Rotation

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

(8) work and power in rotational motion

• We know that when we apply force on any object in direction of the displacement of the object ,work is said to be done
• Similarly force applied to the rotational body does work on it and this work done can be expressed in terms of moment of force (torque) and angular displacement θ
• Consider the figure given below where a force F acts on the wheel of radius R pivoted at point O .so that it can rotate through point O

• This force F rotates the wheel through an angle dθ and dθ is small enough so that we can regard force to be constant during corresponding time interval dt
• Workdone by this force is
dW=Fds
but ds=Rdθ
So
dW=FRdθ
• Now FR is the torque Τ due to force F.so we have
dW=Τdθ ----(19)
• if the torque is constant while angle changes from θ1 to θ2 then
W=Τ(θ21)=ΤΔθ ---(20)
Thus workdone by the constant torque equals the product of the torque and angular displacement
• we know that rate of doing work is the power input of torque so
P=dW/dt=Τ(dθ/dt)=Τω
• In vector notation
P=Τ.ω