Circular motion worksheet with answers

Free printable Circular motion worksheet: 6 problems from easy to exam level, with an answer key. Print it or save it as a PDF.

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1.Easy
r = 140 m27 m/sF = ?

A 1200 kg car drives around a flat curve of radius 140 m at a constant 27 m/s. Find the inward (centripetal) force Fc needed.

m = 1200 kgv = 27 m/sr = 140 m
Fc = N
2.Easy
r = 0.8 mvmax​ = ?

A 0.2 kg stone is whirled in a horizontal circle of radius 0.8 m on a smooth, level table, held by a string that breaks at a tension of 200 N. Find the highest speed vmax the stone can have.

m = 0.2 kgr = 0.8 mTmax = 200 N
vmax = m/s
3.Medium
r = 70 mvmax​ = ?

A car takes a flat curve of radius 70 m. The coefficient of static friction between tyres and road is 0.4. Use g = 9.81 m/s². Find the highest speed vmax at which the car can take the curve without skidding.

r = 70 mμs = 0.4g = 9.81 m/s²
vmax = m/s
4.Medium
r = 30 mvmax​ = ?

A car takes a flat curve of radius 30 m. The coefficient of static friction between tyres and road is 0.8. Use g = 9.81 m/s². Find the highest speed vmax at which the car can take the curve without skidding.

r = 30 mμs = 0.8g = 9.81 m/s²
vmax = m/s
5.Hard
r = 1.3 mv = ?

A ball on a string 1.3 m long is swung in a vertical circle. Use g = 9.81 m/s². What is the smallest speed vmin at the top for which the string just stays tight?

r = 1.3 mg = 9.81 m/s²
vmin = m/s
6.Hard
r = 1.2 mv = 7.5 m/s

A 0.9 kg ball on a string 1.2 m long is swung in a vertical circle. At the lowest point it moves at 7.5 m/s. Use g = 9.81 m/s². Find the tension T in the string at the lowest point.

m = 0.9 kgr = 1.2 mv = 7.5 m/sg = 9.81 m/s²
T = N

Answer key

  1. 1. Fc = 6249 N
  2. 2. vmax = 28.28 m/s
  3. 3. vmax = 16.57 m/s
  4. 4. vmax = 15.34 m/s
  5. 5. vmin = 3.571 m/s
  6. 6. T = 51.02 N

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