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 = 0.3 m0 → 1700 rpm170 kg

A flywheel of mass 170 kg with a radius of gyration of 0.3 m is accelerated uniformly from rest to 1700 rpm. The driving torque is a constant 140 N·m and friction can be ignored. How long t does this take?

m = 170 kgk = 0.3 mn1 = 0 rpmn2 = 1700 rpmT = 140 N·m
t = s
2.Easy
k = 0.25 m0 → 1600 rpm130 kg

A flywheel of mass 130 kg with a radius of gyration of 0.25 m is accelerated uniformly from rest to 1600 rpm. This takes 5 s. Ignoring friction, find the constant torque T needed.

m = 130 kgk = 0.25 mn1 = 0 rpmn2 = 1600 rpmt = 5 s
T = N·m
3.Medium
0 → 1800 rpmI = 28.5 kg·m²

A motor brings a flywheel with a moment of inertia of 28.5 kg·m² from rest up to 1800 rpm in 28 s. Ignoring losses, find the average power Pavg it delivers, in kW.

I = 28.5 kg·m²n = 1800 rpmt = 28 s
Pavg = kW
4.Medium
v = ?L = 1.7 mm = 8.5 kg

A uniform slender rod of mass 8.5 kg and length 1.7 m is pinned at one end and released from rest in a horizontal position. 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 = 8.5 kgL = 1.7 mg = 9.81 m/s²
v = m/s
5.Hard
disc: 18 kgring: 12 kgn1​ = 500 rpm

A uniform solid disc (mass 18 kg, radius 0.45 m) spins freely at 500 rpm about a vertical frictionless shaft. A thin ring of mass 12 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 = 18 kgMring = 12 kgR = 0.45 mn1 = 500 rpm
ΔE = J
6.Hard
240 m/s20 g0.68 mL = 1.2 m6.5 kgφ

A uniform slender rod of mass 6.5 kg and length 1.2 m hangs at rest from a frictionless pivot at its top end. A 20 g bullet moving horizontally at 240 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 = 6.5 kgL = 1.2 mm = 20 gv = 240 m/sd = 0.68 mg = 9.81 m/s²
φ = °

Answer key

  1. 1. t = 19.46 s
  2. 2. T = 272.3 N·m
  3. 3. Pavg = 18.08 kW
  4. 4. v = 7.073 m/s
  5. 5. ΔE = 1428 J
  6. 6. φ = 17.13 °

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