motion in plane

7/29/2019 motion in plane

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Physics/Class XI 1 2012 Vidyamandir Classes Pvt. Ltd.

Physics: Motion in a Plane

Figure (a): Shows two equal

vectors A

and B

Figure (b): Shows two vectors X

and Y

are unequal

though they are of same length because they have

different directions.

1. SCALARS AND VECTORS

Scalars: Physical quantities that have only magnitude but no directions are called scalar quantities.For example, mass,

length, time, energy.

Scalar quantity is specified completely by a single number along with the proper unit.

Scalars can be added, subtracted, multiplied and divided just as the ordinary numbers.Vectors: Physical quantities that have magnitude as well direction and follow vector laws of addition are called vector

quantities.For example; velocity, linear momentums, force, impulse, etc.

We use () type sign over any vector quantity, for example momentum vector is represented by an arrow placed

over a letter, sayp

.

The magnitude of a vector is called its absolute value indicated by p p

.

A vector is represented by an arrow over the lettere.g., by , , , L.......v a p

with respective magnitudes denoted by

v, a,p, L {without arrow}.

Unit Vector

A unit vector has unit magnitude. It is used to denote the direction of a given vector.

It has no units and dimension.

Any vector a

can be expressed in terms of its unit vector a as follows

| |a a a

. Here a is in the direction of ,a a

is read as a cap.

| |

aa

a

So, if a given vector is divided by its magnitude we get a unit vector.

Three Rectangular Unit Vectors

The three rectangular unit vectors , ,i j k are shown as follows:

i denotes the direction of X-axis. j denotes the direction of Y-axis.

k denotes the direction of Z-axis.

These three vectors are collectively known as base vectors.

Equality of Vectors

Two vectors are said to be equal if they have equal magnitude and same direction.

If two vectors A

and B

are represented by two equal parallel lines drawn with same scale having arrow heads in the same

direction then, A

and B

are equal vectors. i.e., A

= B


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Multiplication of Vectors by Real Numbers

1. Multiplying a vector a

with a positive number gives another vector a

whose magnitude is changed by the factor :

if 0 a a

(i) If is positive, then the direction of a

is the same as that of a

.

In the following figure a

is multiplied by 2 and we get another vector b

such that 2b a

The magnitude of b

is twice the magnitude of a

, while the direction of b

is the same as that of a

.

(ii) If is negative, then the direction of a

is opposite to that of a

Again, in the following figure a

is multiplied by 2. The new

vector b

is such that

2b a

The magnitude ofb

is twice the magnitude a

of but direction opposite to that of a

.

2. If a

is multiplied by zero, we get a vector whose magnitude is zero and whose direction is arbitrary. This vector is

called a zero vector ornull vector.

3. If is a pure number and has no units, then the units ofa

are the same as those ofa

. But, generally is scalar with

certain units. In these cases, the units ofa

will be obtained by multiplying the unit of a

by unit of.

The multiplication of a vector by a real number assumes a lot of significance in statements like velocity of car B is

double the velocity of car A.

2. ADDITION AND SUBTRACTION OF VECTORS-GRAPHICAL METHOD

1. Triangle Law of Vector Addition: (Head to Tail Method)

If two non-zero vectors represent the sides of triangle then their resultant is given by the third side of the triangle

taken in reverse order.

Let us consider two vectors A

and B

that lie in a plane as shown in

the adjoining figure.

To find the sum A + B

, we place vector B

so that its tail is at the head

of the vectorA

. Then, we join the tail of A

to the head ofB

. The line

OQ represents a vector R

, which is the sum of the vectors A

and B

.

Since in this procedure of vector addition, vectors are arranged head to tail, this graphical method is called the head-

to-tail method.

Vector addition obeys commutative law.

A B B A

Vectors addition obeys the associative law (A B) C A (B C)

Zero Vector or Null vector: Let us consider two vectors A

and A

as shown in the figure below:

Their sum is A ( A)

. Since, the magnitudes of two vectors are the same, but the directions are

opposite, the resultant vector has zero magnitude and is represented by 0. This is called a null

vector or a zero vector. i.e., A A 0

, 0 0


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(1) When a zero vector is added to any vector A

, we get A

only..

A 0 A

(2) When a zero vector is subtracted from A

we get A

only..

