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 = 0.15 mL = 1.3 m1.45 m2.5 kg1 kg

A pendulum is made of a uniform slender rod (2.5 kg, 1.3 m long) pivoted at its top end O, with a thin solid disc of mass 1 kg and radius 0.15 m fixed to its lower end. The disc swings in its own plane and its centre is 1.45 m below O. Find the mass moment of inertia IO of the pendulum about O.

mrod = 2.5 kgL = 1.3 mmdisc = 1 kgr = 0.15 md = 1.45 m
IO = kg·m²
2.Easy
R = 0.25 mn = 2500 rpm60 kg

A flywheel is a uniform solid disc of mass 60 kg and radius 0.25 m. It spins at 2500 rpm. Find its kinetic energy Ek in kJ.

M = 60 kgR = 0.25 mn = 2500 rpm
Ek = kJ
3.Medium
6 kgaαM = 19 kgR = 0.35 m

A 6 kg block hangs from a light cord wrapped around a pulley, a uniform solid disc of mass 19 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 tension T in the cord.

m = 6 kgM = 19 kgR = 0.35 mg = 9.81 m/s²
T = N
4.Medium
0 → 700 rpmI = 13.5 kg·m²

A motor brings a flywheel with a moment of inertia of 13.5 kg·m² from rest up to 700 rpm in 6 s. Ignoring losses, find the average power Pavg it delivers, in kW.

I = 13.5 kg·m²n = 700 rpmt = 6 s
Pavg = kW
5.Hard
0.65 m0.25 mn1​ = 40 rpm

A person stands on a frictionless turntable and holds a 5 kg dumbbell in each hand, 0.65 m from the axis. The person and turntable alone have a moment of inertia of 3 kg·m² about the axis. They turn at 40 rpm, then pull the dumbbells in to 0.25 m from the axis. Treat the dumbbells as point masses. Find the new speed n2 in rpm.

I0 = 3 kg·m²m = 2 × 5 kgr1 = 0.65 mr2 = 0.25 mn1 = 40 rpm
n2 = rpm
6.Hard
160 m/s25 g0.68 mL = 1.2 m2 kgφ

A uniform slender rod of mass 2 kg and length 1.2 m hangs at rest from a frictionless pivot at its top end. A 25 g bullet moving horizontally at 160 m/s hits the rod 0.68 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 = 2 kgL = 1.2 mm = 25 gv = 160 m/sd = 0.68 mg = 9.81 m/s²
φ = °

Answer key

  1. 1. IO = 3.522 kg·m²
  2. 2. Ek = 64.26 kJ
  3. 3. T = 36.08 N
  4. 4. Pavg = 6.045 kW
  5. 5. n2 = 79.72 rpm
  6. 6. φ = 47.07 °

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