Two Dimensional Motion

 Two Dimensional Motion

Two Dimensional Motion

Two Dimensional Motion

Questions: Deduce the equations below by vector method.

clip_image002

clip_image004

clip_image006

Let, a body moves with uniform accelerationclip_image008, Let, at time ti = 0, then initial position vector

clip_image010, and initial velocityclip_image012, Again time tf = t, then final position vectorclip_image014and final velocity clip_image016

In physics, acceleration is the rate at which the velocity of a body changes with time.

clip_image018

When, t = 0, then clip_image020and when t = t thenclip_image022, in this range, Integrate the above equation,

clip_image024

clip_image026

clip_image028

Let, a body moves with uniform accelerationclip_image008[1], Let, at time ti = 0, then initial position vector

clip_image010[1], and initial velocityclip_image012[1], Again time tf = t, then final position vectorclip_image014[1]and final velocity clip_image016[1]

So, after t second the average velocity of the particle is

clip_image030

clip_image032

clip_image034

clip_image036

clip_image038

clip_image040

clip_image042

Let, a body moves with uniform accelerationclip_image008[2], Let, at time ti = 0, then initial position vector

clip_image010[2], and initial velocityclip_image012[2], Again time tf = t, then final position vectorclip_image014[2]and final velocity clip_image016[2]

In physics, acceleration is the rate at which the velocity of a body changes with time.

clip_image044

clip_image046

When, t = 0, then clip_image020[1]and when tf = t thenclip_image022[1], in this range, Integrate the above equation,

equation we get,

clip_image048

clip_image050

clip_image052

clip_image054

clip_image056

Since, the velocity of a moving body is defined as its rate of displacement so by the definition,

clip_image058

clip_image060

When, ti = 0, then clip_image062 and when, tf = t thenclip_image064, in this range, Integrate the above equation we get,

clip_image066

clip_image068

clip_image070

clip_image072

clip_image074

clip_image076

Let, a body moves with uniform acceleration clip_image078. Let, at time ti=0 then initial position vector clip_image080 and Initial velocityclip_image082, Again time tf = t then final position vector clip_image084and final velocityclip_image086.

In physics, acceleration is the rate at which the velocity of a body changes with time.

clip_image088

clip_image090

When, t= 0, then clip_image062[1] and when, t = t thenclip_image064[1], in this range, Integrate the above equation we get,

clip_image048[1]

clip_image050[1]

clip_image052[1]

clip_image054[1]

clip_image056[1]

By dot product of clip_image092we get,

clip_image094

clip_image096

clip_image098

clip_image100

clip_image102

clip_image104

Two Dimensional Motion

Projectile:

Projectile motion is the act of projecting an object into the air at an angle. A few real life examples of projectile motion are throwing a tennis ball, the motion of football, missile etc.

Two Dimensional Motion

Equation of motion of a projectile:

Let a projectile begin its flight from a point O (Fig: right side) Taking

O as Origin and let the horizontal and vertical direction be considered along

clip_image106X and Y-axis respectively.

Let the initial velocity or

velocity of projection = vo

Angle of Projection = θo

The horizontal acceleration

of the projectile, ax = 0;

The vertical acceleration of

the projectile, ay = -g;

Since ay is negative because

the direction of g is negative,

Since the point of projection

and origin is same, ∴ xo = yo = 0

The horizontal component of initial velocity = vo cosθo

And the vertical component of initial velocity = vo sinθo

The change of motion along X-axis is dependent on acceleration along X-axis.

The change of motion along Y-axis is dependent on acceleration along Y-axis.

At a certain interval of time t, the projectile cross the point P(x, y) where its velocity is = v

Then the acceleration along X-axis is, ax = 0

Displacement along X-axis is = x

x = vocosθot + ½axt2

or, x = vocosθot + 0[ax = 0]

or, x = vocosθot

clip_image108

The acceleration along Y-axis is, ay= g;

Displacement along Y-axis is = y

y = vo sinθot – ½gt2

clip_image110[By putting the value of t]

clip_image112

clip_image114

This is an equation of a parabola.  ∴  Path of motion of projectile is parabolic.

 

Questions: In case of Projectile Find the expression of (i) Required time to reach the maximum height (ii) Maximum height (iii) Time of flight (iv) Range (v) Maximum Range

(i) Required time to reach the maximum height:

Let a projectile begin its flight from a point O (In Fig) Taking O as Origin and let the horizontal and vertical direction be considered along X and Y-axis respectively. At a certain interval of time t, the projectile stay at the point P(x, y) where its velocity is = v

(i) Time to reach maximum height:

Vertical component of vo isvo sinθo

After time t, the final velocity at maximum height,

clip_image115

vy = vosinθogt………………(1)

But the velocity at maximum height, vy= 0…………. (2)

By putting the value, vy= 0, in equation (1)

(1) 0 = vo sinθo – gt

clip_image117

(ii) Maximum height :

Let, the maximum height be H

H = vo sin θo t – ½ gt2

clip_image119

clip_image121

clip_image123

(iii) Time of Flight :

Let, the time of flight T i.e. is time taken from the point of projection and return to the ground is T.

