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.

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1.Easy
Or = 0.15 mL = 0.9 m1.05 m4 kg7.5 kg

A pendulum is made of a uniform slender rod (4 kg, 0.9 m long) pivoted at its top end O, with a solid sphere of mass 7.5 kg and radius 0.15 m fixed to its lower end. The centre of the sphere is 1.05 m below O. Find the mass moment of inertia IO of the pendulum about O.

mrod = 4 kgL = 0.9 mmsphere = 7.5 kgr = 0.15 md = 1.05 m
IO = kg·m²
2.Easy
Or = 0.1 mL = 0.7 m0.8 m3.5 kg8 kg

A pendulum is made of a uniform slender rod (3.5 kg, 0.7 m long) pivoted at its top end O, with a thin solid disc of mass 8 kg and radius 0.1 m fixed to its lower end. The disc swings in its own plane and its centre is 0.8 m below O. Find the mass moment of inertia IO of the pendulum about O.

mrod = 3.5 kgL = 0.7 mmdisc = 8 kgr = 0.1 md = 0.8 m
IO = kg·m²
3.Medium
ω = ?3 kgL = 1.4 mm = 3.5 kg

A uniform slender rod of mass 3.5 kg and length 1.4 m is pinned at one end and released from rest in a horizontal position. A small ball of mass 3 kg is fixed to its free end. Ignore friction and use g = 9.81 m/s². Find the angular velocity ω of the rod as it swings through the vertical position.

m = 3.5 kgL = 1.4 mmball = 3 kgg = 9.81 m/s²
ω = rad/s
4.Medium
0 → 1900 rpmI = 38.5 kg·m²

A motor brings a flywheel with a moment of inertia of 38.5 kg·m² from rest up to 1900 rpm in 26 s. Ignoring losses, find the average power Pavg it delivers, in kW.

I = 38.5 kg·m²n = 1900 rpmt = 26 s
Pavg = kW
5.Hard
260 m/s10 g0.84 mL = 1.1 m5.5 kgφ

A uniform slender rod of mass 5.5 kg and length 1.1 m hangs at rest from a frictionless pivot at its top end. A 10 g bullet moving horizontally at 260 m/s hits the rod 0.84 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 = 5.5 kgL = 1.1 mm = 10 gv = 260 m/sd = 0.84 mg = 9.81 m/s²
φ = °
6.Hard
3.5 m/sω

A hollow sphere of mass 8 kg rolls without slipping at 3.5 m/s. Find its rotational kinetic energy Erot.

m = 8 kgv = 3.5 m/sI = (2/3)·m·r²
Erot = J

Answer key

  1. 1. IO = 9.416 kg·m²
  2. 2. IO = 5.732 kg·m²
  3. 3. ω = 3.997 rad/s
  4. 4. Pavg = 29.31 kW
  5. 5. φ = 15.42 °
  6. 6. Erot = 32.67 J

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