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
k = 0.2 m0 → 1700 rpm30 kg

A flywheel of mass 30 kg with a radius of gyration of 0.2 m is accelerated uniformly from rest to 1700 rpm. This takes 24 s. Ignoring friction, find the constant torque T needed.

m = 30 kgk = 0.2 mn1 = 0 rpmn2 = 1700 rpmt = 24 s
T = N·m
2.Easy
Or = 0.25 mL = 1 m1.25 m2 kg7 kg

A pendulum is made of a uniform slender rod (2 kg, 1 m long) pivoted at its top end O, with a thin solid disc of mass 7 kg and radius 0.25 m fixed to its lower end. The disc swings in its own plane and its centre is 1.25 m below O. Find the mass moment of inertia IO of the pendulum about O.

mrod = 2 kgL = 1 mmdisc = 7 kgr = 0.25 md = 1.25 m
IO = kg·m²
3.Medium
15 kgaαM = 18.5 kgR = 0.3 m

A 15 kg block hangs from a light cord wrapped around a pulley, a uniform solid disc of mass 18.5 kg and radius 0.3 m that turns on a frictionless horizontal axle. The block is released from rest. Use g = 9.81 m/s². Find the tension T in the cord.

m = 15 kgM = 18.5 kgR = 0.3 mg = 9.81 m/s²
T = N
4.Medium
9 kgaαM = 15.5 kgR = 0.1 m

A 9 kg block hangs from a light cord wrapped around a pulley, a uniform solid disc of mass 15.5 kg and radius 0.1 m that turns on a frictionless horizontal axle. The block is released from rest. Use g = 9.81 m/s². Find the tension T in the cord.

m = 9 kgM = 15.5 kgR = 0.1 mg = 9.81 m/s²
T = N
5.Hard
380 m/s25 g0.53 mL = 0.7 m9.5 kgφ

A uniform slender rod of mass 9.5 kg and length 0.7 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 0.53 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 = 9.5 kgL = 0.7 mm = 25 gv = 380 m/sd = 0.53 mg = 9.81 m/s²
φ = °
6.Hard
9 m/sω

A thin hoop of mass 16.5 kg rolls without slipping at 9 m/s. Find its rotational kinetic energy Erot.

m = 16.5 kgv = 9 m/sI = m·r²
Erot = J

Answer key

  1. 1. T = 8.901 N·m
  2. 2. IO = 11.82 kg·m²
  3. 3. T = 56.13 N
  4. 4. T = 40.85 N
  5. 5. φ = 41.26 °
  6. 6. Erot = 668.3 J

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