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 = 0.1 mL = 0.4 m0.5 m4.5 kg8.5 kg

A pendulum is made of a uniform slender rod (4.5 kg, 0.4 m long) pivoted at its top end O, with a thin solid disc of mass 8.5 kg and radius 0.1 m fixed to its lower end. The disc swings in its own plane and its centre is 0.5 m below O. Find the mass moment of inertia IO of the pendulum about O.

mrod = 4.5 kgL = 0.4 mmdisc = 8.5 kgr = 0.1 md = 0.5 m
IO = kg·m²
2.Easy
L = 0.5 mGOd = 0.1 mm = 11 kg

A uniform slender rod of mass 11 kg and length 0.5 m rotates about an axis O perpendicular to the rod, 0.1 m from its centre G. Find its mass moment of inertia IO about that axis.

m = 11 kgL = 0.5 md = 0.1 m
IO = kg·m²
3.Medium
v = ?4 kgL = 1.2 mm = 3 kg

A uniform slender rod of mass 3 kg and length 1.2 m is pinned at one end and released from rest in a horizontal position. A small ball of mass 4 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 = 3 kgL = 1.2 mmball = 4 kgg = 9.81 m/s²
v = m/s
4.Medium
ω = ?0.5 kgL = 1.9 mm = 3 kg

A uniform slender rod of mass 3 kg and length 1.9 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 angular velocity ω of the rod as it swings through the vertical position.

m = 3 kgL = 1.9 mmball = 0.5 kgg = 9.81 m/s²
ω = rad/s
5.Hard
4 m/sω

A solid cylinder of mass 5.5 kg rolls without slipping at 4 m/s. What percentage p of its total kinetic energy is rotational?

m = 5.5 kgv = 4 m/sI = ½·m·r²
p = %
6.Hard
160 m/s50 g0.43 mL = 0.7 m5.5 kgφ

A uniform slender rod of mass 5.5 kg and length 0.7 m hangs at rest from a frictionless pivot at its top end. A 50 g bullet moving horizontally at 160 m/s hits the rod 0.43 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 = 0.7 mm = 50 gv = 160 m/sd = 0.43 mg = 9.81 m/s²
φ = °

Answer key

  1. 1. IO = 2.408 kg·m²
  2. 2. IO = 0.3392 kg·m²
  3. 3. v = 5.089 m/s
  4. 4. ω = 3.711 rad/s
  5. 5. p = 33.33 %
  6. 6. φ = 48.81 °

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