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 = 0.05 mL = 0.8 m0.85 m3.5 kg8 kg

A pendulum is made of a uniform slender rod (3.5 kg, 0.8 m long) pivoted at its top end O, with a solid sphere of mass 8 kg and radius 0.05 m fixed to its lower end. The centre of the sphere is 0.85 m below O. Find the mass moment of inertia IO of the pendulum about O.

mrod = 3.5 kgL = 0.8 mmsphere = 8 kgr = 0.05 md = 0.85 m
IO = kg·m²
2.Easy
L = 1.1 mGOd = 0.45 mm = 2.5 kg

A uniform slender rod of mass 2.5 kg and length 1.1 m rotates about an axis O perpendicular to the rod, 0.45 m from its centre G. Find its mass moment of inertia IO about that axis.

m = 2.5 kgL = 1.1 md = 0.45 m
IO = kg·m²
3.Medium
14.5 kgaαM = 7.5 kgR = 0.2 m

A 14.5 kg block hangs from a light cord wrapped around a pulley, a uniform solid disc of mass 7.5 kg and radius 0.2 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 = 14.5 kgM = 7.5 kgR = 0.2 mg = 9.81 m/s²
T = N
4.Medium
v = ?0.5 kgL = 1.5 mm = 4.5 kg

A uniform slender rod of mass 4.5 kg and length 1.5 m is pinned at one end and released from rest in a horizontal position. A small ball of mass 0.5 kg is fixed to its free end. 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 = 4.5 kgL = 1.5 mmball = 0.5 kgg = 9.81 m/s²
v = m/s
5.Hard
θ = 10°as = 2 m

A thin hoop rolls without slipping down a slope inclined at 10° to the horizontal. Use g = 9.81 m/s². It starts from rest. How long t does it take to roll 2 m down the slope?

θ = 10°I = m·r²g = 9.81 m/s²s = 2 m
t = s
6.Hard
460 m/s40 g0.7 mL = 1.5 m7 kgφ

A uniform slender rod of mass 7 kg and length 1.5 m hangs at rest from a frictionless pivot at its top end. A 40 g bullet moving horizontally at 460 m/s hits the rod 0.7 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 = 7 kgL = 1.5 mm = 40 gv = 460 m/sd = 0.7 mg = 9.81 m/s²
φ = °

Answer key

  1. 1. IO = 6.535 kg·m²
  2. 2. IO = 0.7583 kg·m²
  3. 3. T = 29.23 N
  4. 4. v = 6.361 m/s
  5. 5. t = 2.167 s
  6. 6. φ = 45.89 °

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