A 1000 kg car approaches a flat curve of radius 90 m at a steady 15 m/s. Its tires can supply at most 3000 N of friction.
Will it make the curve?
The question is not which force does the job. It is whether the available force is big enough for the job.
Two numbers decide it. The required centripetal force comes from F = mv²/r: what the circle demands. The maximum available force is stated by the scenario: what the supplier can give. Compare them.
First, the requirement. For the car, F = mv²/r with m = 1000 kg, v = 15 m/s, r = 90 m.
15 × 15 = 225. Then 1000 × 225 = 225000, and 225000 ÷ 90 = 2500. The turn demands 2500 N toward the center.
Next, the supply. The tires can give at most 3000 N of friction.
Now compare. The available 3000 N is at least the required 2500 N, so friction can supply everything the circle demands. The car holds its circular path.
Rain changes the supply, not the demand. On the wet road the tires can give at most 2000 N, and the demand is still 2500 N. The available force falls short, so the car cannot hold the circle.
A 2000 kg dump truck takes a flat curve of radius 50 m at a steady 10 m/s. Its tires can supply at most 3500 N of friction. Does the truck hold its circular path?
Write down the values:
m = 2000 kg, v = 10 m/s, r = 50 m, maximum available force 3500 N, F = ?
Write down the equation:
F = mv²/r
Substitute in the values, and calculate:
F = 2000 × 10² ÷ 50
10 × 10 = 100
2000 × 100 = 200000
200000 ÷ 50 = 4000
F = 4000 N
Compare: the required 4000 N is more than the available 3500 N.
No. The curve demands 4000 N and the tires can give only 3500 N, so the truck cannot hold its circular path.
An 800 kg buggy rounds a flat curve of radius 48 m at a steady 12 m/s. Enter the centripetal force its circle requires, in N.
A 1500 kg camper van takes a flat bend of radius 60 m at a steady 10 m/s. Its tires can supply at most 2200 N of friction. Can the van hold its circular path?
A 0.2 kg metal nut whirls on a test cord at a steady 4 m/s in a circle of radius 0.4 m. The cord can supply at most 25 N of tension. Does the nut hold its circular path?
Write down the values:
m = 0.2 kg, v = 4 m/s, r = 0.4 m, maximum available force 25 N, F = ?
Write down the equation:
F = mv²/r
Substitute in the values, and calculate:
F = 0.2 × 4² ÷ 0.4
4 × 4 = 16
0.2 × 16 = 3.2
3.2 ÷ 0.4 = 8
F = 8 N
Compare: the available 25 N is at least the required 8 N.
Yes. The circle demands 8 N and the cord can give up to 25 N, so the nut holds its circular path.
Minna works out that her go-kart's circle requires 1800 N toward the center, and that its tires can supply at most 2500 N of friction. She says: 'The go-kart keeps its circular path.' Is she correct?
A 1200 kg sports car takes a flat curve of radius 80 m at a steady 20 m/s. Its tires can supply at most 5500 N of friction. Enter the centripetal force the curve requires, in N.
The table gives each vehicle's required centripetal force and the maximum force available to it on its flat curve. Which vehicle keeps its circular path?
A 0.5 kg juggling ball whirls at a steady 12 m/s in a circle of radius 2 m, on a cord that can supply at most 30 N of tension. Enter how many newtons the required centripetal force exceeds the cord's maximum by.