PHYS 165 Module 7 Exam Physics Portage
Learning Rotational Motion Gravitation
Oscillations Official Practice Exam Actual
Exam 2026/2027 with Detailed Rationales |
Complete Exam-Style Questions | Pass
Guaranteed – A+ Graded
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SECTION 1: ANGULAR KINEMATICS & TORQUE Q1 – Q5
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Question 1 of 25
A flywheel on a lathe accelerates uniformly from rest to 120 rpm in 2.0 seconds. What is the
magnitude of its angular acceleration?
A. π rad/s²
B. 2π rad/s² ✓ CORRECT
C. 4π rad/s²
D. 6π rad/s²
Correct Answer: B
Rationale: The angular acceleration is found using α = Δω/Δt, where 120 rpm converts to 4π
rad/s, giving α = 4π/2.0 = 2π rad/s². Choice A results from incorrectly dividing the final
angular velocity by 4.0 seconds instead of 2.0 seconds. Remember to convert rpm to rad/s by
multiplying by 2π/60 before applying kinematic equations.
Question 2 of 25
A mechanic applies a 45 N force perpendicular to the end of a 0.35 m wrench to loosen a
rusted bolt. What torque does she exert on the bolt?
A. 16 N·m ✓ CORRECT
B. 13 N·m
C. 45 N·m
D. 129 N·m
, Correct Answer: A
Rationale: Torque is calculated as τ = rF sinθ, and with the force applied perpendicular to the
wrench, sin90° = 1 yields τ = (0.35 m)(45 N) = 15.75 N·m ≈ 16 N·m. Choice C mistakenly uses
the force magnitude alone and ignores the lever arm distance. Always verify that both the
force and the perpendicular distance from the pivot are included in torque problems.
Question 3 of 25
A potter's wheel starts from rest and accelerates at 4.0 rad/s² for 6.0 seconds. Through what
total angle does the wheel rotate during this interval?
A. 24 rad
B. 36 rad
C. 48 rad
D. 72 rad ✓ CORRECT
Correct Answer: D
Rationale: For constant angular acceleration from rest, the angular displacement is θ = ½αt²
= ½(4.0 rad/s²)(6.0 s)² = 72 rad. Choice B comes from using αt instead of ½αt², which is a
common kinematic error. When starting from rest, the factor of one-half is essential in the
rotational displacement equation.
Question 4 of 25
A uniform rod of length L is pivoted at its center. A force F is applied perpendicular to one
end, and a force 2F is applied perpendicular at the midpoint in the opposite direction. What is
the net torque on the rod?
A. 3FL/2
B. FL/2
C. zero ✓ CORRECT
D. 2FL
Correct Answer: C
Rationale: The torque from the force at the end is τ₁ = FL, while the torque from the force at
the midpoint is τ₂ = (2F)(L/2) = FL in the opposite direction, producing zero net torque.
Choice A incorrectly adds the magnitudes without considering their opposite directions.
Always assign a sign convention to clockwise and counterclockwise torques before
summing.
Question 5 of 25
A car tire of radius 0.30 m rolls without slipping and experiences an angular acceleration of
5.0 rad/s². What is the tangential linear acceleration of a point on the outer edge of the tire?
Learning Rotational Motion Gravitation
Oscillations Official Practice Exam Actual
Exam 2026/2027 with Detailed Rationales |
Complete Exam-Style Questions | Pass
Guaranteed – A+ Graded
══════════════════════════════════════
SECTION 1: ANGULAR KINEMATICS & TORQUE Q1 – Q5
══════════════════════════════════════
Question 1 of 25
A flywheel on a lathe accelerates uniformly from rest to 120 rpm in 2.0 seconds. What is the
magnitude of its angular acceleration?
A. π rad/s²
B. 2π rad/s² ✓ CORRECT
C. 4π rad/s²
D. 6π rad/s²
Correct Answer: B
Rationale: The angular acceleration is found using α = Δω/Δt, where 120 rpm converts to 4π
rad/s, giving α = 4π/2.0 = 2π rad/s². Choice A results from incorrectly dividing the final
angular velocity by 4.0 seconds instead of 2.0 seconds. Remember to convert rpm to rad/s by
multiplying by 2π/60 before applying kinematic equations.
Question 2 of 25
A mechanic applies a 45 N force perpendicular to the end of a 0.35 m wrench to loosen a
rusted bolt. What torque does she exert on the bolt?
A. 16 N·m ✓ CORRECT
B. 13 N·m
C. 45 N·m
D. 129 N·m
, Correct Answer: A
Rationale: Torque is calculated as τ = rF sinθ, and with the force applied perpendicular to the
wrench, sin90° = 1 yields τ = (0.35 m)(45 N) = 15.75 N·m ≈ 16 N·m. Choice C mistakenly uses
the force magnitude alone and ignores the lever arm distance. Always verify that both the
force and the perpendicular distance from the pivot are included in torque problems.
Question 3 of 25
A potter's wheel starts from rest and accelerates at 4.0 rad/s² for 6.0 seconds. Through what
total angle does the wheel rotate during this interval?
A. 24 rad
B. 36 rad
C. 48 rad
D. 72 rad ✓ CORRECT
Correct Answer: D
Rationale: For constant angular acceleration from rest, the angular displacement is θ = ½αt²
= ½(4.0 rad/s²)(6.0 s)² = 72 rad. Choice B comes from using αt instead of ½αt², which is a
common kinematic error. When starting from rest, the factor of one-half is essential in the
rotational displacement equation.
Question 4 of 25
A uniform rod of length L is pivoted at its center. A force F is applied perpendicular to one
end, and a force 2F is applied perpendicular at the midpoint in the opposite direction. What is
the net torque on the rod?
A. 3FL/2
B. FL/2
C. zero ✓ CORRECT
D. 2FL
Correct Answer: C
Rationale: The torque from the force at the end is τ₁ = FL, while the torque from the force at
the midpoint is τ₂ = (2F)(L/2) = FL in the opposite direction, producing zero net torque.
Choice A incorrectly adds the magnitudes without considering their opposite directions.
Always assign a sign convention to clockwise and counterclockwise torques before
summing.
Question 5 of 25
A car tire of radius 0.30 m rolls without slipping and experiences an angular acceleration of
5.0 rad/s². What is the tangential linear acceleration of a point on the outer edge of the tire?
