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 = 0.6 m0 → 1500 rpm60 kg

A flywheel of mass 60 kg with a radius of gyration of 0.6 m is accelerated uniformly from rest to 1500 rpm. This takes 21 s. Ignoring friction, find the constant torque T needed.

m = 60 kgk = 0.6 mn1 = 0 rpmn2 = 1500 rpmt = 21 s
T = N·m
2.Easy
Or = 0.1 mL = 0.9 m1 m0.5 kg9.5 kg

A pendulum is made of a uniform slender rod (0.5 kg, 0.9 m long) pivoted at its top end O, with a thin solid disc of mass 9.5 kg and radius 0.1 m fixed to its lower end. The disc swings in its own plane and its centre is 1 m below O. Find the mass moment of inertia IO of the pendulum about O.

mrod = 0.5 kgL = 0.9 mmdisc = 9.5 kgr = 0.1 md = 1 m
IO = kg·m²
3.Medium
v = ?4.5 kgL = 1.9 mm = 4.5 kg

A uniform slender rod of mass 4.5 kg and length 1.9 m is pinned at one end and released from rest in a horizontal position. A small ball of mass 4.5 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 = 4.5 kgL = 1.9 mmball = 4.5 kgg = 9.81 m/s²
v = m/s
4.Medium
12 kgaαM = 15 kgR = 0.2 m

A 12 kg block hangs from a light cord wrapped around a pulley, a uniform solid disc of mass 15 kg and radius 0.2 m that turns on a frictionless horizontal axle. The block is released from rest. Use g = 9.81 m/s². Find the speed v of the block after it has fallen 1.5 m.

m = 12 kgM = 15 kgR = 0.2 mg = 9.81 m/s²h = 1.5 m
v = m/s
5.Hard
θ = 15°vs = 4.5 m

A solid 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 4.5 m along the slope.

θ = 15°s = 4.5 mI = (2/5)·m·r²g = 9.81 m/s²r = 0.4 m
ω = rad/s
6.Hard
440 m/s40 g1.56 mL = 1.9 m10 kgφ

A uniform slender rod of mass 10 kg and length 1.9 m hangs at rest from a frictionless pivot at its top end. A 40 g bullet moving horizontally at 440 m/s hits the rod 1.56 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 = 10 kgL = 1.9 mm = 40 gv = 440 m/sd = 1.56 mg = 9.81 m/s²
φ = °

Answer key

  1. 1. T = 161.6 N·m
  2. 2. IO = 9.682 kg·m²
  3. 3. v = 6.476 m/s
  4. 4. v = 4.256 m/s
  5. 5. ω = 10.1 rad/s
  6. 6. φ = 48.03 °

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