Chapter 6: Systems of Particles and Rotational Motion

CHAPTER SIX: SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

6.1 INTRODUCTION

  • Single Particle Motion:

    • Earlier chapters focused on the motion of a single particle (point mass, no size).
    • Results applied to bodies of finite size by assuming motion can be described as that of a particle.
  • Limitations of Particle Model:

    • Real bodies have a finite size, and the idealized model is sometimes inadequate for extended bodies.
    • Extended bodies are systems of particles.
  • Key Concept - Centre of Mass:

    • This chapter focuses on the centre of mass of a system of particles, its motion, and significance in understanding extended body motion.
    • Many problems can be simplified by treating extended bodies as rigid bodies.
  • Rigid Bodies:

    • Definition: A rigid body maintains a fixed shape; distances between all particle pairs do not change.
    • Though no real body is perfectly rigid, many situations can approximate this assumption (e.g., wheels, beams, planets).

6.1.1 Types of Motion for Rigid Bodies

  • Translational Motion:

    • Example: Block sliding down an inclined plane.
    • All particles have the same velocity at any given time (pure translational motion).
  • Rolling Motion:

    • Example: Solid cylinder rolling down an incline.
    • Not pure translational because particles have different velocities at any instant.
  • Rotational Motion:

    • A fixed rigid body will exhibit rotational motion around a fixed axis.
    • Examples: Ceiling fan, potter’s wheel.
    • In rotation, each particle moves in a circular path with the center on the fixed axis.
    • Points on the axis remain stationary, while particles not on the axis describe circles of radius equal to their distance from the axis.

6.2 CENTRE OF MASS

  • Definition and Calculation:

    • For a two-particle system with masses m<em>1m<em>1 and m</em>2m</em>2 at distances x<em>1x<em>1 and x</em>2x</em>2, the centre of mass (C) is given by:
      X=m<em>1x</em>1+m<em>2x</em>2m<em>1+m</em>2X = \frac{m<em>1x</em>1 + m<em>2x</em>2}{m<em>1 + m</em>2}
    • If m<em>1=m</em>2=mm<em>1 = m</em>2 = m, then: X=x<em>1+x</em>22X = \frac{x<em>1 + x</em>2}{2}
    • For nn particles, formula becomes:
      X=extstyleextSumof(m<em>ix</em>i)extTotalMassX = \frac{ extstyle ext{Sum of}(m<em>ix</em>i)}{ ext{Total Mass}}
    • In two dimensions, for three particles:
    • X=m<em>1x</em>1+m<em>2x</em>2+m<em>3x</em>3m<em>1+m</em>2+m3X = \frac{m<em>1 x</em>1 + m<em>2 x</em>2 + m<em>3 x</em>3}{m<em>1 + m</em>2 + m_3}
    • Y=m<em>1y</em>1+m<em>2y</em>2+m<em>3y</em>3m<em>1+m</em>2+m3Y = \frac{m<em>1 y</em>1 + m<em>2 y</em>2 + m<em>3 y</em>3}{m<em>1 + m</em>2 + m_3}
  • Generalization for n Particles in 3D:

    • X=extstyleextSumof(m<em>ix</em>i)M,Y=extstyleextSumof(m<em>iy</em>i)M,Z=extstyleextSumof(m<em>iz</em>i)MX = \frac{ extstyle ext{Sum of}(m<em>ix</em>i)}{M}, Y = \frac{ extstyle ext{Sum of}(m<em>iy</em>i)}{M}, Z = \frac{ extstyle ext{Sum of}(m<em>iz</em>i)}{M}
      where M=extTotalMassM = ext{Total Mass}
  • Centre of Mass of a Rigid Body:

    • For a continuous body, approximate integration can yield positions of a centre of mass similarly as in discrete systems.
    • Homogeneous bodies (e.g., rods, spheres) have their centres of mass at their geometric centre due to symmetry.

6.3 MOTION OF CENTRE OF MASS

  • Using equations of motion, the velocity (V) and acceleration (A) of the centre of mass for a system of n particles can be derived:

    • V=extTotalMomentum(P)MV = \frac{ ext{Total Momentum(P)}}{M}
    • A=extTotalExternalForceMA = \frac{ ext{Total External Force}}{M}
  • External Forces:

    • Only external forces affect the movement of the centre of mass, as internal forces cancel out.
    • Equation governing motion of the centre of mass:
    • MA=FextMA = F_{ext}

6.4 LINEAR MOMENTUM OF A SYSTEM OF PARTICLES

  • Momentum (PP) definition for individual particle: p=mvp = mv

  • For n particles:
    P=m<em>1v</em>1+m<em>2v</em>2+…+m<em>nv</em>nP = m<em>1v</em>1 + m<em>2v</em>2 + … + m<em>nv</em>n

  • Total momentum relates to centre of mass as:
    P=MVP = MV
    where M is the total mass of the system.

