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 = 0.75 mv45 rpm

A small object sits 75 cm from the axis of a disc that turns at a steady 45 revolutions per minute. Find its centripetal acceleration ac.

n = 45 rpmr = 0.75 m
ac = m/s²
2.Easy
r = 1.1 mv200 rpm

A small object sits 110 cm from the axis of a disc that turns at a steady 200 revolutions per minute. Find its centripetal acceleration ac.

n = 200 rpmr = 1.1 m
ac = m/s²
3.Medium
r = 3.9 mn = ?

In a fairground rotor, riders stand against the wall of a vertical cylinder of radius 3.9 m. When it spins fast enough the floor drops away and friction holds them up. The coefficient of static friction is 0.7. Use g = 9.81 m/s². Find the smallest rotation rate n in revolutions per minute.

r = 3.9 mμs = 0.7g = 9.81 m/s²
n = rpm
4.Medium
L = 0.5 m35°

A small ball on a string of length 0.5 m moves in a horizontal circle so that the string makes a constant angle of 35° with the vertical (a conical pendulum). Use g = 9.81 m/s². Find the time T for one revolution.

L = 0.5 mθ = 35°g = 9.81 m/s²
T = s
5.Hard
R = 0.5 mh = 2.1 m

A 0.5 kg cart is released from rest at a height of 2.1 m on a frictionless track and runs through a vertical loop of radius 0.5 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 = 0.5 kgh = 2.1 mR = 0.5 mg = 9.81 m/s²
Nbottom = N
6.Hard
r = 0.5 mv = 9 m/s

A 1.6 kg ball on a string 0.5 m long is swung in a vertical circle. At the lowest point it moves at 9 m/s. Use g = 9.81 m/s². Find the tension T in the string at the lowest point.

m = 1.6 kgr = 0.5 mv = 9 m/sg = 9.81 m/s²
T = N

Answer key

  1. 1. ac = 16.65 m/s²
  2. 2. ac = 482.5 m/s²
  3. 3. n = 18.1 rpm
  4. 4. T = 1.284 s
  5. 5. Nbottom = 46.11 N
  6. 6. T = 274.9 N

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