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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

    Two forces acting on stick are equal in magnitude but acting in opposite direction tends to produce the torque

  • 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

    force action of the rigid body pivoted at any point produces a torque

  • 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

    Figure shows the different forces trying to rotate the body in different direction


  • 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

    Figure shows the force action on wheel pivoted at 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

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