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Lesson 15 of 15

System of Particles and Rotational Motion · Lesson 15 of 15

Chapter Summary and Practice

“Connect centre-of-mass motion, equilibrium and rotation through model choice and mixed practice.”

Learning Objectives

• Choose the correct system, reference origin and axis. • Connect particle sums to whole-body laws. • Distinguish force, torque and conservation conditions. • Apply formulas with their assumptions and units.

A connected description of an extended body

Start by deciding whether deformation is negligible. Choose a system boundary, frame, origin and rotation axis. The CM describes overall translation; angular variables describe orientation and rotation. For CM acceleration sum external forces. For angular-momentum change sum external torques about the same reference. Internal forces can redistribute energy and motion even when they cancel from the total-force sum. To describe both translation and rotation, one set of variables is not enough.

QuestionMain relationCondition or caution
Where is CM?R = Σmᵢrᵢ/MUse signed coordinates and actual mass weights
How does CM accelerate?MA = F_extConstant-mass particle system
Is linear momentum conserved?dP/dt = F_extCheck resultant force or relevant component
How large is torque?τ = Fd⊥State origin and perpendicular lever arm
How does angular momentum change?dL/dt = τ_extCentral internal forces cancel pair torques
Is a supported body at rest in equilibrium?ΣF = 0 and Στ = 0Include weights, reactions and friction
How does a fixed-axis rotor accelerate?τ_z = IαConstant I; otherwise d(Iω)/dt
What is rotational energy?K_rot = Iω²/2Use the same axis as I
How does internal redistribution change spin?I_iω_i = I_fω_fNegligible external axial torque

Connections that prevent common mistakes

Mass tells how much material a body contains; I also records where the material is relative to an axis. CM and CG coincide in uniform gravity but express different ideas. Force and torque equilibrium are independent conditions. Torque and energy have equal dimensions yet different meanings. Linear speed differs between points of a rotor even though angular velocity is common. Angular momentum can remain constant while angular speed and kinetic energy change. These distinctions connect the chapter’s many formulas.

For an equilibrium calculation, draw the force diagram first. For a cross product, fix the order and right-hand direction. For a rotation problem, convert revolutions to radians and name the inertia axis. For a conservation calculation, include the complete system and identify the missing external force or torque component. A numerical answer is incomplete if its sign, axis or physical interpretation is unclear.

Practice Problems

Practice Problems
  1. Locate CM for masses 2 kg at x = 0 and 6 kg at x = 4 m. Explain the inverse division of distances.
  2. A uniform L-plate consists of three equal 1 m squares. Derive its CM by component centres.
  3. A uniform disc loses a radius-R/2 hole centred R/2 to the right. Find the remaining CM and explain the sign.
  4. Derive MA = F_ext by differentiating the CM definition and explain which Newton laws are used.
  5. Use a = (3, −4, 5) and b = (−2, 1, −3) to find the dot product, cross product and area of their spanning triangle.
  6. Show that a · (b × c) gives signed parallelepiped volume and explain the absolute value for volume.
  7. Derive the torque law for a particle and show why central internal forces cancel in a system.
  8. Explain how zero resultant force can coexist with nonzero torque, and give a couple example.
  9. Find reactions for the 70 cm loaded bar and the ladder discussed earlier; identify every moment arm.
  10. Use all three equilibrium equations for the angled-string suspended bar. Explain why its CG is 0.72 m from the left end.
  11. Derive I and rotational energy for a thin ring and a pair of equal end masses on a light rod.
  12. Convert 1200 rpm and 3120 rpm to SI angular speeds and obtain the revolutions during the 16 s acceleration interval.
  13. Solve the cord-driven flywheel by torque, work–energy and angular kinematics; compare results.
  14. For a system changing from I to 2I/5 without axial torque, find angular-speed and kinetic-energy ratios. Explain the energy source.
  15. Explain why L_z = Iω remains useful even when the full L vector is not parallel to ω.

Selected answers and reasoning

CM and vector checks

Problem
Check the first and vector-product problems.

  1. 1.The two-mass CM is 6(4)/8 = 3 m from the 2 kg mass, so its distances from the masses are 3 m and 1 m.
  2. 2.For the L-plate the two coordinates are each 5/6 m. The hole problem gives X = −R/6 and Y = 0.
  3. 3.The vector dot product is −25 and cross product is (7, −1, −5). Its magnitude is √75 = 5√3, so triangle area is 5√3/2 in the appropriate squared edge-length units.
Rotation and equilibrium checks

Problem
Connect the numerical applications.

  1. 1.Beam reactions are 54.88 N and 43.12 N. For the ladder, wall reaction and floor friction are each 34.65 N and floor normal is 196 N.
  2. 2.The motor-wheel angular acceleration is 4π rad/s² and its angular displacement is 576 revolutions.
  3. 3.The 20 kg, 0.20 m flywheel has I = 0.40 kg m², α = 12.5 rad/s² and final rotational energy 50 J after 2 m of cord unwinds.
  4. 4.Reducing total I to 2I/5 multiplies angular speed and kinetic energy by 5/2. Angular momentum stays fixed and internal work supplies the energy rise.

Quiz

Quick check

Which pair of equilibrium conditions is required for static rigid-body balance?

Explanation

Static equilibrium requires both translational and rotational balance.

Quick check

Which conservation condition is correct?

Explanation

The torque law specifies the conservation condition.

Quick check

What must accompany a tabulated I value?

Explanation

Different axes through the same body can give different I.

Quick check

Which body may have CM in empty space?

Explanation

The ring’s symmetry places its CM at its empty centre.

Quick check

If a driving torque balances friction at steady ω, what is true?

Explanation

The engine can supply energy dissipated by resistance without changing ω.

Key Takeaways

Key Takeaways

• Separate CM translation from motion relative to CM. • Reference origin, axis and system boundary belong in the solution. • Force, torque and energy relations answer different questions. • Use conservation and standard formulas only with their stated conditions.