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 = 0.6 mvmax​ = ?

A 1.8 kg stone is whirled in a horizontal circle of radius 0.6 m on a smooth, level table, held by a string that breaks at a tension of 90 N. Find the highest speed vmax the stone can have.

m = 1.8 kgr = 0.6 mTmax = 90 N
vmax = m/s
2.Easy
r = 60 m15 m/sac​ = ?

A 1500 kg car drives around a flat curve of radius 60 m at a constant 15 m/s. Find its centripetal acceleration ac.

m = 1500 kgv = 15 m/sr = 60 m
ac = m/s²
3.Medium
r = 100 mvmax​ = ?

A car takes a flat curve of radius 100 m. The coefficient of static friction between tyres and road is 0.5. Use g = 9.81 m/s². Find the highest speed vmax at which the car can take the curve without skidding.

r = 100 mμs = 0.5g = 9.81 m/s²
vmax = m/s
4.Medium
r = 150 mvmax​ = ?

A car takes a flat curve of radius 150 m. The coefficient of static friction between tyres and road is 0.5. Use g = 9.81 m/s². Find the highest speed vmax at which the car can take the curve without skidding.

r = 150 mμs = 0.5g = 9.81 m/s²
vmax = m/s
5.Hard
r = 0.7 mv = ?

A ball on a string 0.7 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 = 0.7 mg = 9.81 m/s²
vmin = m/s
6.Hard
r = 0.7 mv = 4 m/s

A 1.2 kg ball on a string 0.7 m long is swung in a vertical circle. At the lowest point it moves at 4 m/s. Use g = 9.81 m/s². Find the tension T in the string at the lowest point.

m = 1.2 kgr = 0.7 mv = 4 m/sg = 9.81 m/s²
T = N

Answer key

  1. 1. vmax = 5.477 m/s
  2. 2. ac = 3.75 m/s²
  3. 3. vmax = 22.15 m/s
  4. 4. vmax = 27.12 m/s
  5. 5. vmin = 2.62 m/s
  6. 6. T = 39.2 N

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