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 = 0.2 mL = 1 m1.2 m3 kg9.5 kg

A pendulum is made of a uniform slender rod (3 kg, 1 m long) pivoted at its top end O, with a solid sphere of mass 9.5 kg and radius 0.2 m fixed to its lower end. The centre of the sphere is 1.2 m below O. Find the mass moment of inertia IO of the pendulum about O.

mrod = 3 kgL = 1 mmsphere = 9.5 kgr = 0.2 md = 1.2 m
IO = kg·m²
2.Easy
Or = 0.15 mL = 1.2 m1.35 m2.5 kg4.5 kg

A pendulum is made of a uniform slender rod (2.5 kg, 1.2 m long) pivoted at its top end O, with a solid sphere of mass 4.5 kg and radius 0.15 m fixed to its lower end. The centre of the sphere is 1.35 m below O. Find the mass moment of inertia IO of the pendulum about O.

mrod = 2.5 kgL = 1.2 mmsphere = 4.5 kgr = 0.15 md = 1.35 m
IO = kg·m²
3.Medium
L = 2.2 mm = 18.5 kgα

A uniform slender rod of mass 18.5 kg and length 2.2 m is pinned at one end and held horizontal. It is released from rest. Use g = 9.81 m/s². Find the acceleration atip of the free end just after release.

m = 18.5 kgL = 2.2 mg = 9.81 m/s²
atip = m/s²
4.Medium
6.5 kgaαM = 16 kgR = 0.5 m

A 6.5 kg block hangs from a light cord wrapped around a pulley, a uniform solid disc of mass 16 kg and radius 0.5 m that turns on a frictionless horizontal axle. The block is released from rest. Use g = 9.81 m/s². Find the speed v of the block after it has fallen 1.5 m.

m = 6.5 kgM = 16 kgR = 0.5 mg = 9.81 m/s²h = 1.5 m
v = m/s
5.Hard
θ = 35°as = 7.5 m

A solid sphere rolls without slipping down a slope inclined at 35° to the horizontal. Use g = 9.81 m/s². It starts from rest. How long t does it take to roll 7.5 m down the slope?

θ = 35°I = (2/5)·m·r²g = 9.81 m/s²s = 7.5 m
t = s
6.Hard
100 m/s45 g1.68 mL = 1.7 m3.5 kgφ

A uniform slender rod of mass 3.5 kg and length 1.7 m hangs at rest from a frictionless pivot at its top end. A 45 g bullet moving horizontally at 100 m/s hits the rod 1.68 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 = 3.5 kgL = 1.7 mm = 45 gv = 100 m/sd = 1.68 mg = 9.81 m/s²
φ = °

Answer key

  1. 1. IO = 14.83 kg·m²
  2. 2. IO = 9.442 kg·m²
  3. 3. atip = 14.71 m/s²
  4. 4. v = 3.632 m/s
  5. 5. t = 1.932 s
  6. 6. φ = 43.36 °

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