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 = 0.45 m0 → 1700 rpm160 kg

A flywheel of mass 160 kg with a radius of gyration of 0.45 m is accelerated uniformly from rest to 1700 rpm. The driving torque is a constant 1900 N·m and friction can be ignored. How long t does this take?

m = 160 kgk = 0.45 mn1 = 0 rpmn2 = 1700 rpmT = 1900 N·m
t = s
2.Easy
k = 0.45 m1000 → 2300 rpm190 kg

A flywheel of mass 190 kg with a radius of gyration of 0.45 m is accelerated uniformly from 1000 rpm to 2300 rpm. The driving torque is a constant 580 N·m and friction can be ignored. How long t does this take?

m = 190 kgk = 0.45 mn1 = 1000 rpmn2 = 2300 rpmT = 580 N·m
t = s
3.Medium
8.5 kgaαM = 8 kgR = 0.15 m

An 8.5 kg block hangs from a light cord wrapped around a pulley, a uniform solid disc of mass 8 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 tension T in the cord.

m = 8.5 kgM = 8 kgR = 0.15 mg = 9.81 m/s²
T = N
4.Medium
0.5 kgaαM = 16 kgR = 0.5 m

A 0.5 kg block hangs from a light cord wrapped around a pulley, a uniform solid disc of mass 16 kg and radius 0.5 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 3 m.

m = 0.5 kgM = 16 kgR = 0.5 mg = 9.81 m/s²h = 3 m
v = m/s
5.Hard
0.7 m0.2 mn1​ = 50 rpm

A person stands on a frictionless turntable and holds a 4.5 kg dumbbell in each hand, 0.7 m from the axis. The person and turntable alone have a moment of inertia of 2.5 kg·m² about the axis. They turn at 50 rpm, then pull the dumbbells in to 0.2 m from the axis. Treat the dumbbells as point masses. Find the new speed n2 in rpm.

I0 = 2.5 kg·m²m = 2 × 4.5 kgr1 = 0.7 mr2 = 0.2 mn1 = 50 rpm
n2 = rpm
6.Hard
340 m/s15 g1.7 mL = 2 m1.5 kg

A uniform slender rod of mass 1.5 kg and length 2 m hangs at rest from a frictionless pivot at its top end. A 15 g bullet moving horizontally at 340 m/s hits the rod 1.7 m below the pivot and stays embedded. Find the angular velocity ω of the rod just after the impact.

M = 1.5 kgL = 2 mm = 15 gv = 340 m/sd = 1.7 m
ω = rad/s

Answer key

  1. 1. t = 3.036 s
  2. 2. t = 9.031 s
  3. 3. T = 26.68 N
  4. 4. v = 1.861 m/s
  5. 5. n2 = 120.8 rpm
  6. 6. ω = 4.243 rad/s

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