Source: Final Exam Practice Problems and Solutions
Tags: angular momentum, moment of inertia, conservation of angular momentum, torque, angular acceleration, rotational kinetic energy, flywheel, simple pendulum, damped oscillation, underdamped, critically damped, overdamped, ENGR 216, PHYS 216
Angular momentum is the rotational analogue of linear momentum: L = Iω for spinning objects, L = mvd for point masses about an axis. It is conserved when no external torque acts. Oscillation problems deal with pendulums and mass-spring-damper systems, where the damping coefficient relative to the critical value tells you the system's behaviour.
Angular momentum (L)
For a point mass: L = mvd (mass × velocity × perpendicular distance to the axis). For a rigid body: L = Iω (moment of inertia × angular velocity).
Moment of inertia (I)
The rotational equivalent of mass. Depends on both the mass and how it is distributed relative to the axis of rotation. Units: kg·m².
Conservation of angular momentum
When net external torque is zero: I₁ω₁ = I₂ω₂. If I decreases, ω increases proportionally.
Torque (τ)
The rotational equivalent of force: τ = Iα, or equivalently τ = dL/dt. Measured in N·m.
Angular acceleration (α)
The rate of change of angular velocity: α = Δω/Δt. Units: rad/s².
Rotational kinetic energy
KE_rot = ½Iω². Analogous to ½mv² for translation.
Simple pendulum
A mass on a massless string swinging under gravity. Period: T = 2π√(L/g), independent of mass.
Damping coefficient (c or b)
Quantifies the resistance to motion in a damped oscillator. Units: N·s/m.
Critical damping coefficient
c_crit = 2√(km). The boundary between oscillatory and non-oscillatory decay.
Underdamped
c < c_crit. The system oscillates with gradually decreasing amplitude.
Critically damped
c = c_crit. The system returns to equilibrium as fast as possible without oscillating.
Overdamped
c > c_crit. The system returns to equilibrium without oscillating, but more slowly than critical damping.
For a point mass moving in a straight line, its angular momentum about a point A is:
L = m × v × d
where d is the perpendicular distance from point A to the velocity vector.
Worked Example: Two Objects About Point A
Object 1: m₁ = 6.0 kg, v₁ = 2.0 m/s, d₁ = 1.5 m (moving right, above A)
Object 2: m₂ = 3.0 kg, v₂ = 3.5 m/s, d₂ = 2.5 m (moving left, to the right of A)
From the diagram, both objects produce angular momentum in the same direction (counterclockwise about A, using the right-hand rule).
L_total = m₁v₁d₁ + m₂v₂d₂ = (6.0)(2.0)(1.5) + (−)(3.0)(3.5)(2.5)
Looking at the geometry: Object 1 contributes +18 and Object 2 contributes −9.75 (or the signs work out differently depending on orientation).
L_total = 8.25 kg·m²/s
Use the right-hand rule or cross product to determine the sign of each contribution.
Given: ω₁ = 1.2 rev/s, I₁ = 6.0 kg·m², I₂ = 2.0 kg·m² (arms pulled in).
(a) New angular speed:
I₁ω₁ = I₂ω₂
ω₂ = (6.0 × 1.2) / 2.0 = 3.6 rev/s
(b) Ratio of new KE to old KE:
KE₁ = ½I₁ω₁² = ½(6.0)(1.2)² = 4.32 (in appropriate units)
KE₂ = ½I₂ω₂² = ½(2.0)(3.6)² = 12.96
KE₂ / KE₁ = 12.96 / 4.32 = 3.0
The kinetic energy triples even though angular momentum is conserved. The extra energy comes from the person's muscles doing work to pull the bricks inward.
Given: I = 0.140 kg·m², L₁ = 3.00 kg·m²/s, L₂ = 0.800 kg·m²/s, Δt = 1.50 s.
(a) Average torque:
τ = ΔL/Δt = (0.800 − 3.00) / 1.50 = −1.47 N·m
Negative because angular momentum is decreasing (decelerating).
