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 = 2.1 mGOd = 0.95 mm = 3.5 kg

A uniform slender rod of mass 3.5 kg and length 2.1 m rotates about an axis O perpendicular to the rod, 0.95 m from its centre G. Find its mass moment of inertia IO about that axis.

m = 3.5 kgL = 2.1 md = 0.95 m
IO = kg·m²
2.Easy
Or = 0.15 mL = 0.7 m0.85 m1 kg10 kg

A pendulum is made of a uniform slender rod (1 kg, 0.7 m long) pivoted at its top end O, with a thin solid disc of mass 10 kg and radius 0.15 m fixed to its lower end. The disc swings in its own plane and its centre is 0.85 m below O. Find the mass moment of inertia IO of the pendulum about O.

mrod = 1 kgL = 0.7 mmdisc = 10 kgr = 0.15 md = 0.85 m
IO = kg·m²
3.Medium
n = 400 rpmTb​ = 50 N·mI = 9 kg·m²

A flywheel with a moment of inertia of 9 kg·m² spins at 400 rpm. A brake applies a constant friction torque of 50 N·m. How many revolutions N does the flywheel make before it stops?

I = 9 kg·m²n = 400 rpmTb = 50 N·m
N = rev
4.Medium
v = ?3.5 kgL = 2.1 mm = 9.5 kg

A uniform slender rod of mass 9.5 kg and length 2.1 m is pinned at one end and released from rest in a horizontal position. A small ball of mass 3.5 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 = 9.5 kgL = 2.1 mmball = 3.5 kgg = 9.81 m/s²
v = m/s
5.Hard
θ = 25°a

A solid cylinder rolls without slipping down a slope inclined at 25° to the horizontal. Use g = 9.81 m/s². Find the smallest coefficient of static friction μs that prevents it from slipping.

θ = 25°I = ½·m·r²g = 9.81 m/s²
μs =
6.Hard
500 m/s15 g0.61 mL = 1.4 m10 kgφ

A uniform slender rod of mass 10 kg and length 1.4 m hangs at rest from a frictionless pivot at its top end. A 15 g bullet moving horizontally at 500 m/s hits the rod 0.61 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 = 10 kgL = 1.4 mm = 15 gv = 500 m/sd = 0.61 mg = 9.81 m/s²
φ = °

Answer key

  1. 1. IO = 4.445 kg·m²
  2. 2. IO = 7.501 kg·m²
  3. 3. N = 25.13 rev
  4. 4. v = 7.141 m/s
  5. 5. μs = 0.1554
  6. 6. φ = 12.39 °

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