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 = 0.25 m0 → 1900 rpm90 kg

A flywheel of mass 90 kg with a radius of gyration of 0.25 m is accelerated uniformly from rest to 1900 rpm. The driving torque is a constant 51 N·m and friction can be ignored. How long t does this take?

m = 90 kgk = 0.25 mn1 = 0 rpmn2 = 1900 rpmT = 51 N·m
t = s
2.Easy
Or = 0.1 mL = 0.8 m0.9 m3.5 kg8.5 kg

A pendulum is made of a uniform slender rod (3.5 kg, 0.8 m long) pivoted at its top end O, with a solid sphere of mass 8.5 kg and radius 0.1 m fixed to its lower end. The centre of the sphere is 0.9 m below O. Find the mass moment of inertia IO of the pendulum about O.

mrod = 3.5 kgL = 0.8 mmsphere = 8.5 kgr = 0.1 md = 0.9 m
IO = kg·m²
3.Medium
n = 1000 rpmTb​ = 110 N·mI = 29 kg·m²

A flywheel with a moment of inertia of 29 kg·m² spins at 1000 rpm. A brake applies a constant friction torque of 110 N·m. How many revolutions N does the flywheel make before it stops?

I = 29 kg·m²n = 1000 rpmTb = 110 N·m
N = rev
4.Medium
1.5 kgaαM = 10.5 kgR = 0.15 m

A 1.5 kg block hangs from a light cord wrapped around a pulley, a uniform solid disc of mass 10.5 kg and radius 0.15 m that turns on a frictionless horizontal axle. The block is released from rest. Use g = 9.81 m/s². Find the acceleration a of the block.

m = 1.5 kgM = 10.5 kgR = 0.15 mg = 9.81 m/s²
a = m/s²
5.Hard
θ = 40°as = 9.5 m

A solid cylinder rolls without slipping down a slope inclined at 40° to the horizontal. Use g = 9.81 m/s². It starts from rest. How long t does it take to roll 9.5 m down the slope?

θ = 40°I = ½·m·r²g = 9.81 m/s²s = 9.5 m
t = s
6.Hard
100 m/s50 g0.99 mL = 1.1 m6.5 kgφ

A uniform slender rod of mass 6.5 kg and length 1.1 m hangs at rest from a frictionless pivot at its top end. A 50 g bullet moving horizontally at 100 m/s hits the rod 0.99 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 = 6.5 kgL = 1.1 mm = 50 gv = 100 m/sd = 0.99 mg = 9.81 m/s²
φ = °

Answer key

  1. 1. t = 21.94 s
  2. 2. IO = 7.666 kg·m²
  3. 3. N = 230.1 rev
  4. 4. a = 2.18 m/s²
  5. 5. t = 2.126 s
  6. 6. φ = 29.43 °

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