Unit 7 Torque and Rotation Workbook Answers
Unit 7 Torque and Rotation Workbook Answers provides detailed solutions for every problem in the typical high school physics workbook, helping students master torque calculations, rotational dynamics, and angular motion concepts. This guide walks you through each exercise, explains the underlying science, and offers step‑by‑step reasoning so you can check your work, understand where mistakes happen, and build confidence in solving rotational problems.
Introduction
The seventh unit of most physics workbooks focuses on torque and rotation. The workbook usually contains ten to twelve distinct problems, ranging from simple torque calculations to more complex scenarios involving combined translational and rotational motion. Students encounter a mix of conceptual questions and quantitative problems that require applying Newton’s second law for rotation (τ = Iα), the relationship between linear and angular quantities, and conservation principles. Having a reliable answer key is essential for self‑study, as it lets you verify each step and reinforce the mathematical relationships that govern rotating systems That's the part that actually makes a difference..
Key Concepts Covered
- Torque (τ) – The rotational equivalent of force, calculated as τ = r F sin θ, where r is the lever arm, F is the applied force, and θ is the angle between them.
- Moment of Inertia (I) – A measure of an object’s resistance to angular acceleration. For common shapes:
- Solid cylinder: I = ½ m r²
- Hollow cylinder: I = m r²
- Solid sphere: I = (2/5) m r²
- Angular Acceleration (α) – The rate of change of angular velocity, related to torque by τ = Iα.
- Rotational Kinematics – Equations analogous to linear motion: ω = ω₀ + αt, θ = ω₀t + ½αt², ω² = ω₀² + 2αθ.
- Rotational Kinetic Energy – KE_rot = ½ I ω².
- Angular Momentum (L) – L = I ω; conserved when net external torque is zero.
- Power in Rotation – P = τ ω.
Understanding these principles is crucial for solving the workbook’s problems accurately.
Step‑by‑Step Solutions
Below are the complete answers for each problem, presented in the same order they typically appear. Numbers in bold highlight the final result Small thing, real impact..
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Torque Calculation
Problem: A force of 20 N is applied perpendicular to a lever arm of length 0.5 m. Find the torque.
Solution: τ = r F sin θ = (0.5 m)(20 N)(sin 90°) = 10 N·m That's the whole idea.. -
Moment of Inertia of a Solid Cylinder
Problem: Determine I for a solid cylinder with mass 5 kg and radius 0.4 m.
Solution: I = ½ m r² = ½ (5 kg)(0.4 m)² = 0.4 kg·m². -
Angular Acceleration from Torque
Problem: A torque of 15 N·m acts on a wheel with I = 3 kg·m². What is the angular acceleration?
Solution: α = τ / I = 15 N·m / 3 kg·m² = 5 rad/s² It's one of those things that adds up.. -
Rotational Equilibrium
Problem: Two forces act on a seesaw. F₁ = 10 N at a distance of 2 m from the pivot (counter‑clockwise). Find the balancing force F₂ applied 1 m from the pivot (clockwise).
Solution: Στ = 0 → (10 N)(2 m) = F₂(1 m) → F₂ = 20 N (clockwise) That's the part that actually makes a difference.. -
Rotational Kinematics
Problem: Starting from ω₀ = 2 rad/s, a wheel accelerates at α = 3 rad/s² for 4 s -
Rotational Kinematics
Problem: Starting from ω₀ = 2 rad/s, a wheel accelerates at α = 3 rad/s² for 4 s. Find the final angular velocity and the angular displacement.
Solution:
Final angular velocity: ω = ω₀ + αt = 2 rad/s + (3 rad/s²)(4 s) = 14 rad/s.
Angular displacement: θ = ω₀t + ½αt² = (2 rad/s)(4 s) + ½(3 rad/s²)(4 s)² = 8 rad + 24 rad = 32 rad The details matter here.. -
Rotational Kinetic Energy
Problem: A solid sphere (m = 2 kg, r = 0.3 m) rotates about its center at 5 rad/s. Calculate its rotational kinetic energy.
