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
k = 0.15 m0 → 500 rpm180 kg

A flywheel of mass 180 kg with a radius of gyration of 0.15 m is accelerated uniformly from rest to 500 rpm. The driving torque is a constant 9.2 N·m and friction can be ignored. How long t does this take?

m = 180 kgk = 0.15 mn1 = 0 rpmn2 = 500 rpmT = 9.2 N·m
t = s
2.Easy
Or = 0.25 mL = 1.2 m1.45 m4.5 kg5.5 kg

A pendulum is made of a uniform slender rod (4.5 kg, 1.2 m long) pivoted at its top end O, with a thin solid disc of mass 5.5 kg and radius 0.25 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 = 4.5 kgL = 1.2 mmdisc = 5.5 kgr = 0.25 md = 1.45 m
IO = kg·m²
3.Medium
n = 900 rpmP = 104.5 kW

A shaft transmits 104.5 kW at 900 rpm. Find the torque T in the shaft.

P = 104.5 kWn = 900 rpm
T = N·m
4.Medium
L = 0.9 mm = 11 kgα

A uniform slender rod of mass 11 kg and length 0.9 m is pinned at one end and held horizontal. It is released from rest. Use g = 9.81 m/s². Find its angular acceleration α just after release.

m = 11 kgL = 0.9 mg = 9.81 m/s²
α = rad/s²
5.Hard
140 m/s45 g1.18 mL = 1.7 m9.5 kgφ

A uniform slender rod of mass 9.5 kg and length 1.7 m hangs at rest from a frictionless pivot at its top end. A 45 g bullet moving horizontally at 140 m/s hits the rod 1.18 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 = 9.5 kgL = 1.7 mm = 45 gv = 140 m/sd = 1.18 mg = 9.81 m/s²
φ = °
6.Hard
3.5 m/sω

A solid cylinder of mass 18 kg rolls without slipping at 3.5 m/s. What percentage p of its total kinetic energy is rotational?

m = 18 kgv = 3.5 m/sI = ½·m·r²
p = %

Answer key

  1. 1. t = 23.05 s
  2. 2. IO = 13.9 kg·m²
  3. 3. T = 1109 N·m
  4. 4. α = 16.35 rad/s²
  5. 5. φ = 15.76 °
  6. 6. p = 33.33 %

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