A 0 A

(3) When a zero vector is multiplied by a non-zero scalar we get zero vector

0 0

(4) When a vector B

is multiplied by zero, we get zero vector

0(B) 0

PROPERTIES OF A ZERO VECTOR

Physical Significance of Zero Vector ( 0

)

The physical significance of zero vector can be clearly understood from the following examples:

(i) The displacement of a ball thrown up and received back by the thrower is a zero vector.(ii) The velocity vector of a stationary body is zero vector.

(iii) The acceleration vector of a body in a uniform motion is a zero vector.

Subtraction of vector can be defined in terms of

addition of vectors.

We define the difference of two vectors A

and

B as the sum of two vectors A and B .

A B A ( B)

. It is shown in figure. The

Vector B

is added to vector A

to get

2R A ( B) A B

, vector1R A B

is also shown in the same figure for comparison.

2. Parallelogram Law of Vectors

Parallelogram law of vectors is another useful law for the addition of two vectors.

It states that: If two vectors acting simultaneously at a point, can be represented both in magnitude and

direction by the two adjacent sides of a parallelogram drawn from a point, then the resultant is representedcompletely both in magnitude and direction by the diagonal of the parallelogram passing through that point.

In the adjoining figure P

vectors and Q

are completely represented by the two sides OA and OB respectively

of a parallelogram.

Then, according to parallelogram law of vectors, the diagonal OC of the parallelogram will give the resultant R

such that R P Q


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Analytical Method

1. In Two Dimensions: It is much easier to add vectors by combining their respective components. Consider two

vectors A

and B

inx-yplane with componentsAx,A

yandB

x,B

y:

A = AAx i + AAy j and B = Bx i + By j

Let R

be their sum. We have R

= A

+ B

= (Ax

+ AAy) i + (Bx+ By)

j

Since vectors obey the commutative and associative laws, we can arrange and regroup the vectors.

R

= (Ax+B

x) i + (A

y+B

y)

j

Since, R

=Rxi +R

y

j

Thus, each component of the resultant vector R

is the sum of the corresponding components ofA and B.

2. In Three Dimensions: We have,

A

=A

x i

+ Ay

j + Azk

and B

=Bx i

+ By

j + Bzk

R

= A

+ B

=Rxi + R

y

j +Rzk with R

x= AA

x+ B

x, R

y= AA

y+B

y, R

z= AA

z+B

z

This method can be extended to addition and subtraction of any number of vectors.

Let us analytically calculate the magnitude and direction of the resultant vector

R

.

Let be the angle between two given vectors P

and Q

. From C, draw a

perpendicular CN on OA (produced) as shown in the adjoining figure.

R A B

=Rx i + Ry

j +Rzk withR

x=A

x+B

x, R

y=A

y+B

y, R

z=A

z+B

z

This method can be extended to addition and subtraction of any number of vectors.

Let us analytically calculate the magnitude and direction of the resultant vectorR .

Let be the angle between two given vectors P

and Q

. From C, draw a perpendicular CN on OA (produced) as shown

in the adjoining figure.

In the right-angled ANC,

CNsin or CN ACsin

AC [AC = OB = Q]

Or CN = Q sin (1)

Also,AN

cos or AN AC cos AC

Or AN Q cos

Now, ON = OA + AN = P + Q cos (2)

Consider the right-angled ONC,

2 2 2OC ON CN

Or 2 2 2R (P Q cos ) (Qsin ) [From (2) and (1)]


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Or 2 2 2 2 2 2R P Q cos 2PQcos Q sin

2 2 2 2 2P Q cos Q sin 2PQcos

2 2 2 2

P Q (cos sin ) 2PQ cos

2 2 2R P Q 2PQcos [2 2sin cos 1 ]

2 2R P Q 2PQcos (3)

The equation above is the required expression for the magnitude of the resultant of two vectorsP and Q

inclined to each

other at an angle .

Let be the angle which the resultant R

makes with P

.

Then,

CN

tan ON (in rt. dONC)

OrQsin

tanP Qcos

[from (2) and (1)] (4)

1 QsintanP Qcos

The above equation gives the direction of the resultant vector.

Special Cases

Case (i): When the given vectorsP

and

Q

act in the same direction.

