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 m200 → 1400 rpm40 kg

A flywheel of mass 40 kg with a radius of gyration of 0.15 m is accelerated uniformly from 200 rpm to 1400 rpm. The driving torque is a constant 6 N·m and friction can be ignored. How long t does this take?

m = 40 kgk = 0.15 mn1 = 200 rpmn2 = 1400 rpmT = 6 N·m
t = s
2.Easy
Or = 0.2 mL = 1.2 m1.4 m3 kg6 kg

A pendulum is made of a uniform slender rod (3 kg, 1.2 m long) pivoted at its top end O, with a solid sphere of mass 6 kg and radius 0.2 m fixed to its lower end. The centre of the sphere is 1.4 m below O. Find the mass moment of inertia IO of the pendulum about O.

mrod = 3 kgL = 1.2 mmsphere = 6 kgr = 0.2 md = 1.4 m
IO = kg·m²
3.Medium
0 → 600 rpmI = 11.5 kg·m²

A motor brings a flywheel with a moment of inertia of 11.5 kg·m² from rest up to 600 rpm in 10 s. Ignoring losses, find the average power Pavg it delivers, in kW.

I = 11.5 kg·m²n = 600 rpmt = 10 s
Pavg = kW
4.Medium
v = ?3 kgL = 2.3 mm = 9.5 kg

A uniform slender rod of mass 9.5 kg and length 2.3 m is pinned at one end and released from rest in a horizontal position. A small ball of mass 3 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 = 9.5 kgL = 2.3 mmball = 3 kgg = 9.81 m/s²
v = m/s
5.Hard
disc: 10 kgring: 19 kgn1​ = 750 rpm

A uniform solid disc (mass 10 kg, radius 0.2 m) spins freely at 750 rpm about a vertical frictionless shaft. A thin ring of mass 19 kg and the same radius is dropped onto it from rest, centred on the shaft, and the two quickly spin together. How much kinetic energy ΔE is lost in the process?

Mdisc = 10 kgMring = 19 kgR = 0.2 mn1 = 750 rpm
ΔE = J
6.Hard
θ = 15°vs = 6.5 m

A hollow sphere starts from rest and rolls without slipping down a slope inclined at 15° to the horizontal. Use g = 9.81 m/s². Its radius is 0.4 m. Find its angular velocity ω after it has rolled 6.5 m along the slope.

θ = 15°s = 6.5 mI = (2/3)·m·r²g = 9.81 m/s²r = 0.4 m
ω = rad/s

Answer key

  1. 1. t = 18.85 s
  2. 2. IO = 13.3 kg·m²
  3. 3. Pavg = 2.27 kW
  4. 4. v = 7.531 m/s
  5. 5. ΔE = 488.3 J
  6. 6. ω = 11.13 rad/s

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