Unit 7 Torque And Rotation Workbook Answers

Unit 7 Torque and RotationWorkbook Answers: A Comprehensive Guide

Understanding torque and rotational motion is essential for mastering mechanics, and Unit 7 of most physics workbooks focuses on translating linear concepts into their rotational counterparts. This article walks through the core ideas, outlines a step‑by‑step problem‑solving framework, provides detailed solutions to representative workbook questions, highlights frequent pitfalls, and answers common student queries. By the end, you’ll have a clear roadmap for tackling any torque‑and‑rotation exercise and the confidence to check your work against the workbook’s answer key.

1. Why Torque and Rotation Matter

Torque ((\tau)) is the rotational analogue of force; it quantifies how effectively a force causes an object to spin about an axis. Rotation introduces new quantities—angular displacement ((\theta)), angular velocity ((\omega)), angular acceleration ((\alpha)), moment of inertia ((I)), and rotational kinetic energy ((K_{rot})). Mastering these concepts lets you analyze everything from opening a door to the dynamics of galaxies.

Main keyword: unit 7 torque and rotation workbook answers
Semantic keywords: torque, moment of inertia, angular acceleration, rotational equilibrium, conservation of angular momentum, rotational kinetic energy, problem‑solving strategy.

2. Core Concepts and Formulas

Important notes

• The lever arm (r) is the perpendicular distance from the axis to the line of action of the force.
• Sign convention: counter‑clockwise torques are usually taken as positive.
• For a system of particles, (I) adds linearly; for continuous bodies, use integration or tabulated values (e.g., (I_{solid;cylinder}= \frac12 MR^2) about its central axis).

3. Problem‑Solving Strategy for Torque and Rotation

1. Identify the axis of rotation and draw a clear free‑body diagram (FBD).
2. List all forces acting on the object, noting their points of application and directions.
3. Compute each torque using (\tau = rF\sin\theta) (or (\tau = r\times F) in vector form). Assign signs according to the chosen rotation direction.
4. Sum the torques to obtain (\tau_{net}). 5. Apply Newton’s second law for rotation: (\tau_{net}=I\alpha). Solve for the unknown (often (\alpha), (\omega), or (\theta)).
5. If needed, use kinematic equations for constant (\alpha) to find angular displacement or time.
6. Check energy or momentum conservation when external torques are zero (e.g., collisions, spinning ice skater).
7. Verify units and reasonableness of the answer (does the direction make sense? Is the magnitude plausible?).

4. Sample Workbook Problems with Detailed Solutions

Below are five representative questions that mirror the style and difficulty found in Unit 7 workbooks. Each solution follows the strategy above and explains every algebraic step.

Problem 1 – Basic Torque Calculation

A 10 N force is applied at the end of a 0.5 m wrench, making a 30° angle with the wrench handle. What is the magnitude of the torque about the bolt?

Solution

1. Lever arm (r = 0.5) m.
2. Force component perpendicular to the wrench: (F_{\perp}=F\sin\theta = 10\sin30° = 10 \times 0.5 = 5) N.
3. Torque magnitude: (\tau = rF_{\perp}=0.5 \times 5 = 2.5) N·m.
4. Direction: Using the right‑hand rule, the torque tends to rotate the bolt counter‑clockwise (positive).

Answer: (\boxed{2.5\ \text{N·m (counter‑clockwise)}})

Problem 2 – Net Torque and Angular Acceleration

A uniform solid disk of mass 4 kg and radius 0.2 m is free to rotate about its central axis. Two forces act tangentially at the rim: (F_1 = 12) N clockwise and (F_2 = 8) N counter‑clockwise. Find the angular acceleration of the disk.

Solution

1. Moment of inertia for a solid disk: (I = \frac12 MR^2 = \frac12 (4)(0.2)^2 = 0.5 \times 4 \times 0.04 = 0.08) kg·m².
2. Torques:
   • (\tau_1 = rF_1 = 0.2 \times 12 = 2.4) N·m (clockwise → negative).
   • (\tau_2 = rF_2 = 0.2 \times 8 = 1.6) N·m (counter‑clockwise → positive).
3. Net torque: (\tau_{net}= \tau_2 - \tau_1 = 1.6 - 2.4 = -0.8) N·m (negative indicates clockwise).
4. Angular acceleration: (\alpha = \tau_{net}/I = (-0.8)/0.08 = -10) rad/s².

Answer: (\boxed{\alpha = -10\ \text{rad/s}^2}) (10 rad/s² clockwise).

Problem 3 – Rotational Kinetic Energy

A thin hoop of mass 3 kg and radius 0.15 m spins at 20 rad/s. Calculate its rotational kinetic energy.

Solution

1. Moment of inertia for a thin hoop about its central axis: (I =

Problem 3 – Rotational Kinetic Energy (continued)

1. Moment of inertia for a thin hoop: (I = MR^2 = 3 \times (0.15)^2 = 3 \times 0.0225 = 0.0675) kg·m².
2. Rotational kinetic energy: (K_{\text{rot}} = \frac{1}{2} I \omega^2 = \frac{1}{2} \times 0.0675 \times (20)^2 = 0.03375 \times 400 = 13.5) J.

Answer: (\boxed{13.5\ \text{J}})

Problem 4 – Torque, Angular Acceleration, and Kinematics

A 2 kg block is attached to a 0.3 m long string wrapped around a solid cylindrical pulley (mass 1 kg, radius 0.1 m). The block is released from rest. Find (a) the angular acceleration of the pulley and (b) the linear acceleration of the block. Assume no slipping and negligible friction at the pulley’s axle.

Solution

1. Free‑body diagram for block: Tension (T) upward, weight (mg = 2 \times 9.8 = 19.6) N downward. Linear acceleration (a) downward.
   Newton’s second law: (mg - T = ma) → (19.6 - T = 2a).
2. Torque on pulley: Only tension provides torque (weight and normal force act through axis).
   (\tau_{\text{net}} = Tr = I\alpha), with (r = 0.1) m, (I = \frac{1}{2} M r^2 = \frac{1}{2} \times 1 \times (0.1)^2 = 0.005) kg·m².
   No slipping ⇒ (a = r\alpha).
3. Substitute (\alpha = a/r) into torque equation:
   (T \times 0.1 = 0.005 \times (a/0.1)) → (0.1T = 0.005 \times 10a = 0.05a) → (T = 0.5a).
4. Plug into block equation: (19.6 - 0.5a = 2a) → (19.6 = 2.5a) → (a = 7.84) m/s².
5. Angular acceleration: (\alpha = a/r = 7.84 / 0.1 = 78.4) rad/s².

Answers:
(a) (\boxed{\alpha = 78.4\ \text{rad/s}^2})
(b) (\boxed{a = 7.84\ \text{m/s}^2})

Problem 5 – Conservation of Angular Momentum

A figure skater initially spins at 2 rad/s with arms extended, modeling her as a cylinder (mass 50 kg, radius 0.5 m) plus two rods (each mass 5 kg, length 1 m) extended horizontally. She pulls her arms in, reducing her radius to 0.2 m. What is her new angular speed?

Solution

1. Initial moment of inertia:
   • Cylinder (torso): (I_{\text{cyl}} = \frac{1}{2} M R^2 = \frac{1}{2} \times 50 \times (0.5)^2 = 6.25) kg·m².
   • Two arms (modeled as rods rotating about one end): (I_{\text{arm}} = 2 \times \left( \frac{1}{3} m L^2 \right) = 2 \times \frac{1}{3} \times 5 \times (0.5)^2 = 2 \times \frac{1}{3} \times 5 \times 0.25 = 2 \times 0.