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 = 0.4 m100 → 1100 rpm110 kg

A flywheel of mass 110 kg with a radius of gyration of 0.4 m is accelerated uniformly from 100 rpm to 1100 rpm. The driving torque is a constant 77 N·m and friction can be ignored. How long t does this take?

m = 110 kgk = 0.4 mn1 = 100 rpmn2 = 1100 rpmT = 77 N·m
t = s
2.Easy
k = 0.5 mn = 900 rpm250 kg

A flywheel of mass 250 kg has a radius of gyration of 0.5 m about its axis. It spins at 900 rpm. Find its kinetic energy Ek in kJ.

M = 250 kgk = 0.5 mn = 900 rpm
Ek = kJ
3.Medium
v = ?3.5 kgL = 1.7 mm = 3 kg

A uniform slender rod of mass 3 kg and length 1.7 m is pinned at one end and released from rest in a horizontal position. A small ball of mass 3.5 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 = 3 kgL = 1.7 mmball = 3.5 kgg = 9.81 m/s²
v = m/s
4.Medium
4 kgaαM = 6.5 kgR = 0.45 m

A 4 kg block hangs from a light cord wrapped around a pulley, a uniform solid disc of mass 6.5 kg and radius 0.45 m that turns on a frictionless horizontal axle. The block is released from rest. Use g = 9.81 m/s². Find the angular acceleration α of the pulley.

m = 4 kgM = 6.5 kgR = 0.45 mg = 9.81 m/s²
α = rad/s²
5.Hard
θ = 10°vs = 5.5 m

A solid cylinder starts from rest and rolls without slipping down a slope inclined at 10° to the horizontal. Use g = 9.81 m/s². Its radius is 0.05 m. Find its angular velocity ω after it has rolled 5.5 m along the slope.

θ = 10°s = 5.5 mI = ½·m·r²g = 9.81 m/s²r = 0.05 m
ω = rad/s
6.Hard
240 m/s30 g0.52 mL = 0.8 m2.5 kgφ

A uniform slender rod of mass 2.5 kg and length 0.8 m hangs at rest from a frictionless pivot at its top end. A 30 g bullet moving horizontally at 240 m/s hits the rod 0.52 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 = 2.5 kgL = 0.8 mm = 30 gv = 240 m/sd = 0.52 mg = 9.81 m/s²
φ = °

Answer key

  1. 1. t = 23.94 s
  2. 2. Ek = 277.6 kJ
  3. 3. v = 6.088 m/s
  4. 4. α = 12.03 rad/s²
  5. 5. ω = 70.69 rad/s
  6. 6. φ = 107.4 °

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