Solution:
Moment of inertia: I = (2/5) m r² = (2/5)(2 kg)(0.3 m)² = 0.072 kg·m².
KE_rot = ½ I ω² = ½(0.072 kg·m²)(5 rad/s)² = 0.9 J Worth keeping that in mind. Which is the point.. -
Conservation of Angular Momentum
Problem: A figure skater spins with initial angular velocity ω₀ = 4 rad/s and moment of inertia I₀ = 5 kg·m². When she pulls her arms in, her moment of inertia becomes I = 2 kg·m². Assuming no external torques, find her new angular velocity.
Solution:
Conservation of angular momentum: I₀ω₀ = Iω → (5 kg·m²)(4 rad/s) = (2 kg·m²)ω → ω = 10 rad/s That alone is useful.. -
Rolling Without Slipping
Problem: A solid cylinder rolls down an incline of height 2 m. Find its speed at the bottom, assuming it started from rest.
Solution:
Using energy conservation: mgh = ½mv² + ½Iω².
For rolling without slipping, v = rω and I = ½ m r². Substituting gives:
mgh = ½mv² + ½(½ m r²)(v/r)² = ½mv² + ¼mv² = (3/4)mv².
Solving for v: v = √(4gh/3) = √(4×9.8×2/3) = 5.14 m/s. -
Power in Rotation
Problem: An engine produces a torque of 50 N·m at an angular velocity of 120 rad/s. What is the rotational power output?
Solution: P = τ ω = (50 N·m)(120 rad/s) = 6000 W (or 6 kW) Still holds up.. -
Combined Rotational and Translational Motion
Problem: A block of mass 3 kg hangs from a pulley of mass 2 kg and radius 0.2 m. The system is released from rest. Find the acceleration of the block.
Solution:
For the block: mg - T = ma.
For the pulley: TR = Iα, where I = ½ m_pulley R² = ½(2 kg)(0.2 m)² = 0.04 kg·m².
Since a = Rα, we get T = Iα/R = Ia/R².
Substituting T into the block equation: mg - Ia/R² = ma → a = mg / (m + I/R²) = (3×9.8) / (3 + 0.04/0.04) = 7.35 m/s². -
Angular Momentum Vector Direction
Problem: Using the right-hand rule, determine the direction of the angular momentum vector for a disk rotating counterclockwise when viewed from above.
Solution: Curl the fingers of the right hand in the direction of rotation (counterclockwise); the thumb points upward. That's why, the angular momentum vector points vertically upward. -
Critical Thinking Challenge
Problem: Explain why a hollow cylinder and a solid cylinder of the same mass and radius, when released simultaneously down an incline, will not reach the bottom at the same time.
Solution: The acceleration depends on the moment of inertia. A solid cylinder (I = ½ m r²) has less rotational inertia than a hollow cylinder (I = m r²), meaning more of its gravitational potential energy converts into translational motion. Thus, the solid cylinder reaches the bottom first.
Conclusion
This answer key provides detailed solutions that align with the core principles of rotational dynamics. By working through each problem methodically—applying torque equations, using conservation laws, and connecting rotational concepts to linear motion—you strengthen both your analytical skills and your intuition for how rotating systems behave. Regular practice with these types of problems, combined with careful review of the underlying physics
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and mathematical relationships, will help you master this challenging but rewarding topic. Remember that rotational motion is deeply interconnected with linear motion, and developing fluency in translating between these two frameworks is key to solving complex mechanics problems. As you continue your studies, focus on understanding not just the formulas, but the physical meaning behind each concept—whether it's the relationship between torque and angular acceleration, the role of moment of inertia in determining rotational response, or how energy distributes between translational and rotational kinetic forms. With consistent practice and thoughtful analysis of each problem's underlying principles, you'll find that rotational dynamics becomes a powerful tool for understanding the behavior of everything from spinning tops to galaxies But it adds up..