Vertical displacement at time t, y = vo sin θ ot – ½gt2. Let the time of flight be T. In time T the projectile returns to the ground. I.e. t = T and y = 0,

0 = vosin θ oT – ½gT2

or, ½ gT2= vo sin θoT

clip_image125

(iv) Range :

Let, Range =R, R is the distance travelled along the horizontal direction in the time of flight T.

R = ( vocos θ o ) × T

clip_image127 [By putting the value of T from Eq. (5)]

clip_image129

clip_image131

Maximum Range:

Let Rmax be the maximum range, R will be maximum when sin2θ 0 = 1

or, Sin2θ 0 = Sin900

or, 0 = 900

θ0 = 450 so, if a body is thrown at an angle 450 with the horizontal direction, the range will be maximum.

∴Maximum range, clip_image133

clip_image135

clip_image137

clip_image139

Angular acceleration and proof of clip_image141:

Angular acceleration of an object in circular motion is defined as the time rate of change of its angular velocity. it is generally denoted by α.

Letclip_image143be the change in angular velocity of the object in time interval t and clip_image145while moving on a circular path, then angular acceleration of the object will be, clip_image147

S.I unit of angular acceleration is rad s-2.

We know that linear velocity v is denoted with angular velocity clip_image149 by by the relation clip_image151

Differentiatingclip_image153with respect to time, we have

clip_image155

clip_image157

a=rα where a is the linear acceleration and α is the angular acceleration.

In terms of vector rotation, we have clip_image141[1]

 

Linear velocity and Angular velocity and Relation of clip_image159:

 

Linear velocity:

The rate of change of displacement is called linear velocity. Velocity is the speed of an object and a specification of its direction of motion. Speed describes only how fast an object is moving, whereas linear velocity gives both how fast and in what direction the object is moving. The linear velocity of a moving body is defined as its rate of displacement. The linear velocity may be uniform or variable. It is donated by clip_image161. It is a vector quantity and it’s unit is ms-1. Velocity,clip_image163

Angular velocity:

angular velocity of an object in circular motion is defined as the time rate of change of its angular displacement. It is denoted byclip_image149[1].

Let, clip_image166 be the angle through which the body moves in time clip_image168Angular Velocity; clip_image170.

The unit of angular velocity is rad s-1

The Dimension of angular velocity is clip_image172

Relation:

Let us consider an object is moving along a circular path of radius r = OC=OB, in a anti clock wise direction with a uniform angular velocity. If the object rotted one turn in T second then angular displacement clip_image174

clip_image176

clip_image177 Naw in T Sec. the object travels clip_image179

clip_image181

clip_image183

clip_image185

From equation (1) and (2) we gate

clip_image187

clip_image189

clip_image191 Linear velocity = angular velocity × radius of the circle.

Again, We know clip_image193 is a vector quantity and radius vector is also a vector quantity. Therefore, the cross product of two vectors will also be a vector. Let the resultant of the cross productclip_image195

clip_image197

The magnitude of clip_image199is

clip_image201

clip_image203

clip_image205

In addition, the direction clip_image207are the same (according to the rule of cross-product)

clip_image209

So, from equation (4) and (5)

clip_image211

 

 

Difference between Angular Velocity and Linear Velocity:

 

Sl.no Angular velocity Linear velocity
1. The rate of change of angular displacement is known as angular velocity. The rate of change of linear displacement is known as angular velocity.
2. It is denoted by clip_image213 It is denoted by v
3. Unit of it is rads-1 Unit of it is ms-1
4. Dimensional equation of it is [T-1] Dimensional equation of it is [LT-1]
5. It does not depend on displacement. It depends on displacement.

 

Centripetal Force:

The external force required to make a body moves along circular path with uniform speed is called centripetal force. Mathematically, the centripetal force required to move a body of massm along a circular path of radius r is given by,clip_image215.

Centrifugal Force :

An object traveling in a circle behaves as if it is experiencing an outward force. This force, known as the centrifugal force, depends on the mass of the object, the speed of rotation, and the distance from the center. The more massive the object, the greater the force; the greater the speed of the object, the greater the force; and the greater the distance from the center, the greater the force. Mathematically, the centrifugal force required to move a body of mass m along a circular path of radius r is given by,clip_image215[1].

Expression for centripetal acceleration and centripetal force:

Let us consider an object of mass m is moving along a circular path of radius r, in anticlockwise direction with a uniform speed v, and an angular velocity clip_image213[1] with at center O. When the body is come from A to B in small time variation t. The velocity at A is the Tangent along AC.

Now, in rectangular OAEB, we get

clip_image218∠AEB+∠AOB =180°

Again, ∠AEB+∠BEC = 180°

Spouse, ∠AOB = ∠BEC = θ

The horizontal component of velocity at A, vy = 0

and the vertical component, vx = v

The vertical component of velocity along AC, vy = vsinθ

and the horizontal component, vx = vcosθ

when t is very small then θ is also very small

sin θ = θ, and cos θ = 1

the vertical component of velocity at B, vy = vθ

and the horizontal component, vx = v

It appears that, there is no change of velocity along horizontally,

So, centripetal acceleration,

clip_image220

clip_image222

clip_image224 clip_image226

clip_image228 clip_image230

clip_image232

and Centripetal Force = clip_image234

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Two Dimensional Motion

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Two Dimensional Motion

Two Dimensional Motion

Two Dimensional Motion

Two Dimensional Motion

Two Dimensional Motion

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