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
L = 0.4 mGOd = 0.15 mm = 2 kg

A uniform slender rod of mass 2 kg and length 0.4 m rotates about an axis O perpendicular to the rod, 0.15 m from its centre G. Find its mass moment of inertia IO about that axis.

m = 2 kgL = 0.4 md = 0.15 m
IO = kg·m²
2.Easy
k = 0.1 m500 → 1800 rpm200 kg

A flywheel of mass 200 kg with a radius of gyration of 0.1 m is accelerated uniformly from 500 rpm to 1800 rpm. The driving torque is a constant 11 N·m and friction can be ignored. How long t does this take?

m = 200 kgk = 0.1 mn1 = 500 rpmn2 = 1800 rpmT = 11 N·m
t = s
3.Medium
L = 1.7 mm = 1 kgα

A uniform slender rod of mass 1 kg and length 1.7 m is pinned at one end and held horizontal. It is released from rest. Use g = 9.81 m/s². Find the vertical force R that the pin exerts on the rod just after release.

m = 1 kgL = 1.7 mg = 9.81 m/s²
R = N
4.Medium
0 → 700 rpmI = 9 kg·m²

A motor brings a flywheel with a moment of inertia of 9 kg·m² from rest up to 700 rpm in 4 s. Ignoring losses, find the average power Pavg it delivers, in kW.

I = 9 kg·m²n = 700 rpmt = 4 s
Pavg = kW
5.Hard
0.85 m0.25 mn1​ = 60 rpm

A person stands on a frictionless turntable and holds a 2.5 kg dumbbell in each hand, 0.85 m from the axis. The person and turntable alone have a moment of inertia of 5 kg·m² about the axis. They turn at 60 rpm, then pull the dumbbells in to 0.25 m from the axis. Treat the dumbbells as point masses. Find the new speed n2 in rpm.

I0 = 5 kg·m²m = 2 × 2.5 kgr1 = 0.85 mr2 = 0.25 mn1 = 60 rpm
n2 = rpm
6.Hard
380 m/s25 g0.9 mL = 1.5 m8 kgφ

A uniform slender rod of mass 8 kg and length 1.5 m hangs at rest from a frictionless pivot at its top end. A 25 g bullet moving horizontally at 380 m/s hits the rod 0.9 m below the pivot and stays embedded. Use g = 9.81 m/s². Find the largest angle φ from the vertical through which the rod swings.

M = 8 kgL = 1.5 mm = 25 gv = 380 m/sd = 0.9 mg = 9.81 m/s²
φ = °

Answer key

  1. 1. IO = 0.07167 kg·m²
  2. 2. t = 24.75 s
  3. 3. R = 2.453 N
  4. 4. Pavg = 6.045 kW
  5. 5. n2 = 97.27 rpm
  6. 6. φ = 26.2 °

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