Circular or Angular motion=Circular Motion
When a body moves in a circle then it move through a certain angle . This type of Motion Is called. Angular or circular motion. It is two dimensional motion.
Angular displacement :- The angle that a body traces while moving in a circle is called angular displacement angular displacement is denoted by
Explanation :- Consider a particle is moving in acircal of radius r . let at any instant its position is op1making θ1with x-axis. At later time let its position be op2 making angle θ2with x-axis
Its shown in figure
The angular displacement ∆θ traces by particle can be written as
∆θ = θ2 – θ1
The angular displacements is considered as positive when rotation is anti clock wise and negative when rotation is clock wise.
Unit :- The S-I unit of angular displacement is radian (rad), but it is also measured in degree (deg) and revolution (rev)
Angular displacement is vector quantity and its direction is determined by Right hand rule
Right hand rule:- Grasp the axis of rotation in Right hand Then curl the fingers in the Direction of rotation the erect thumb will indicate of angular displacement,
1 radian:- If the length of arc is equal to the length of radius then angular displacement is said to be 1 radian.
Relation B/w linear and angular displacement consider a body is moving in a circle of radius r As shown in figure
Where S is length of arc Ac and θ is angle subtended by an arc Ac . Now take another arc AB whose length is equal to the radius of circle r .
In this case arc AB subtend an angle of 1 radian. The length of arcs AC and AB. Is proportional to the angle subtended At o by both arcs.
So =
Where <AOC =θ
<AOB = 1 radian
Arc Ac = S
Arc AB = r
So =
Or θ = × 1rad
Or S= rθ
Relation between degree and radian . The circum ference of a circle is divided into 360° equal parts each part is 1 degree
Since s= rθ
Or θ =
For complete trip S= 2r
So θ =
. θ = 2 radian
So we can write
2 radian = 360degree
1 radian = degree
1 radian = 57.3 degree
Angular velocity :- The rate of change of Angular displacement is known as angular velocity denoted by w(moega)
Mathematic ally
W͞ =
Unit :- The SI unit of angular velocity is Radian per second ()
Angular velocity is a sector quantity rand rule
Right Hand rule:- Grasp the axis of rotation in your right hand then curl your fingers along the direction of rotation the erect thumb will point in The direction of angular velocity w.
Relation between linear and angular velocity . consider a body is moving in a circle of radius r . let in time interval ∆t body moves from point P1 to P2 . In this care body cover linear displacement ∆s and angular displacement ∆θ as shown in figure
As we known that
S = r θ
Or ∆s = r ∆θ
Dividing both sides by ∆t
So =
Where is linear velocity v and is angular velocity w
So The above equation becomes
V= rw
In vector form
V͞ = w͞ × r͞
Angular acceleration :- The rate of change of angular velocity is known as angular acceleration . It is denoted by (alpha)
Mathematically
͞ =
͞ =
Unit :- The SI unit of angular velocity is radian per second-2 (rad s-2)
Relation b/w linear and angular acceleration
We know that
V= r w
Or ∆v = r ∆w
Dividing both sides by ∆t
= r
Where is linear acceleration a and is angular acceleration
So the above equation becomes
A= r
In vector form
A= ×r
Linear equation Angular equation
Vf = vi +at wf = wi + t
S= vit + at2 θ = wit + t2
2as = vf2- vi2 2θ =wf2-wi2
Centripetal acceleration:- when a body is moving in a circle then its direction of velocity is changing on every point. The rate at which the direction of velocity changes is known as centripetal Acceleration. It is denoted by ac.
The direction of centripetal acceleration is always towards the centre of circle
Explanation:- Let us consider a body of mass m is moving in a circle of radius r with constant speed v. Let v1 represent linear velocity of body at point A and V2 at point B. The magnitude of velocity is same on every point of circle but its direction is changing from point to point.
As shown in figure
In order to calculate ∆v we join the tail of v1and v2then we get two isosceles triangles AOB and PQR
Using the property of isosceles triangle
=
Where = v and = ac
So =
Rearranging ac =
Since centripetal acceleration is always directed towards the centre so we assign – ve sign
A͞ = r˄ ( vector form ) ………….(1)
Since v= rw
Putting value of v in eq (1)
A͞ = (rw) r˄
A͞ = – r˄
Or a͞ = rw2 r˄
Now v=
For complete rotation s= 2r and t=T (time period)
So v=
Putting the value of v in eq (1)
. a͞ = - ()2 r˄
. a͞ = - r˄
. a͞ = - r˄
Centripetal force = The force required to keep the moving body in a circular path is known as centripetal force.
Centripetal force is denoted by Fc According to new ton’s second 2aw
F= ma
Or Fc = mac
Putting value of ac in above equation
Fc = m , F= mrw2
Examples of circular motion :- the normal reaction of a road surface in equal to the weight of vehicle if the surface is flat. If the road curves up or down the normal reaction changes. Here we will discuss two cases.
When road curved upward :- When a vehicle is moving on a road curved upward then it experiences a normal reaction less then its weight . The difference provides the centripetal force . as shown in fig
So mg – N1 =
When road curved down ward :- when a vehicle is moving on a road curved down ward the it experiences a normal reaction greater then its weight then its weight . The difference provide the centripetal force . As shown in figure
So N2 – mg =
The conical pendulum:- Consider a bob of mass m is attached with a string and moves in a horizontal circle of radius r then. Such pendulum.
