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
L = 3 mGOd = 1.35 mm = 7.5 kg

A uniform slender rod of mass 7.5 kg and length 3 m rotates about an axis O perpendicular to the rod, 1.35 m from its centre G. Find its mass moment of inertia IO about that axis.

m = 7.5 kgL = 3 md = 1.35 m
IO = kg·m²
2.Easy
Or = 0.15 mL = 0.6 m0.75 m3.5 kg9 kg

A pendulum is made of a uniform slender rod (3.5 kg, 0.6 m long) pivoted at its top end O, with a solid sphere of mass 9 kg and radius 0.15 m fixed to its lower end. The centre of the sphere is 0.75 m below O. Find the mass moment of inertia IO of the pendulum about O.

mrod = 3.5 kgL = 0.6 mmsphere = 9 kgr = 0.15 md = 0.75 m
IO = kg·m²
3.Medium
ω = ?L = 1.7 mm = 2 kg

A uniform slender rod of mass 2 kg and length 1.7 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 angular velocity ω of the rod as it swings through the vertical position.

m = 2 kgL = 1.7 mg = 9.81 m/s²
ω = rad/s
4.Medium
4 kgaαM = 17 kgR = 0.35 m

A 4 kg block hangs from a light cord wrapped around a pulley, a uniform solid disc of mass 17 kg and radius 0.35 m that turns on a frictionless horizontal axle. The block is released from rest. Use g = 9.81 m/s². Find the angular acceleration α of the pulley.

m = 4 kgM = 17 kgR = 0.35 mg = 9.81 m/s²
α = rad/s²
5.Hard
θ = 30°vs = 6 m

A solid sphere starts from rest and rolls without slipping down a slope inclined at 30° to the horizontal. Use g = 9.81 m/s². Its radius is 0.3 m. Find its angular velocity ω after it has rolled 6 m along the slope.

θ = 30°s = 6 mI = (2/5)·m·r²g = 9.81 m/s²r = 0.3 m
ω = rad/s
6.Hard
140 m/s40 g0.66 mL = 1.4 m6.5 kgφ

A uniform slender rod of mass 6.5 kg and length 1.4 m hangs at rest from a frictionless pivot at its top end. A 40 g bullet moving horizontally at 140 m/s hits the rod 0.66 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 = 6.5 kgL = 1.4 mm = 40 gv = 140 m/sd = 0.66 mg = 9.81 m/s²
φ = °

Answer key

  1. 1. IO = 19.29 kg·m²
  2. 2. IO = 5.564 kg·m²
  3. 3. ω = 4.161 rad/s
  4. 4. α = 8.969 rad/s²
  5. 5. ω = 21.61 rad/s
  6. 6. φ = 15.35 °

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