- Rotational Motion
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- Angular velocity
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- Angular acceleration
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- Rotation with constant angular acceleration
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- Kinetic energy of Rotation
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- Calculation of moment of inertia
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- Parallel Axis Theorem|Theorems of Moment of Inertia
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- Perpendicular Axis Theorem
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- Torque
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- work and power in rotational motion
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- Angular acceleration
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- Relationship between Angular momentum and torque
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- Conservation of Angular momentum
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- Radius of gyration
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- Rolling Motion|Kinetic Energy of rolling bodies
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- Rotational Motion problems with solutions

- 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

- 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=90
^{0}this means**r**is perpendicular to**F**then,

- FRom equation 8 magnitude or resultant of torque is given by

|**τ**|=rF

and if θ=0^{0}this means**r**is parallel to**F**then,

|**τ**|=0

- Unit of torque is Dyne-cm or Newton-m

- 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 F
_{i}acting on the ith particle of mass m_{i}of the rigid body pivoted about an axis through point O as shown below in the figure

- This force F
_{i}as discussed earlier has two components one parallel to the radius vector**r**and one perpendicular to the_{i}**r**_{i}

- Component of force parallel to radius vector does not have any effect on the rotation of the body

- Component of force F
_{i}perpendicular does affect the rotation of the body and produces torque about point O through which the body is pivoted which is given by

τ_{i}=F_{i⊥}r_{i}---(21) - if F
_{i⊥}is the resultant force acting on the ith particle ,then from Newton's second law of motion

F_{i⊥}=m_{i}a_{i⊥}= m_{i}r_{i}α ----(22)

where a_{i⊥}is the tangential acceleration of the body

- From equation (21) and (22)

τ_{i}=m_{i}r_{i}^{2}α

And taking sum over all the particles in the body we have

∑τ_{i}=∑(m_{i}r_{i}^{2}α)=α∑(m_{i}r_{i}^{2}) ---(23)

as angular acceleration is same for all the particles of the body

- we know that

∑(m_{i}r_{i}^{2}) =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**=I**α**---(25)

- Alternatively equation (24) which is rotational analogue of Newton second law of motion ( ∑
**F**=m**a**) can be written as

∑**τ**=I**α**= I(d**ω**/dt)=d(I**ω**)/dt ---(26)

which is similar to the equation

**F**=d(m**v**/dt=d**p**/dt

where**p**is the linear momentum - The quantity I
**ω**is defined as the angular momentum of the system of particles

Angular momentum =I**ω**

**L**=I**ω** - 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**=d**L**/dt

Class 11 Maths Class 11 Physics Class 11 Chemistry

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