# Rotation

## (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

• 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 Newtons second law of motion
Fi⊥=miai⊥ = miriα ----(22)
where ai⊥ is the tangential acceleration of the body
• From equation (21) and (22)
τi=miri2α
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
F=d(mv/dt=dp/dt
where p is the linear momentum
• The quantity Iω is defined as the angular momentum of the system of particles
Angular momentum =Iω
L=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
τ=dL/dt

## (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 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.

Therefore
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.

• 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
|τ|=rF
and if θ=00 this menas r is parallel to F then,
|τ|=0
• 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.

Confused and have questions?
Head over to Chegg and use code “CS5OFFBTS18” (exp. 11/30/2018) to get \$5 off your first month of Chegg Study, so you can get step-by-step solutions to your questions from an expert in the field. Plus get a free 30 minute tutoring session.