(b) Angular displacement (constant α):
ω₁ = L₁/I = 3.00/0.140 = 21.43 rad/s
ω₂ = L₂/I = 0.800/0.140 = 5.714 rad/s
θ = ½(ω₁ + ω₂) × t = ½(21.43 + 5.714)(1.50) = 20.4 rad
(c) Work done:
W = ΔKE = ½I(ω₂² − ω₁²) = ½(0.140)(5.714² − 21.43²)
= ½(0.140)(32.65 − 459.2) = ½(0.140)(−426.6) = −29.9 J
Negative work means energy was removed from the flywheel.
(d) Average power:
P = |W| / t = 29.9 / 1.50 = 19.9 W
T = 2π√(L/g)
The period depends only on the string length and the local gravitational acceleration, not on mass.
Moon (g = 1.62 m/s²), L = 1 m:
T = 2π√(1/1.62) = 2π(0.786) = 4.94 s
Mars (g = 3.721 m/s²), L = 1 m:
T = 2π√(1/3.721) = 2π(0.518) = 3.26 s
For a mass-spring-damper system, compute the critical damping coefficient and compare:
c_crit = 2√(km)
Given: m = 0.5 kg, k = 60 N/m, c = 10 N·s/m.
c_crit = 2√(60 × 0.5) = 2√30 = 10.95 N·s/m
Since c = 10 < 10.95 = c_crit, the system is underdamped.
It will oscillate, but each successive peak will be smaller than the last.
Angular momentum (point mass): L = mvd (d = perpendicular distance to axis)
Angular momentum (rigid body): L = Iω
Conservation of angular momentum: I₁ω₁ = I₂ω₂ (when τ_ext = 0)
Torque from angular momentum: τ = ΔL / Δt
Rotational kinetic energy: KE = ½Iω²
Angular displacement (constant α): θ = ½(ω₁ + ω₂) × t
Work-energy (rotation): W = ΔKE = ½I(ω₂² − ω₁²)
Power: P = W / t
Simple pendulum period: T = 2π√(L/g)
Critical damping coefficient: c_crit = 2√(km)
⚠️ When using conservation of angular momentum with rev/s, you can keep those units throughout as long as you are consistent. But for KE calculations, you often need rad/s (multiply rev/s by 2π).
⚠️ In the spinning platform problem, KE is not conserved even though angular momentum is. The person's muscles supply the additional energy.
⚠️ For the flywheel, the sign of torque and work matters. A decelerating flywheel has negative torque and negative work done on it.
⚠️ The simple pendulum formula T = 2π√(L/g) is only valid for small angles of oscillation.
⚠️ To classify damping, compare c to c_crit = 2√(km). Close values (like 10 vs 10.95) can easily be misjudged, so compute carefully.
⚠️ For angular momentum of a point mass, the distance d is the perpendicular distance from the reference point to the line of the velocity vector, not the distance to the object.
Q: A person on a frictionless platform has I = 6.0 kg·m² at ω = 1.2 rev/s. They pull their arms in, reducing I to 2.0 kg·m². What is the new ω?
A: I₁ω₁ = I₂ω₂, so ω₂ = (6.0)(1.2)/2.0 = 3.6 rev/s.
Q: In that same scenario, does kinetic energy increase, decrease, or stay the same?
A: It increases by a factor of 3. The person's muscles do work to pull the masses inward.
Q: A flywheel with I = 0.140 kg·m² slows from L = 3.00 to L = 0.800 kg·m²/s in 1.50 s. What is the average torque?
A: τ = (0.800 − 3.00)/1.50 = −1.47 N·m.
Q: What is the period of a 1 m pendulum on the Moon (g = 1.62 m/s²)?
A: T = 2π√(1/1.62) = 4.94 s. Slower than on Earth because gravity is weaker.
Q: A mass-spring-damper has m = 0.5 kg, k = 60 N/m, c = 10 N·s/m. Classify the damping.
A: c_crit = 2√(60 × 0.5) = 10.95. Since c = 10 < 10.95, the system is underdamped.
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