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.

New worksheetPractise online with hints

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
L = 1.8 mGOd = 0.8 mm = 3.5 kg

A uniform slender rod of mass 3.5 kg and length 1.8 m rotates about an axis O perpendicular to the rod, 0.8 m from its centre G. Find its mass moment of inertia IO about that axis.

m = 3.5 kgL = 1.8 md = 0.8 m
IO = kg·m²
2.Easy
k = 0.6 m1000 → 1900 rpm110 kg

A flywheel of mass 110 kg with a radius of gyration of 0.6 m is accelerated uniformly from 1000 rpm to 1900 rpm. The driving torque is a constant 150 N·m and friction can be ignored. How long t does this take?

m = 110 kgk = 0.6 mn1 = 1000 rpmn2 = 1900 rpmT = 150 N·m
t = s
3.Medium
v = ?5 kgL = 1.4 mm = 1 kg

A uniform slender rod of mass 1 kg and length 1.4 m is pinned at one end and released from rest in a horizontal position. A small ball of mass 5 kg is fixed to its free end. Ignore friction and use g = 9.81 m/s². Find the speed v of the free end as the rod swings through the vertical position.

m = 1 kgL = 1.4 mmball = 5 kgg = 9.81 m/s²
v = m/s
4.Medium
L = 2.4 mm = 13.5 kgα

A uniform slender rod of mass 13.5 kg and length 2.4 m is pinned at one end and held horizontal. It is released from rest. Use g = 9.81 m/s². Find its angular acceleration α just after release.

m = 13.5 kgL = 2.4 mg = 9.81 m/s²
α = rad/s²
5.Hard
100 m/s50 g1.24 mL = 1.6 m6.5 kg

A uniform slender rod of mass 6.5 kg and length 1.6 m hangs at rest from a frictionless pivot at its top end. A 50 g bullet moving horizontally at 100 m/s hits the rod 1.24 m below the pivot and stays embedded. Find the angular velocity ω of the rod just after the impact.

M = 6.5 kgL = 1.6 mm = 50 gv = 100 m/sd = 1.24 m
ω = rad/s
6.Hard
θ = 30°a

A solid sphere rolls without slipping down a slope inclined at 30° to the horizontal. Use g = 9.81 m/s². Find the acceleration a of its centre.

θ = 30°I = (2/5)·m·r²g = 9.81 m/s²
a = m/s²

Answer key

  1. 1. IO = 3.185 kg·m²
  2. 2. t = 24.88 s
  3. 3. v = 5.322 m/s
  4. 4. α = 6.131 rad/s²
  5. 5. ω = 1.103 rad/s
  6. 6. a = 3.504 m/s²

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