TORQUE, ANGULAR, MOMENTUM AND EQUIVALENCE
Torque, Angular, Momentum and EquilibriumTorque or Moment of
ForceDefinitionIf a body is capable of rotating about an axis, then
force applied properly on this body will rotate it about the axis
(axis of rotating). This turning effect of the force about the axis of
rotation is called torque.Torque is the physical quantity which
produces angular acceleration in the body.ExplanationConsider a body
which can rotate aboutO (axis of rotation). A force F acts on point P
whose position vector w.r.t O is r.Diagram Coming SoonF is resolved
into F1 and F2. θ is the angle between F and extended line of r.The
component of F which produces rotation in the body is F1.The magnitude
of torgue (π) is the product of the magnitudes of r and F1.Equation
(1) shows that torque is the cross-product of displacement r and force
F.Torque → positive if directed outward from paperTorque → negative if
directed inward from paperThe direction of torque can be found byusing
Right Hand Rule and is always perpendicular to the plane containing r&
F.ThusClockwise torque → negativeCounter-Clockwise torque →
positive@import
"/extensions/GoogleAdSense/GoogleAdSense.css";Alternate Definition of
Torqueπ = r x F|π| = r F sin θ|π| = F x r sin θBut r sin θ = L
(momentum arm) (from figure)Therefore,|π| = F LMagnitude of Torque =
Magnitude of force x Moment ArmNoteIf line of action of force passes
through the axis of rotation then this force cannot produce torque.The
unit of torque is N.m.CoupleTwo forces are said to constitute a couple
if they have1. Same magnitudes2. Opposite directions3. Different lines
of actionThese forces cannot produces transiatory motion, but produce
rotatory motion.Moment (Torque) of a CoupleConsider a couple composed
of two forces F and -F acting at points A and B (on a body)
respectively, having position vectors r1 & r2.If π1 is the torque due
to force F, thenπ1 = r1 x FSimilarly if π2 is the torque due to force-
F, thenπ2 = r2 x (-F)The total torque due to the two forces isπ = π1 +
π2π = r1 x F + r2 x (-F)π = r1 x F - r2 x (-F)π = (r1 - r2) x Fπ = r x
Fwhere r is the displacement vector fromB to A.The magnitude of torque
isπ = r F sin (180 - θ)π = r F sin θ .................... {since sin
(180 - θ) = sin θ}Where θ is the angle between r and -F.π = F (r sin
θ)But r sin θ is the perpendicular distancebetween the lines of action
of forces F and -F is calledmoment armof the couple denoted by d.π =
FdThus[Mag. of the moment of a couple] = [Mag. of any of the forces
forming the couple] x [Moment arm of the couple]Moment (torque) of a
given couple is independent of the location of origin.Centre of
MassDefinitionThe centre of mass of a body, or a system of particles,
is a point on the body that moves in the same way that asingle
particle would move under the influence of the same external forces.
The whole mass of the body is supposed to be concentrated at this
point.ExplanationDuring translational motion each point of a body
moves in the same manner i.e., different particles of the body do not
change their position w.r.t each other. Each point on the body
undergoes the same displacement as any other point as time goes on. So
the motion of one particle represents the motion of the whole body.
But in rotating or vibrating bodies different particles move in
different manners except one point called centre of mass. The centre
of mass of a body or a system of particle is a point which represents
the movement of the entire system. It moves in the same way that a
single particle would move under the influence of same external
forces.Centre of Mass and Centre of GravityIn a completely uniform
gravitational field, the centre of mass and centre of gravity of an
extended body coincides. But if gravitational field is not uniform,
these points are different.Determination of the Centre of MassConsider
a system of particles having masses m1, m2, m3, ................. mn.
Suppose x1, and z1, z2, z3 are their distances on z-axis, all measured
from origin.EquilibriumA body is said to be in equilibrium if it is1.
At rest, or2. Moving with uniform velocityA body in equilibrium
possess no acceleration.Static EquilibriumThe equilibrium of bodies at
rest is called static equilibrium. For example,1. A book lying on a
table2. A block hung from a stringDynamic EquilibriumThe equilibrium
of bodies moving with uniform velocity is called dynamic equilibrium.
For example,1. The jumping of a paratrooper by a parachute is an
example of uniform motion. In this case, weight is balanced by the
reaction of the air on the parachute acts in the vertically upward
direction.2. The motion of a small steel ball through a viscous
liquid. Initially the ball has acceleration but after covering a
certain distance, its velocity becomes uniform because weight of the
ball is balanced by upward thrust and viscous force of the liquid.
Therefore, ball is in dynamic equilibrium.Angular
MomentumDefinitionThe quantity of rotational motion in a body is
called its angular momentum. Thus angular momentum plays same role in
rotational motion as played by linear momentum in translational
motion.Mathematically, angular momentum is the cross-product of
position vector and the linear momentum, both measured in an inertial
frame of reference.ρ = r x PExplanationConsider a mass 'm' rotating
anti-clockwise in an inertial frame of reference. At any point, let P
be the linear momentum and r be the position vector.ρ = r x Pρ = r P
sinθ ........... (magnitude)ρ = r m V sinθ .......... {since P = m
v)where,V is linear speedθ is the angle between r and Pθ = 90º in
circular motion (special case)The direction of the angular momentum
can be determined by the Righ-Hand Rule.Alsoρ = r m (r ω) sin θρ = m
r2 ω sin θUnits of Angular MomentumThe units of angular momentum in
S.I system are kgm2/s orJs.1. ρ = r m V sin θ= m x kg x m/s= kg.m2/s2.
ρ = r P sin θ= m x Ns= (Nm) x s= J.sDimensions of Angular Momentum[ρ]
= [r] [P]= [r] [m] [V]= L . M . L/T= L2 M T-1Relation Between Torque
and Angular MomentumORProve that the rate of change of angular
momentum is equal to the external torque acting on the body.ProofWe
know that rate of change of linear momentum is equal to the applied
force.F = dP / dtTaking cross product with r on both sides, we getR x
F = r x dP / dtτ = r x dP / dt ............................. {since r
x P= τ}Now, according to the definition of angular momentumρ = r x
PTaking derivative w.r.t time, we getdρ / dt = d / dt (r x P)=> dρ /
dt = r x dP / dt + dr / dt x P=> dρ / dt = τ + V x P
.................. {since dr / dt = V}=> dρ / dt = τ + V x mV=> dρ /
dt = τ + m (V x V)=> dρ / dt = τ + 0 ................. {since V x V =
0}=> dρ / dt = τOr, Rate of change of Angular Momentum = External
Torque ................. (Proved)
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