(9) Torque and angular acceleration

  • While discussing and defining torque or moment of force ,we found that necessary condition for a body not to rotate is that resultant torque about any point should be zero
  • However this condition is necessary but not sufficient for a rigid body to be static for example in absence of resultant torque a body once set in rotation will continue to rotate with constant angular velocity
  • Analogous to translation motion when torque acts on a rigid body rotating about a point with constant angular velocity then angular velocity of the body does not remain constant but changes with angular acceleration α which is proportional to the externally applied torque
  • Consider a force Fi acting on the ith particle of mass mi of the rigid body pivoted about an axis through point O as shown below in the figure

    Force acting on the ith particle at point P

  • This force Fi as discussed earlier has two components one parallel to the radius vector ri and one perpendicular to the ri
  • Component of force parallel to radius vector does not have any effect on the rotation of the body
  • Component of force Fi perpendicular does affect the rotation of the body and produces torque about point O through which the body is pivoted which is given by
    τi=Fi⊥ri ---(21)
  • if Fi⊥ is the resultant force acting on the ith particle ,then from Newton’s second law of motion
    Fi⊥=miai⊥ = miriα ----(22)
    where ai⊥ is the tangential acceleration of the body
  • From equation (21) and (22)
    And taking sum over all the particles in the body we have
    ∑τi=∑(miri2α)=α∑(miri2) ---(23)
    as angular acceleration is same for all the particles of the body
  • we know that
    ∑(miri2) =I
    where I is the moment of inertia of the rigid body .Hence in terms of moment of inertia equation 23 becomes
    ∑τ=Iα                      ---(24)
    we have denoted resultant torque acting on the body ∑τsub>i as ∑τ
  • Both the torque and angular acceleration are vector quantities so in vector form
    τ=Iα                ---(25)
  • Alternatively equation (24) which is rotational analogue of Newton second law of motion ( ∑F=ma) can be written as
    τ=Iα = I(dω/dt)=d(Iω)/dt                      ---(26)
    which is similar to the equation
    where p is the linear momentum
  • The quantity Iω is defined as the angular momentum of the system of particles
    Angular momentum =Iω
  • From equation 26 we see that resultant torque acting on a system of particles equal to the rate of change of the angular momentum

(10) Angular momentum and torque as vector product

(A) Angular momentum
  • In any inertial frame of refrance the moment of linear momentum of a particle is known as angular momentum or, angular momentum of a particle is defined as the moment of its linear momentul.
  • In rotational motion angular momentum has the same significance as linear momentum have in the linear motion of a particle.
  • Value of angular momentum of angular momentum is equal to the product of linear momentum and p(=mv) and the position vector r of the particle from origin of axis of rotation.

    Angular momentum

  • Angular mmomentum vector is usually represented by L.
  • If the linear momentum of any particle is p=mv and its position vector from any constant point be r then abgular momentum of the particle is given by
    L = r×p = m(r×v)               (1)
  • Angular momentum is a vector quantity and its direction is perpandicular to the direction of r and p and could be found out by right hand screw rule.
  • From equation 1 scalar value or magnitude of angular momentum is given as
    |L|=rpsinθ                    (2)
    where V is the angle between r and p.
  • For a particle moving in a circular path
    v=ω×r;                    (3)
    where ω is the angular velocity.

    Angular momentum and angular velocity

    L=m[r×(ω×r)] = m{ω(r.r)-r(r.ω)} = mr2ω=Iω;                    (4)
    (r.ω)=0 because in circular motion r and ω are perpandicular to each other. Here I is the moment of inertia of the particle about the given axis also the direction of L and ω is same and this is a axial vector.
    writing equation 1 in the component form we get

  • Writing angular momentum in component form we get

    writing equation 5 again we get

    Comparing unit vectors on both the sides we get

  • Unit of angular momentum in CGS is gm.cm2/sec and in MKS system it is Kgm.m2/sec or Joule/sec.
(B) Torque
  • The turning effect of the force about the axis of rotation is called the moment of force or torque..
  • In rotational motion torque has same importance as that of force in the linear motion.
  • Torque due to a force F is measured as a vector product of force F and position vector r of line of action of force from the axis of rotation.
  • We already know that orque is denoted by letter τ.

    Torque and angular momentum

  • If F is the force acting on the particle and r is the position vector of particle with respect to constant point then the torque acting on the particle is given by
    τ=r×F                         (8)
  • FRom equation 8 magnitude or resultant of torque is given by
    |τ|=rfsinθ                         (9)
    where θ is the angle between r and F.
  • From equation 9 if θ=900 this menas r is perpandicular to F then,
  • FRom equation 8 magnitude or resultant of torque is given by
    and if θ=00 this menas r is parallel to F then,
  • Unit of torque is Dyne-cm or Newton-m
(C) Relation between angular momentum and torque
  • Differentiating equation 1 w.r.t. t we get

  • But from Newton's second law of motion we have

    Hence rate of change of angular momentum with time is equal to the torque of the force.

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