← Centripetal force

Solve for the speed

The friction between a car's tires and a flat curve can supply only so much inward force. That maximum inward force sets a maximum steady speed for the curve.

An 800 kg car approaches a flat circular curve of radius 50 m. The road can supply at most 6400 N of inward force on the tires. What is the car's maximum steady speed for the curve?

You have seen the equation: F = mv²/r. This time the unknown is the speed, v.

Getting v out takes one extra step, because the speed is squared.

Step 1: the values, unknown markedquantityvalueF6400 Nr50 mm800 kgv?
Step 1: write down the values, and mark the unknown.

Step 1: write down the values in the question, and mark the unknown.

Step 2: write down the equation, and make v² the subject: v² = Fr/m.

Step 3: substitute in the values, and calculate v².

Step 4: take the square root: v = √(Fr/m). The result line carries its unit, m/s.

Steps 2 to 4: subject, substitution, square rootsteplinemake v² the subjectv² = Fr/msubstitutev² = (6400 × 50) / 800calculatev² = 400square rootv = 20 m/s
The extra step: calculate v², then take the square root.
The question gives the available inward force directly. Put that force in as F, and the v that comes out is the fastest steady speed that force can hold.

An 800 kg car approaches a flat circular curve of radius 50 m. The road can supply at most 6400 N of inward force on the tires. Calculate the car's maximum steady speed for the curve.

Write down the values:

F = 6400 N, r = 50 m, m = 800 kg, v = ?

Make v² the subject:

v² = Fr/m

Substitute in the values, and calculate, then take the square root:

v² = (6400 × 50) / 800 = 400, so v = √400 = 20

The car's maximum steady speed for the curve is 20 m/s.

Rain lowers the maximum inward force on the same curve to 1600 N for the same 800 kg car. Calculate its maximum steady speed now.

Write down the values:

F = 1600 N, r = 50 m, m = 800 kg, v = ?

Make v² the subject:

v² = Fr/m

Substitute in the values, and calculate, then take the square root:

v² = (1600 × 50) / 800 = 100, so v = √100 = 10

The car's maximum steady speed is now 10 m/s. A quarter of the force, half the speed.

maximum steady speed (m/s)20 m/sdry, 6400 N10 m/swet, 1600 N
A quarter of the available force, half the maximum speed.

A 1.5 kg tetherball swings on its rope in a horizontal circle of radius 2 m. The rope can supply at most 27 N of inward force. Enter the ball's maximum steady speed in m/s.

A 2 kg remote-control car rounds a flat circle of radius 4 m. The floor can supply at most 32 N of inward friction force. Enter the car's maximum steady speed in m/s.

A 2 kg model plane flies a horizontal circle on an 18 m control line. The line can supply at most 16 N of inward force. Enter the plane's maximum steady speed in m/s.

Your turn

A 2 kg instrument package rides a spinning test arm of radius 10 m. The arm can supply at most 45 N of inward force. Enter the package's greatest steady speed in m/s.

A 3 kg training ball swings on a 2.5 m rope. The rope can supply at most 30 N of inward force. Enter the ball's greatest steady speed in m/s.

A delivery van of mass 800 kg approaches a flat circular loop road of radius 100 m. On the day's rain-soaked surface, the road can supply at most 5000 N of inward force on the tires. Enter the van's maximum steady speed for the loop in m/s.