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 = 1.8 mGOd = 0.8 mm = 2 kg

A uniform slender rod of mass 2 kg and length 1.8 m rotates about an axis O perpendicular to the rod, 0.8 m from its centre G. Find its mass moment of inertia IO about that axis.

m = 2 kgL = 1.8 md = 0.8 m
IO = kg·m²
2.Easy
Or = 0.15 mL = 0.4 m0.55 m3.5 kg3.5 kg

A pendulum is made of a uniform slender rod (3.5 kg, 0.4 m long) pivoted at its top end O, with a solid sphere of mass 3.5 kg and radius 0.15 m fixed to its lower end. The centre of the sphere is 0.55 m below O. Find the mass moment of inertia IO of the pendulum about O.

mrod = 3.5 kgL = 0.4 mmsphere = 3.5 kgr = 0.15 md = 0.55 m
IO = kg·m²
3.Medium
n = 2000 rpmT = 1920 N·m

A motor delivers a torque of 1920 N·m at 2000 rpm. Find its power output P in kW.

T = 1920 N·mn = 2000 rpm
P = kW
4.Medium
ω = ?4 kgL = 0.9 mm = 2 kg

A uniform slender rod of mass 2 kg and length 0.9 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 angular velocity ω of the rod as it swings through the vertical position.

m = 2 kgL = 0.9 mmball = 4 kgg = 9.81 m/s²
ω = rad/s
5.Hard
0.75 m0.15 mn1​ = 35 rpm

A person stands on a frictionless turntable and holds a 5 kg dumbbell in each hand, 0.75 m from the axis. The person and turntable alone have a moment of inertia of 4.5 kg·m² about the axis. They turn at 35 rpm, then pull the dumbbells in to 0.15 m from the axis. Treat the dumbbells as point masses. Find the new speed n2 in rpm.

I0 = 4.5 kg·m²m = 2 × 5 kgr1 = 0.75 mr2 = 0.15 mn1 = 35 rpm
n2 = rpm
6.Hard
500 m/s50 g1.16 mL = 1.9 m5.5 kgφ

A uniform slender rod of mass 5.5 kg and length 1.9 m hangs at rest from a frictionless pivot at its top end. A 50 g bullet moving horizontally at 500 m/s hits the rod 1.16 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 = 1.9 mm = 50 gv = 500 m/sd = 1.16 mg = 9.81 m/s²
φ = °

Answer key

  1. 1. IO = 1.82 kg·m²
  2. 2. IO = 1.277 kg·m²
  3. 3. P = 402.1 kW
  4. 4. ω = 4.833 rad/s
  5. 5. n2 = 75 rpm
  6. 6. φ = 102.3 °

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