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 = 2.8 mGOd = 0.55 mm = 2.5 kg

A uniform slender rod of mass 2.5 kg and length 2.8 m rotates about an axis O perpendicular to the rod, 0.55 m from its centre G. Find its mass moment of inertia IO about that axis.

m = 2.5 kgL = 2.8 md = 0.55 m
IO = kg·m²
2.Easy
Or = 0.05 mL = 1.5 m1.55 m2.5 kg3.5 kg

A pendulum is made of a uniform slender rod (2.5 kg, 1.5 m long) pivoted at its top end O, with a thin solid disc of mass 3.5 kg and radius 0.05 m fixed to its lower end. The disc swings in its own plane and its centre is 1.55 m below O. Find the mass moment of inertia IO of the pendulum about O.

mrod = 2.5 kgL = 1.5 mmdisc = 3.5 kgr = 0.05 md = 1.55 m
IO = kg·m²
3.Medium
v = ?3.5 kgL = 1.3 mm = 1 kg

A uniform slender rod of mass 1 kg and length 1.3 m is pinned at one end and released from rest in a horizontal position. A small ball of mass 3.5 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 = 1 kgL = 1.3 mmball = 3.5 kgg = 9.81 m/s²
v = m/s
4.Medium
n = 2100 rpmTb​ = 70 N·mI = 21.5 kg·m²

A flywheel with a moment of inertia of 21.5 kg·m² spins at 2100 rpm. A brake applies a constant friction torque of 70 N·m. How many revolutions N does the flywheel make before it stops?

I = 21.5 kg·m²n = 2100 rpmTb = 70 N·m
N = rev
5.Hard
120 m/s40 g0.72 mL = 1.8 m6 kgφ

A uniform slender rod of mass 6 kg and length 1.8 m hangs at rest from a frictionless pivot at its top end. A 40 g bullet moving horizontally at 120 m/s hits the rod 0.72 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 kgL = 1.8 mm = 40 gv = 120 m/sd = 0.72 mg = 9.81 m/s²
φ = °
6.Hard
θ = 35°a

A thin hoop rolls without slipping down a slope inclined at 35° to the horizontal. Use g = 9.81 m/s². Find the smallest coefficient of static friction μs that prevents it from slipping.

θ = 35°I = m·r²g = 9.81 m/s²
μs =

Answer key

  1. 1. IO = 2.39 kg·m²
  2. 2. IO = 10.29 kg·m²
  3. 3. v = 5.159 m/s
  4. 4. N = 1182 rev
  5. 5. φ = 10.66 °
  6. 6. μs = 0.3501

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