Source: ENGR/PHYS 216 Final Exam Practice Problems
Tags: center of mass, moment of inertia, parallel axis theorem, angular acceleration, flywheel, simple harmonic motion, SHM, pendulum, spring, engineering ethics, ENGR 216, Texas A&M
This section covers three remaining exam topics: rotational dynamics (moment of inertia, angular acceleration, the parallel axis theorem), simple harmonic motion (spring-board systems and pendulums in non-standard gravity), and engineering ethics true/false questions. The rotational problems require knowing moment-of-inertia formulas, and the harmonic motion problems require identifying the effective spring constant or effective gravity.
Center of mass
The mass-weighted average position of all particles in a system. Its location depends only on the masses and their positions. It is a geometric/mass property, independent of velocity or forces.
Moment of inertia (I)
A measure of how much an object resists angular acceleration. Depends on both the mass distribution and the choice of rotation axis. Units: kg·m².
Parallel axis theorem
Relates the moment of inertia about any axis to the moment about a parallel axis through the centre of mass: I = I_cm + m·d²
where d is the distance between the two parallel axes. The centre-of-mass axis always gives the minimum moment of inertia.
Angular acceleration (α)
The rate of change of angular velocity: α = τ / I, where τ is the net torque. Analogous to a = F/m for linear motion.
Simple harmonic motion (SHM)
Oscillation where the restoring force (or torque) is proportional to the displacement. Characterised by: a = −ω²x (linear) or α = −ω²θ (angular)
The period is T = 2π/ω, independent of amplitude for small oscillations.
Effective gravity (g_eff)
In an accelerating frame (like an elevator), the effective gravitational acceleration changes. For upward acceleration a: g_eff = g + a. For downward acceleration: g_eff = g − a.
The centre of mass is defined such that its location depends only on the masses of the particles and their locations. Answer: E.
The other options describe conditions that apply only in special circumstances (no external forces, constant net force, etc.), not the definition itself.
A thin rectangular plate of width a and length b, with I_cm = (1/12)m(a² + b²), rotates about an axis z through one corner.
Step 1 – Find the distance from the centre of mass to the corner: d = √[(a/2)² + (b/2)²] = (1/2)√(a² + b²) d² = (a² + b²)/4
Step 2 – Apply the parallel axis theorem: I_z = I_cm + m·d² = m(a² + b²)/12 + m(a² + b²)/4
I_z = m(a² + b²) × (1/12 + 3/12) = m(a² + b²)/3
Step 3 – Take the ratio: I_z / I_cm = (1/3) / (1/12) = 4
A solid disk flywheel (I = ½mr²) accelerated by a constant torque.
Given: m = 30 kg, r = 0.25 m, τ = 30 N·m, t = 15 s, starts from rest.
Step 1 – Moment of inertia: I = ½ × 30 × 0.25² = 0.9375 kg·m²
Step 2 – Angular acceleration: α = τ/I = 30 / 0.9375 = 32.0 rad/s²
Step 3 – Final angular velocity: ω = α·t = 32.0 × 15 = 480 rad/s
Step 4 – Convert to rev/s: f = ω / (2π) = 480 / 6.283 = 76.4 rev/s
A uniform board pinned at one end, with a spring at the other. This is rotational SHM.
Given: m = 12 kg, k = 1300 N/m, board length L (cancels out).
For small angle θ, the spring deflection is L·θ, giving a restoring torque: τ = −k·L·θ × L = −kL²θ
Using Newton's second law for rotation: I·α = −kL²θ (1/3)mL²·α = −kL²θ α = −(3k/m)θ
This gives ω² = 3k/m.
T = 2π/ω = 2π √(m / 3k) = 2π √(12 / 3900) = 2π × 0.05547 = 0.35 s
Note that the board length cancels. The period depends only on mass and spring constant.
Given: L = 1.5 m, elevator accelerating upward at 2.0 m/s².
In an upward-accelerating elevator, the effective gravity increases: g_eff = g + a = 9.81 + 2.0 = 11.81 m/s²
The pendulum period formula with effective gravity: T = 2π √(L / g_eff) = 2π √(1.5 / 11.81) = 2π × 0.3564 = 2.24 s
This is shorter than the period at rest (2.46 s), which makes physical sense: stronger effective gravity means faster swings.
Problem 29: "Engineers shall at all times strive to serve their employer."
This is False. The fundamental canon of engineering ethics states that engineers shall hold paramount the safety, health, and welfare of the public. Serving the employer is secondary to public welfare.
Problem 30: "Bad news, criticism, questions, and information outside expectations are regarded as negatives by the organization. This is an example of suppression of open communication."
This is True. When an organisation treats dissenting information or critical feedback as unwelcome, it creates a culture where important safety and quality concerns go unreported. This is a recognised failure mode in engineering ethics and organisational behaviour.
Parallel axis theorem: I = I_cm + m·d²
Moment of inertia, solid disk/cylinder: I = ½mr²
Moment of inertia, uniform rod about one end: I = (1/3)mL²
Moment of inertia, rectangular plate about CM: I_cm = (1/12)m(a² + b²)
Rotational Newton's second law: τ = I·α
Angular kinematics (constant α): ω = ω₀ + α·t
SHM angular frequency from spring-board system: ω = √(3k/m)
Simple pendulum period: T = 2π √(L/g_eff)
Effective gravity in accelerating elevator: g_eff = g + a (upward acceleration), g_eff = g − a (downward)
⚠️ The parallel axis theorem only works in one direction: from the CM axis to a parallel axis at distance d. You cannot apply it between two arbitrary parallel axes without going through the CM first.
⚠️ For the flywheel problem, the question asks for revolutions per second, not rad/s. Divide ω by 2π to convert.
⚠️ In the spring-board oscillation, the board length L cancels out of the period formula. If you end up with L in your final answer, something has gone wrong.
⚠️ For pendulum problems, upward acceleration increases g_eff and shortens the period. Downward acceleration (or being on the Moon) decreases g_eff and lengthens it. Free-fall (g_eff = 0) means the pendulum does not oscillate.
⚠️ Engineering ethics: "serve the public" is the paramount duty, always above loyalty to an employer or client. This is a near-guaranteed exam question.
Q: What is the moment of inertia of a 5 kg solid disk of radius 0.3 m?
A: I = ½ × 5 × 0.3² = 0.225 kg·m².
Q: A torque of 10 N·m is applied to a wheel with I = 2 kg·m² for 5 seconds. If it starts from rest, what is the final angular velocity?
A: α = 10/2 = 5 rad/s². ω = 5 × 5 = 25 rad/s.
Q: A pendulum has period 2.0 s on Earth. What is its period in an elevator accelerating downward at 3 m/s²?
A: g_eff = 9.81 − 3 = 6.81 m/s². New period = 2.0 × √(9.81/6.81) = 2.0 × 1.20 = 2.40 s.
Q: An engineer discovers a safety flaw in a product but their employer tells them to stay quiet. What should the engineer do?
A: The engineer's primary obligation is to public safety. They should report the flaw through appropriate channels, even if it conflicts with the employer's wishes.
Q: What does the parallel axis theorem tell you about which axis gives the smallest moment of inertia?
A: The axis through the centre of mass always gives the minimum moment of inertia. Any parallel axis at distance d away adds m·d² to the total.
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