Is known as conical pendulum. Tension T is making an angle θ with vertical line while its weight is acting downward as shown in fig
Resolving tension into its components
Horizontal component (T sinθ)
Vertical component (T cosθ)
The horizontal component of tension provide the centripetal force
So = Tsinθ …………..(1)
The vertical component of force balances the weight there fore
Mg = T cosθ ……….(2)
Dividing equation (1) by (2) we get
=
Or =
Tanθ =
Or θ = tan-1()
Worked example :- a ball of mass m is tied to on e end of a chord and is whirled from the other end in a vertical circle of radius r with a constant speed v
Find the tension in the chord. When the ball is at
the highest point
The lowest point
Solution :- At highest point . at highest point tension and weight are in same direction so their sum provide centripetal force i.e
T +mg =
Or T = - mg
At lowest point :- At lowest point tension and weight are in opposite direction therefore the difference provide centripetal force
T- mg =
T= + mg
Moment of inertia :- It is tendency of a body due to which it resist a change in its angular velocity or
The product of mass and square of radius is called moment of inertia.
Explanation:- consider a body of mass m attached to the end of a mass less rod as shown in figure
Consider a force F act on the body perpendicular to the rod then according to new ton’s 2nd 2aw.
F= ma …….(1)
The applied force produce torque in the rod about o
So ϥ = r͞ × F͞
Ϥ= rFsin
, = 90°
Ϥ = rF(1) sin90° = 1
Putting the value of F we have
Ϥ = r m a
Now a= r
So ϥ = rm (r)
Ϥ = r2m
Where m r2 = I (moment of inertia )
So ϥ = I ……..(2)
Equation (1) and (2) are an alogus . In eq (2) F is replaced by ϥ , a by and m by mr2. The moment of inertia Depends not only on mass but also on r2. Moment of inertia is a scalar
Quantity
Unit :- The SI unit of moment of inertia is kg m2 moment of inertia of a disc.
Let us consider a solid disc rotating about its axis. In order to calculate moment of inertia of disc we divide the disc into many particles of masses (m1 , m2 , m3 …….) located at distances (r1 , r2 , r3 ……………) from the axis of rotation .
As shown in figure
Moment of inertia of
The fist particle
I1 = m1 r12
Moment of inertia of the second particle
I2 = m2r22
Moment of inertia of the Nth particle
In = mn rn2
So the total moment of inertia of the disc
I = I1+I2+I3+……………….+In
Or I = m1r12+ m2r22+ m3r32+…………..+mnrn2
In summition form
I = mi ri2
I=1
Angular momentum :- The rotational momentum of a body Is called angular momentum .
Or
The cross product of linear momentum and distance of object from axis of rotation. It is denoted by L͞
Mathematically L͞ = r͞ × P͞
L = rp sin
Where is the angle between r͞ and P͞ the direction of angular momentum is perpendicular to both r͞ and P͞ . Its direction is determined by right hand rule.
If particle is moving in a circle the
L͞ = r͞ × P͞
Or L= rP sin
L = rmv sin
L = rmv sin . = 90°
L = mvr sin 90°
L = mvr
Since v= rw
So l= m rw r
L = mr2w
Where mr2= I
So L= Iw or L͞ = Iw͞
From above equation angular momentum can also be defined as the product of the moment of inertia and angular velocity .
Unit = The S-I unit of angular momentum
Is kg m2 s-1 or Js .
Law of conservation of angular momentum . This law state that in the absence of external momentum of the system remains constant .
If ϥext = 0
Ltotal = L1 + L2 + ……………constant
Or L1 = L2
I1w1 = I2w2
According to law of conservation of angular momentum, the angular momentum Iw must remains constant . Hence if Iderrease w must increase in the same ratio so that the product Iw remains constant. And It w decrease then I must increase such that product remains constant.
Rotational K.E :- let a body of mass m is rotating about its axis o .This body is made up of small elements of masses (m1 , m2 , m3 …….)At a distances (r1 , r2 , r3 ……………)repectively.
Each element has a linear velocity (v1,v2v3…….)
As shown in figure
The K.E of first
Element K.E1 = m1 v12
Since v1= r1w
So K.E1 = m1r12w2
K.E of second element
K.E2 = m1r12w2
K.E of nth element
K.En = mn rn2 w2
So the total K.E
Total K.E = K.E1 + K.E2 + K.E3 +,………………+ K.En
Or total K.E = m1 r12 w2+ m2 r22 w2 +…………..+ mn rn2 w2
Total K.E = (m1 r12 + m2 r22 + m3 r32 + mn rn2) w2
Total K.E = ( ᵹN miri2 = I (moment of inertia )
I = 1
So Total K.E = Iw2
Artificial satellites:- satellites are objects that orbit around the earth . The moon is the natural satellite of the earth. Artificial satellites are man made satellites and the are put into orbit by rocketes at a certain height from the surface of earth. Any satellite is kept in orbit by gravitational attraction of earth.
Let a satellite of mass m I revolving distance r from centre of earth. As shown in figure
The centripetal force is provided by gravitational force
There fore
Fg =Fc
=
V2 =
Or v =
Where r = re+h
If the height of satellite is small compare to radius of earth then
R = Re
So v = ……………..(1)
Since above equation have no mass of satellite hence orbital velocity is independe of the mass of satellite
As G = 6.67 × 10-11 N-
Me = 6 × 1024kg
Re = 6.4×106m
Putting these values in eq (1)
V =
Or v= 8.3
This is the minimum velocity necessary to put a satellite in a orbit .
Time period :- The time required to complete 1 revolution is know as time period
As v=
Or t =
For one complete trip s= 2πRe and t = T (time period)
T =
Putting values of re and v
T = 83 min
Geostationary satellites:- A type of satellite whose period of rotation around the earth is equal to the period of rotation of earth about its axis. Is known as geostationary satellites .
These satellites complete one revolution in 24 hours.
Let a geostationary satellite of mass m is revolving around the earth at a instance r from the centre of earth.
As shown in figure
