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 = 160 m30 m/sac​ = ?

A 1800 kg car drives around a flat curve of radius 160 m at a constant 30 m/s. Find its centripetal acceleration ac.

m = 1800 kgv = 30 m/sr = 160 m
ac = m/s²
2.Easy
r = 2 mvmax​ = ?

A 0.5 kg stone is whirled in a horizontal circle of radius 2 m on a smooth, level table, held by a string that breaks at a tension of 10 N. Find the highest speed vmax the stone can have.

m = 0.5 kgr = 2 mTmax = 10 N
vmax = m/s
3.Medium
L = 1.1 m30°

A small ball on a string of length 1.1 m moves in a horizontal circle so that the string makes a constant angle of 30° with the vertical (a conical pendulum). Use g = 9.81 m/s². Find the time T for one revolution.

L = 1.1 mθ = 30°g = 9.81 m/s²
T = s
4.Medium
r = 125 mvmax​ = ?

A car takes a flat curve of radius 125 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 = 125 mμs = 0.5g = 9.81 m/s²
vmax = m/s
5.Hard
r = 0.9 m12 m/svtop​ = ?

A 0.7 kg ball on a light string 0.9 m long is swung in a vertical circle. At the lowest point its speed is 12 m/s. Ignore air resistance and use g = 9.81 m/s². Find the tension Ttop in the string at the highest point.

m = 0.7 kgr = 0.9 mvbottom = 12 m/sg = 9.81 m/s²
Ttop = N
6.Hard
R = 9.5 mh = 30.4 m

A 2 kg cart is released from rest at a height of 30.4 m on a frictionless track and runs through a vertical loop of radius 9.5 m. Use g = 9.81 m/s². Find the normal force Nbottom of the track on the cart at the bottom of the loop.

m = 2 kgh = 30.4 mR = 9.5 mg = 9.81 m/s²
Nbottom = N

Answer key

  1. 1. ac = 5.625 m/s²
  2. 2. vmax = 6.325 m/s
  3. 3. T = 1.958 s
  4. 4. vmax = 24.76 m/s
  5. 5. Ttop = 77.66 N
  6. 6. Nbottom = 145.2 N

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