 # torque

## Torque

• Consider two forces F1 and F2 having equal magnitude and opposite direction acting on a stick placed on a horizontal table as shown below in the figure • Here note that line of action of forces F1 and F2 is not same .So they tend to rotate the stick in clockwise direction
• This tendency of the force to rotate an object about some axis is called torque
• Torque is the rotational counterpart of force. torque tends to rotate an body in the same way as force tends to change the state of motion of the body
• Figure below shows a rigid body pivoted at point O so that point O is fixed in space and the body is free to rotate • Let P be the point of application of force. This force acting at point P makes an angle θ with the radius vector r from point O to P
• This force F can be resolved into two components
F=Fsinθ
F||=Fcosθ
as they are perpendicular and parallel to r
• Parallel component of force does not produce rotational motion of body around point O as it passes through O
• Effect of perpendicular components producing rotation of rigid body through point O depends on magnitude of the perpendicular force and on its distance r from O
• Mathematically ,torque about point O is defined as product of perpendicular component of force and r i.e.
τ=Fr=Fsinθr=F(rinθ)=Fd              ---(18)
where d is the perpendicular distance from the pivot point ) to the line of action of force F
• Quantity d=rinθ is called moment arm or liner arm of force F .If d=0 the there would be no rotation
• Torque can either be anticlockwise or clockwise depending on the sense of rotation it tends to produce
• Unit of torque is Nm
• Consider the figure given below where a rigid body pivoted at point O is acted upon by the two force F1 and F2
• d1 is the moment arm of force F1 and d2 is the moment arm of force F2 • Force F2 has the tendency to rotate rigid body in clockwise direction and F1 has the to rotate it in anti clockwise direction
• Here we adopt a convention that anticlockwise moments are positive and clockwise moment are negative
• hence moment τ1 of force F1 about the axis through O is
τ1=F1d1
And that of force F2 would be
τ2=-F2d2
• Hence net torque about O is
τtotal= τ1+ τ2
=F1d1-F2d2
• Rotation of the body can be prevented if
τtotal=0
or τ1=-τ2
• We earlier studied that when a body is in equilibrium under the action of several coplanar forces ,the vector sum of these forces must be zero i.e.
ΣFx=0 and ΣFy=0
• we know state our second condition for static equilibrium of rigid bodies that is
" For static equilibrium of rigid body net torque in clockwise direction must be equal to net torque in anticlockwise direction w.r.t some specified axis i.e.
Στ=0
"
• Thus for static equilibrium of an rigid body
i) The resultant external force must be zero
ΣF=0
ii) The resultant external torque about any point or axis of rotation must be zero i.e.
Στ=0

## Torque in Vector Notation

• 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 torque 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 means r is perpendicular to F then,
• FRom equation 8 magnitude or resultant of torque is given by
|τ|=rF
and if θ=00 this means r is parallel to F then,
|τ|=0
• Unit of torque is Dyne-cm or Newton-m

## (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 Newton's 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