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PHYS 165 Module 7 Exam – Physics: Rotational Motion, Gravitation & Oscillations – 2026–2027 Edition – 25-Question Actual Exam

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This document contains a 25-question Module 7 exam for PHYS 165 Physics covering rotational motion, gravitation, and oscillations for the 2026–2027 academic year. It includes key physics principles involving torque, angular motion, rotational dynamics, gravitational forces, orbital concepts, and harmonic motion. The material is designed to support exam preparation and reinforce important physics concepts and problem-solving skills.

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PHYS 165 Module 7 Exam 2026–2027 Edition
Physics: Rotational Motion, Gravitation & Oscillations
25-Question Actual Exam


Abstract
This document presents the PHYS 165 Module 7 Examination for the 2026–2027 academic year,
comprising 25 actual exam questions organized across four core domains of university-level physics. The
examination assesses the critical application of mathematical principles, physical laws, and analytical
problem-solving skills required to understand complex phenomena in rotational kinematics and
dynamics, including angular displacement, velocity, acceleration, torque, moments of inertia, and the
parallel axis theorem. Angular momentum and static equilibrium domains address conservation of
angular momentum, rotational collisions, and the conditions for translational and rotational
equilibrium in rigid bodies. Universal gravitation and orbital mechanics domains evaluate Newton’s law
of gravitation, gravitational potential energy, Kepler’s laws of planetary motion, and escape velocity
calculations. Oscillations and simple harmonic motion domains examine mass-spring systems, the
simple pendulum, energy conservation in SHM, equations of motion, and damped oscillatory behavior.
Each question includes a detailed rationale, distractor analysis, and reference to the 2026 PHYS 165
Module 7 course material and standard university physics textbooks, ensuring alignment with the
Portage Learning physics curriculum and current scientific standards.

Content Area Overview
Content Area Questions Key Topics Weight

Rotational Kinematics & 1–8 Angular displacement, velocity & acceleration; 30%
Dynamics torque; moment of inertia; parallel axis
theorem; rolling without slipping; static
equilibrium

Angular Momentum & Static 9–13 Conservation of angular momentum; rotational 20%
Equilibrium collisions; skater problem; ladder problems;
cable-supported beams; angular momentum of
particles

Universal Gravitation & 14–19 Newton’s law of gravitation; gravitational 25%
Orbital Mechanics acceleration at altitude; gravitational potential
energy; Kepler’s third law; escape velocity

Oscillations & Simple 20–25 Mass-spring period; simple pendulum; SHM 25%
Harmonic Motion equations; energy in SHM; amplitude,
frequency & phase; damped oscillations



Examination Questions

Domain: Rotational Kinematics & Dynamics
Q1. A solid disk of radius 0.50 m rotates with an angular velocity of 12 rad/s. If the angular
velocity increases uniformly to 28 rad/s in 4.0 s, what is the angular acceleration of the
disk?
A. 4.0 rad/s²
B. 10.0 rad/s²
C. 7.0 rad/s²

, D. 16.0 rad/s²
Correct Answer: A. 4.0 rad/s²
Rationale: Angular acceleration is defined as the rate of change of angular velocity: α = (ω₂ − ω₁)/Δt =
(28 − 12)/4.0 = 16/4.0 = 4.0 rad/s². This constant angular acceleration indicates the disk is speeding up
its rotation at a uniform rate, analogous to constant linear acceleration in translational kinematics.
Why Wrong: B: 10.0 rad/s² has no correct derivation from the given data. C: 7.0 rad/s² is the average
angular velocity, not the angular acceleration. D: 16.0 rad/s² is the change in angular velocity (Δω)
without dividing by the time interval.
Reference: Halliday, Resnick & Walker, Fundamentals of Physics, 13th Ed., Ch. 10; PHYS 165 Module 7,
Lesson 1: Rotational Kinematics

Q2. A wheel starts from rest and undergoes constant angular acceleration of 3.0 rad/s² for
5.0 seconds. Through what total angle does the wheel rotate during this interval?
A. 15.0 rad
B. 37.5 rad
C. 75.0 rad
D. 45.0 rad
Correct Answer: B. 37.5 rad
Rationale: Using the kinematic equation θ = ω₀t + ½αt², with ω₀ = 0, α = 3.0 rad/s², and t = 5.0 s: θ =
0 + ½(3.0)(5.0)² = 0.5(3.0)(25.0) = 37.5 rad. This equation is the rotational analog of the translational
displacement equation, reflecting the one-to-one correspondence between linear and angular kinematics
when acceleration is constant.
Why Wrong: A: 15.0 rad results from α × t, which gives the final angular velocity, not the angular
displacement. C: 75.0 rad omits the ½ factor, using θ = αt² instead of ½αt². D: 45.0 rad does not
correspond to any correct application of the kinematic equations with these values.
Reference: Young & Freedman, University Physics, 16th Ed., Ch. 9; PHYS 165 Module 7, Lesson 1:
Angular Displacement

Q3. A uniform solid sphere and a uniform solid cylinder of equal mass and radius are
released from rest at the top of an incline. Which object reaches the bottom first, and why?
A. The sphere, because it has a smaller moment of inertia and less rotational kinetic energy to develop
B. The cylinder, because its mass is distributed closer to the axis of rotation
C. Both reach the bottom simultaneously because they have equal mass
D. The cylinder, because it converts more gravitational potential energy into translational kinetic energy
Correct Answer: A. The sphere, because it has a smaller moment of inertia and less
rotational kinetic energy to develop
Rationale: The acceleration of a rolling object down an incline is a = g sinθ/(1 + I/(mr²)). For a solid
sphere, I = (2/5)mr², giving a = (5/7)g sinθ. For a solid cylinder, I = (1/2)mr², giving a = (2/3)g sinθ.
Since 5/7 ≈ 0.714 > 2/3 ≈ 0.667, the sphere accelerates faster. The sphere diverts less gravitational
potential energy into rotational kinetic energy because its mass is distributed closer to the axis on
average, leaving more energy available for translational motion.
Why Wrong: B: The cylinder has a larger I/(mr²) ratio, meaning its mass is effectively distributed
farther from the axis. C: Equal mass does not guarantee equal descent time; the mass distribution
determines the outcome. D: The cylinder actually converts less energy into translation and more into
rotation compared to the sphere.
Reference: Halliday, Resnick & Walker, Ch. 11; PHYS 165 Module 7, Lesson 2: Rolling Motion

Q4. A force of 40 N is applied tangentially to the rim of a flywheel of radius 0.25 m. What is
the magnitude of the torque produced about the axis of the flywheel?
A. 160 N·m

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