Rigid-body rotation worksheet with answers

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

For teachers: free to print, copy and use in class. Every new worksheet has new values and its own answer key. A link to ExamProblems on your course page is appreciated.

Name: ____________________   Date: ____________

1.Easy
k = 0.35 m0 → 1100 rpm20 kg

A flywheel of mass 20 kg with a radius of gyration of 0.35 m is accelerated uniformly from rest to 1100 rpm. This takes 6 s. Ignoring friction, find the constant torque T needed.

m = 20 kgk = 0.35 mn1 = 0 rpmn2 = 1100 rpmt = 6 s
T = N·m
2.Easy
k = 0.15 m600 → 1400 rpm160 kg

A flywheel of mass 160 kg with a radius of gyration of 0.15 m is accelerated uniformly from 600 rpm to 1400 rpm. This takes 28 s. Ignoring friction, find the constant torque T needed.

m = 160 kgk = 0.15 mn1 = 600 rpmn2 = 1400 rpmt = 28 s
T = N·m
3.Medium
n = 900 rpmTb​ = 40 N·mI = 5 kg·m²

A flywheel with a moment of inertia of 5 kg·m² spins at 900 rpm. A brake applies a constant friction torque of 40 N·m. How long t does it take to stop?

I = 5 kg·m²n = 900 rpmTb = 40 N·m
t = s
4.Medium
4.5 kgaαM = 13 kgR = 0.45 m

A 4.5 kg block hangs from a light cord wrapped around a pulley, a uniform solid disc of mass 13 kg and radius 0.45 m that turns on a frictionless horizontal axle. The block is released from rest. Use g = 9.81 m/s². Find the speed v of the block after it has fallen 5 m.

m = 4.5 kgM = 13 kgR = 0.45 mg = 9.81 m/s²h = 5 m
v = m/s
5.Hard
0.85 m0.3 mn1​ = 35 rpm

A person stands on a frictionless turntable and holds a 3 kg dumbbell in each hand, 0.85 m from the axis. The person and turntable alone have a moment of inertia of 4.5 kg·m² about the axis. They turn at 35 rpm, then pull the dumbbells in to 0.3 m from the axis. Treat the dumbbells as point masses. Find the new speed n2 in rpm.

I0 = 4.5 kg·m²m = 2 × 3 kgr1 = 0.85 mr2 = 0.3 mn1 = 35 rpm
n2 = rpm
6.Hard
7.5 m/sω

A thin hoop of mass 17 kg rolls without slipping at 7.5 m/s. What percentage p of its total kinetic energy is rotational?

m = 17 kgv = 7.5 m/sI = m·r²
p = %

Answer key

  1. 1. T = 47.04 N·m
  2. 2. T = 10.77 N·m
  3. 3. t = 11.78 s
  4. 4. v = 6.335 m/s
  5. 5. n2 = 61.35 rpm
  6. 6. p = 50 %

Full step-by-step solutions: practise this topic on examproblems.com.