- Introduction
- Angular velocity and angular acceleration
- |
- Rotation with constant angular acceleration
- |
- Kinetic energy of Rotation
- |
- Calculation of moment of inertia
- |
- Theorems of Moment of Inertia
- |
- Torque
- |
- work and power in rotational motion
- |
- Torque and angular acceleration
- |
- Angular momentum and torque as vector product
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- Angular momentum and torque of the system of particles
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- Angular momentum of the system of particles with respect to the center of mass of the system
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- Radius of gyration
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- Kinetic Energy of rolling bodies (rotation and translation combined)
- |
- Solved examples

- Consider two forces F
_{1}and F_{2}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 F
_{1}and F_{2}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=F_{⊥}r=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 F
_{1}and F_{2}

- d
_{1}is the moment arm of force F_{1}and d_{2}is the moment arm of force F_{2}

- Force F
_{2}has the tendency to rotate rigid body in clockwise direction and F_{1}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 τ
_{1}of force F_{1}about the axis through O is

τ_{1}=F_{1}d_{1}

And that of force F_{2}would be

τ_{2}=-F_{2}d_{2}

- Hence net torque about O is

τ_{total}= τ_{1}+ τ_{2}

=F_{1}d_{1}-F_{2}d_{2}

- Rotation of the body can be prevented if

τ_{total}=0

or τ_{1}=-τ_{2} - 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.

ΣF_{x}=0 and ΣF_{y}=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

- 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 θ
_{1}to θ_{2}then

W=Τ(θ_{2}-θ_{1})=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**

Class 11 Maths page Class 11 Physics page

- Principles of Physics Extended (International Student Version) (WSE)
- university physics with modern physics by Hugh D. Young 13th edition
- NCERT Exemplar Problems: Solutions Physics Class 11
- H.C. Verma Concepts of Physics - Vol. 1
- CBSE All in One Physics Class 11 by Arihant
- NCERT Solutions: Physics Class 11th
- New Simplified Physics: A Reference Book for Class 11 (Set of 2 Parts)
- Pradeep's A Text Book of Physics with Value Based Questions - Class XI (Set of 2 Volumes)
- Oswaal CBSE Sample Question Papers For Class 11 Physics (For 2016 Exams)