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
Or = 0.1 mL = 1.5 m1.6 m0.5 kg3.5 kg

A pendulum is made of a uniform slender rod (0.5 kg, 1.5 m long) pivoted at its top end O, with a thin solid disc of mass 3.5 kg and radius 0.1 m fixed to its lower end. The disc swings in its own plane and its centre is 1.6 m below O. Find the mass moment of inertia IO of the pendulum about O.

mrod = 0.5 kgL = 1.5 mmdisc = 3.5 kgr = 0.1 md = 1.6 m
IO = kg·m²
2.Easy
L = 2.2 mGOd = 0.85 mm = 11.5 kg

A uniform slender rod of mass 11.5 kg and length 2.2 m rotates about an axis O perpendicular to the rod, 0.85 m from its centre G. Find its mass moment of inertia IO about that axis.

m = 11.5 kgL = 2.2 md = 0.85 m
IO = kg·m²
3.Medium
5.5 kgaαM = 19 kgR = 0.1 m

A 5.5 kg block hangs from a light cord wrapped around a pulley, a uniform solid disc of mass 19 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 = 5.5 kgM = 19 kgR = 0.1 mg = 9.81 m/s²
T = N
4.Medium
v = ?L = 1.2 mm = 7.5 kg

A uniform slender rod of mass 7.5 kg and length 1.2 m is pinned at one end and released from rest in a horizontal position. 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 = 7.5 kgL = 1.2 mg = 9.81 m/s²
v = m/s
5.Hard
8 m/sω

A thin hoop of mass 1 kg rolls without slipping at 8 m/s. Find its rotational kinetic energy Erot.

m = 1 kgv = 8 m/sI = m·r²
Erot = J
6.Hard
460 m/s45 g0.69 mL = 1.1 m6 kg

A uniform slender rod of mass 6 kg and length 1.1 m hangs at rest from a frictionless pivot at its top end. A 45 g bullet moving horizontally at 460 m/s hits the rod 0.69 m below the pivot and stays embedded. Find the angular velocity ω of the rod just after the impact.

M = 6 kgL = 1.1 mm = 45 gv = 460 m/sd = 0.69 m
ω = rad/s

Answer key

  1. 1. IO = 9.353 kg·m²
  2. 2. IO = 12.95 kg·m²
  3. 3. T = 34.17 N
  4. 4. v = 5.943 m/s
  5. 5. Erot = 32 J
  6. 6. ω = 5.85 rad/s

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