In this case, = 0

R = 2 2P Q 2PQcos 0 [From equation (3)]

= 2 2P Q 2PQ [ cos 0 = 1]

= 2(P Q) P + Q

Or R

= P Q

So, the magnitude of the resultant vector is equal to the sum of the magnitudes of the given vectors.

Also, tan =Qsin0

P Qcos0

[From equation (4)]

Or tan = 0 [ sin 0 = 0]

= 0

So, the resultant vector points in the direction of the given vectors.


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Case (ii): When the given vectors P and Q act at right angles to each other..

In this case, = 90

R = 2 2P Q 2PQcos90

Or R = 2 2P Q [ cos 90 = 0]

Or | R |

= 2 2| P | | Q |

Also, tan =Qsin90

P Qcos90

Or tan =Q

P[ sin 90 = 1]

Or = 1 Qtan P [ cos 90 = 0]

If P = Q, then R = 2 2P P

Or R = 22P 2P

Also, in this case, tan =P

1P

= 45

Case (iii):When the given vectors P and Q act in opposite directions.

In this case, = 180

R = 2 2P Q 2PQ cos180

Also, in this case, tan =P

1P

= 45

Case (iii):When the given vectors P and Q act in opposite directions.

In this case, = 180

R = 2 2P Q 2PQ cos180

Or = 2 2P Q 2PQ [ cos 180 = 1]

= 2(P Q)

R = (P Q) P Q or Q P

Or | R |

= | P | ~ | Q |

[ | P | ~ | Q |

implies positive difference between | P |

and | Q |

.]

So, the magnitude of the resultant vector is equal to the positive difference of the magnitudes of the given vectors.


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Also, tan =Qsin180

P Qcos180

Or tan = 0 = 0 or 180

When | P | | Q | , then = 0

When | P | | Q |

, then = 180

Clearly, the resultant vector acts in the direction of the bigger of the two vectors.

3. RESOLUTION OF VECTOR

The process of splitting a vector is called resolution of a vector.

The parts obtained after resolution are known as components of the given vector.

Consider two non-zero vectors a

and b

in a plane.

Let A

be any other vector in this plane. Through the tail (P) of A

, draw a straight line parallel to a

.

Similarly, draw a straight line parallel to b

through the terminal point (Q) of

A

. Let both the lines intersect at C.

Applying triangle law of vectors, A PC CQ

(1)

As per the geometrical construction, PC a

and CQ b

, where

are real numbers.

Thus, from eq. (1) A a b

. So, A

has been resolved along a

and b

.

Using this method one can resolve a given vector into two component

vectors along a set of two vectors all the three lie in the same plane.

4. RECTANGULAR COMPONENTS OF VECTOR

If the components of a given vector are perpendicular to each other, then they are called rectangular components. These

are the most important components of a vector.

Let us consider a vector A

represented by OP

. Through the point O, draw two mutually perpendicular axes-X-axis and

Y-axis. Let the vector A

make an angle with the X-axis. From the point P, drop a perpendicular PN on X-axis.

Now ON( A )x

is the resolved part of A

alongx-axis.

It is also known as thex-components of A

or the horizontal component of A

.

Ax

may be regarded as the projection of A

on X-axis.

OM( A )y

is the resolved part of A

along Y-axis.

It is also known as they-component of A

or the vertical component of A

.

Ay

may be regarded as the projection of A

on Y-axis.

So, Ax

and Ay

are the rectangular components of A

.

Applying triangle law of vectors to the vector triangle ONP, we get

A A Ax y

A

Ax

A


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In right-angled triangle ONP,

Acos or A Acos

A

xx (1)

Asin or A Asin

A

y

y (2)

Squaring and adding (1) and (2), we get

2 2 2 2 2 2A A A cos A sin x y

Or2 2 2 2 2A A A (cos sin )x y

2 2 2A A Ax y 2 2[ cos sin 1]

Or2 2A A Ax y

This equation gives the magnitude of the given vector in terms of the magnitudes of the components of the given vector.

5. DOT PRODUCT OF TWO VECTORS

Dot product of two vectors is defined as the product of the magnitude of one vector & the magnitude of the component

of other vector in the direction of first vector.

