horizontal component, pointing toward the centre of the circle, supplies the centripetal force.

Worked example

Example 4: The rotor ride

In a rotor ride, riders stand against the wall of a cylinder of radius 2.5 m. When it spins fast enough, the floor drops away and they stay stuck to the wall. If μs = 0.40 between a rider and the wall, find the minimum speed and the longest period.

The wall's normal force points toward the centre and supplies the centripetal force: FN = mv² ÷ r.
Static friction (up the wall) must hold the rider's weight: μs FN ≥ mg, so μs mv² ÷ r ≥ mg.
v ≥ √(gr ÷ μs) = √(9.8 × 2.5 ÷ 0.40) = √61.25 = 7.8 m/s.
T = 2πr ÷ v = 2π(2.5) ÷ 7.83 = 2.0 s.

Speed at least 7.8 m/s, so the period must be 2.0 s or less. The rider's mass cancels, so everyone sticks at the same rate.

Just for fun

What did the string say to the ball on the end of it?

"Stick with me and we'll go places." (Around and around, mostly.)

Hands-on activity

The coin on the spinning plate

You need: a lazy Susan, a turntable, or a sturdy plate on an upturned bowl that can spin, three identical coins, masking tape, a ruler and a stopwatch.

Put coins at 4 cm, 8 cm and 12 cm from the centre.
Spin very slowly and steadily, then a little faster each time. Which coin slides off first? Explain using a = 4π²r ÷ T²: at the same period, the farthest coin needs the most friction per kilogram.
Find the slowest period at which the outer coin slides: count turns in 10 s just as it starts to slip. Then μs ≈ 4π²r ÷ (gT²).
Check your μs another way: tilt the plate (without spinning) and find the slipping angle, μs = tan θ (from Inclined Planes). Do the two values agree?
Stack two coins at the outer position. Does it change when they slide? (It should not: the mass cancels.)

Your turn

A 1200 kg car rounds a flat curve of radius 40 m at 15 m/s. What centripetal force is needed, and what supplies it?

Find the maximum speed on a flat curve of radius 40 m if μs = 0.50.

A 0.20 kg ball on a 0.80 m string moves in a horizontal circle on a frictionless table with a period of 0.50 s. Find the tension.

Name what supplies the centripetal force for: (a) the Moon orbiting Earth, (b) clothes in a washing machine's spin cycle, (c) a ball whirled on a string.

A coin 10 cm from the centre of a turntable just starts to slide when the turntable reaches 60 rpm. Find μs.

A car's speed on a curve doubles. By what factor must the friction force increase?

More practice: warm-up

Should you draw a separate "centripetal force" arrow on a free-body diagram?

Write Fc in terms of m, v and r.

Which force keeps a car on a flat curve?

If the string breaks on a whirled ball, which way does the ball go?

Does a heavier car have a lower maximum speed on the same flat curve?

More practice: on your own

A 0.10 kg conical pendulum has a 0.60 m string at 20° to the vertical. Find the radius, the speed, the period and the tension.

A rotor ride has a radius of 3.0 m and μs = 0.35. Find the minimum speed and the longest period.

On a wet day, μs = 0.45 on a flat curve of radius 80 m. Find the maximum safe speed in km/h.

A string breaks at 50 N. A 0.40 kg ball on 0.90 m of it is whirled in a horizontal circle on a frictionless table. Find the greatest speed and frequency.

A car travels at 20 m/s on a flat road with μs = 0.60. What is the smallest radius of curve it can take?

Challenge

Show that the period of a conical pendulum is T = 2π√(L cos θ ÷ g). Check it with Example 3.

A 0.25 kg puck slides in a circle of radius 0.50 m on a frictionless table. Its string passes through a hole at the centre to a 1.0 kg mass hanging at rest below the table. Find the puck's speed.

A car rounds a flat curve of radius 50 m at 15 m/s while braking at 3.0 m/s². What is the smallest μs that prevents skidding?

Check your understanding

Quiz

A car rounds a flat curve at a steady speed. The force that keeps it on the curve is:

The maximum speed on a flat curve depends on:

A 2.0 kg object moves in a circle of radius 0.50 m at 3.0 m/s. The net force on it is:

In a conical pendulum, the centripetal force is supplied by:

In a rotor ride, what holds the riders up against gravity?

The speed on a flat curve is halved. The friction force needed becomes:

For parents and teachers

Big idea: Centripetal force is the net force toward the centre, mv²/r (B2.6, B3.3), always supplied by real forces: tension, friction, normal force or gravity.

How to help: For any circular-motion situation (a car ramp, a playground spinner, a washing machine), ask your teen: "What is pulling or pushing it toward the centre?" If the answer is "centripetal force", ask them to name the real force behind it.

Common difficulties:

Adding an extra "Fc" or an outward "centrifugal force" arrow to the FBD.
Using kinetic friction for a rolling car on a curve (it is static).
Using the string length instead of the radius for a conical pendulum (r = L sin θ).
Forgetting that the mass cancels in many results.