  • Conservation of Momentum:

    • If external forces are zero, the total linear momentum remains constant.

6.5 VECTOR PRODUCT OF TWO VECTORS

  • The vector (cross) product of vectors aa and bb is defined to yield a vector cc such that:
    1. ∣c∣=∣a∣∣b∣extsinheta|c| = |a||b| ext{sin} heta
    2. cc is perpendicular to the plane formed by aa and bb.
  • The right hand rule for determining direction: curl fingers in direction from aa to bb, thumb points in the direction of cc.

6.6 ANGULAR VELOCITY AND ITS RELATION WITH LINEAR VELOCITY

  • Angular Velocity (extω\boldsymbol{ ext{ω}}):
    • Defined as the time rate of change of angular displacement:
      extω=dhetadt\boldsymbol{ ext{ω}} = \frac{d heta}{dt}
  • Relation between linear velocity (vv) and angular velocity: v=extωrv = ext{ω}r,
    • where rr is the radius from the axis of rotation.

6.7 TORQUE AND ANGULAR MOMENTUM

  • Torque ($ au$):
    • Defined as the moment of force around a point:
      au=rimesFau = r imes F
  • Also, angular momentum (LL) of a particle is given by:
    L=rimespL = r imes p.
  • For multi-particle systems:
    L=extSumofAngularMomentumofeachParticleL = ext{Sum of Angular Momentum of each Particle}.

6.8 EQUILIBRIUM OF A RIGID BODY

  • A rigid body is said to be in mechanical equilibrium when:
    1. Total forces = 0
    2. Total torques = 0

6.9 MOMENT OF INERTIA

  • Definition: Moment of Inertia (II) is defined by I=extSumof(m<em>ir</em>i2)I = ext{Sum of} \bigg(m<em>ir</em>i^2 \bigg) where rir_i is the distance from the rotational axis.

    • This influences the angular acceleration ( ext{with torque as the cause}) of a rigid body.
  • Moment of Inertia of Common Shapes:

    • Thin Circle: I=MR2I = MR^2
    • Solid Sphere: I=25MR2I = \frac{2}{5}MR^2
    • Thin Rod (axis through centre): I=112ML2I = \frac{1}{12}ML^2

6.10 KINEMATICS OF ROTATIONAL MOTION ABOUT A FIXED AXIS

  • Analogous Kinematic Equations for Rotational Motion:
    • Similar to linear motion equations.
    • For uniform angular acceleration, equations now involve angular displacement ($ heta$), angular velocity ($ ext{ω}$), and angular acceleration ($ ext{α}$):
    1. extω=extω0+extαtext{ω} = ext{ω}_0 + ext{α}t
    2. heta=extω0t+12extαt2heta = ext{ω}_0 t + \frac{1}{2} ext{α}t^2
    3. extω2=extω02+2extαhetaext{ω}^2 = ext{ω}_0^2 + 2 ext{α} heta

6.11 DYNAMICS OF ROTATIONAL MOTION ABOUT A FIXED AXIS

  • Analysis similar to Force and Mass in Linear Motion, with Torque and Moment of Inertia:
    • The torque is computed as needed:
    • au=Iextαau = I ext{α}
  • This parallels Newton's Second Law of Motion for dynamic systems:
    • F=maF = ma for linear systems.

6.12 ANGULAR MOMENTUM IN CASE OF ROTATION ABOUT A FIXED AXIS

  • Angular Momentum (LL) can be expressed regarding motion about the fixed axis:
    • extL=Iextωext{L} = I ext{ω}
    • Consistent with conservation laws: if no external torque ($ au_{ext} = 0$), then momentum and angular momentum are conserved.

SUMMARY

  1. Rigid body characteristics include constant particle distances.
  2. Types of rigid body motion include pure rotation and translational motion.
  3. Angular velocity provides a vectorial description of rotational motion.
  4. Linear and angular momentum have consistent definitions
  5. The conservation of momentum applies to isolated systems in both translations and rotations.

POINTS TO PONDER

  1. No internal forces need for determining center of mass.
  2. Center of gravity coincides with center of mass in uniform gravity fields, diverging otherwise.
  3. Examining rigid body's equilibrium involves available forces and torques for analysis.

EXERCISES

  • A variety of exercises are provided to reinforce concepts of centers of mass, angular momentum, equilibrium, and inertial calculations.