Properties

1. . . . 1i i j j k k 2. . . . 0i j j k k i 3. A.B B.A

4.2

A.A A

Examples : -

1. W = F.S

2. P = F.v

3. F.S

6. VECTOR PRODUCT OR CROSS PRODUCT OF TWO VECTORS

The vector product or cross product of two vectors A

& B

is another vector C

. Whose magnitude is equal to the

product of the magnitude of two vectors & sine of the smaller angle between them.

A B A B sin c

where c is a unit vector in the direction of c

. The direction of c

or c is perpendicular to the plane containing A B

&

direction is given by right hand thumb rule

Properties

1.

0i j j j k k

2. i j k j k i k i j

3. j i k k j i i k j

A B

= B A

Examples

1. = r F

2. L = r P

3. v = r

4. a = r

5. ca =


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7. POSITION AND DISPLACEMENT VECTOR

Position Vector: Position vector r

of a particle P located in a plane with reference to the origin of anx-y reference frame

(Fig. (a)) is given by

OP

= r xi yj

In magnitude, 2 2| |r x y

Where x and y are components of r along x- and y- axes (coordinates of the

object).

Position vector gives the position of a point with reference to any point at any

instant of time.

Similarly for three dimensional co-ordinate system, the position vector is given

by

r xi yj zk

In magnitude, 2 2 2| |r x y z

Displacement Vector: Suppose a particle moves along the curve shown

by the thick line and is at point P at time tand P at time t2.

Then, the displacement is: 'r r r

and is directed from P to P

We can write above equation in a component form:

' 'r x i y j x i y j

= i x j y

Where x =xx, y =yy

8. VELOCITY & ACCELERATION

Velocity: The average velocity of an object is the ratio of displacement

and corresponding time interval

x y z

r x y z v i j k v i v j v k

dt dt dt dt

Then x y zv v i v j v k

Since

rv

t

, the direction of average velocity v

is same as that of r

.

In magnitude

2 2 2| | x y zv v v v

The instantaneous velocity is given by the limiting value of the average velocity as the time interval approaches zero.

0

lim

t

r drv

t dt

Now we can express v in a component form

0

lim t

d r x y z v i j k

dt t t t

= 0 0 0

lim lim lim t t t

x y zi j k

t t t


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Or

x y z

dx dy dz v i j k v i v j v k

dt dt dt

where x y ydx dy dz

v v vdt dt dt

Magnitude ofv is then

v = 2 2 2x y zv v v

And the direction ofv is given by the angle :

tan y

x

v

v

1 tany

x

v

v

The direction of velocity at any point on the path of an object is tangential to the path at that point and is in the direction

of motion.

Acceleration: Average acceleration of an object is equal to ratio of the change in velocity to the time interval.

( )

x y zv i v j v k va

t t

=

yx zvv v

i j kt t t

x y za a i a j a k

The instantaneous velocity is given by the limiting value of the average velocity as the time interval approaches zero.

0limt

vat

0

lim

yx z

t

vv vdra i j k

dt t t t

= 0 0 0

lim lim lim

yx z

t t t

vv vi j k

t t t

Or yx z

x y z

dvdv dva i j k a i a j a k

dt dt dt

wherex

x

dva

dt ,

y

y

dva

dt and

z

z

dva

dt

In one dimension, the velocity and the acceleration of an object are always along the same straight line (either in the

same direction or in the opposite direction).

However, for motion in two or three dimensions, velocity and acceleration vectors may have any angle between 0 and

180.

NOTE


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9. MOTION IN A PLANE WITH CONSTANT ACCELERATION

Motion in a plane (two-dimension) can be t reated as two separate simultaneous one-dimensional motions with constant

acceleration along two perpendicular directions.

For an object moving with constant acceleration in a plane, the following equation of motion holds true:

Here a

= constant acceleration, r

= position vector of object at any time t,

r

= position vector of object at time t = 0, v

= velocity at any time t,

v

= velocity at t= 0.

1. v v at

In terms of component

x ox xv v a t (along x-axis)

y oy yv v a t (along y-axis)

2.21

2r r v t at

In terms of component

21

2ox xx x v t a t (alongx-axis)

21

2oy yy y v t a t (alongy-axis)

10. RELATIVE VELOCITY IN TWO DIMENSION

Suppose that two objects A and B are moving with velocitiesAv

and

Bv

(each with respect to some common frame of reference, say ground.).

Then, velocity of an object A relative to that of B is:AB A Bv v v

and, the velocity of object B relative to that of A is:BA B Av v v

Therefore,

AB BAv v

and AB BAv v

Consider two cars A and B moving with velocities A Bandv v

respectively in directions

inclined to each other at an angle . Let us find the relative velocity BAv

of B w.r.t. A.A.

Clearly,

BAv

= B A( )v v

Now2

BAv = 2 2

A B A B2 cos (180 )v v v v


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But cos(180 ) = cos

BAv =2 2

A B A B2 cosv v v v (i)

Also, tan =B

A B

sin(180 )

cos(180 )

v

v v

Or tan =B

A B

sin

cos

v

v v (ii)

Let s now find the relative velocityABv

of A w.r.t. B.

ABv

=Av

+ (

Bv

)

Now 2AB

v = 2 2A B A B

2 cos(180 )v v v v

Or ABv =2 2

A B A B2 cosv v v v (iii)

Also, tan =B B

A B A B

sin(180 ) sin

cos(180 ) cos

v v

v v v v

(iv)

It is clear from equations (i) and (iii) BA ABv v

It is further clear from equations (ii) and (iv), =

Conclusion: The magnitude of relative velocity in both the cases is the same. But the direction is reversed.

Special Cases

1. When two bodies are moving along parallel lines in the same direction i.e., when = 0,

ABv = 2 2

BA A B A B2 cos0v v v v v

Or ABv =2

BA A B A B( ) ( )v v v v v

Or ABv = BA A B B A( ) or ( )v v v v v

Conclusion: The relative speed between two bodies moving in the same direction is equal to the difference of the

individual speeds of two bodies.

2. When two bodies are moving along parallel lines in opposite directions i.e., when = 180.

ABv =2 2

BA A B A B2 cos180v v v v v

Or ABv =2 2 2

BA A B A B( ) 2v v v v v ( cos 180 = 1)

Or ABv = ABv

Conclusion: The relative speed between two bodies moving in opposite directions is equal to the sum of the individual

speeds of the two bodies.


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Have you ever noticed that when it is raining you need to hold umbrella depending upon the direction of the rain. Go

out when it is raining. First you stand still and watch out the the way in which you have to hold the umbrella. Now start

running and watch out the direction in which you have to hold the umbrella. Perform this activity and try to apply the

relative velocity concepts.

ACTIVITY

11. PROJECTILE MOTION

A body that is in flight after being thrown or projected through the atmosphere but is not being propelled by any fuel

is called a projectile.

Example:

(i) A bomb released from an aeroplane in level flight.(ii) A bullet fired from a gun.

(iii) A javelin thrown by an athlete.

(iv) An arrow released from a bow.

The path followed by a projectile is called trajectory.

The motion of a projectile is two-dimensional.

When we consider the motion of a projectile, the following assumptions are made:

(i) There is no resistance due to air.

(ii) The effect due to curvature of earth is negligible.

(iii) The effect due to rotation of earth is negligible.

(iv) For all points of the trajectory, the acceleration due to gravity g is constant in magnitude and direction.

Trajectory of Projectile

Consider a projectile thrown with velocity v at an angle with thex-axis(horizontal).

The velocity v can be resolved into two rectangular components

(i) v cos alongx-axis

(ii) v sin alongy-axis.

The motion of the projectile is two-dimensional motion.

It is supposed to be made up of two motions horizontal motion (alongx-

axis) and vertical motion (alongy-axis) occurring simultaneously.

Horizontal motion: The horizontal motion of the projectile is uniform in nature as the net external force acting on the

body is zero.

Thus, the equations of motion of the projectile for horizontal direction are simply the equations of uniform motion in

a straight line.

The horizontal motion takes place with constant velocity v cos .

Vertical motion: The vertical motion of the projectile is an uniformly accelerated motion controlled by the force of

gravity. This is because the only force acting on the projectile is the force of gravity, acting vertically downward.


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Mathematical expression: Let us take initial position of the object to be origin of the reference frame, therefore,

0 & 0x y . Letx and y be the horizontal & vertical distance covered in time t, then

( cos ) or cos

xx v t t

v

21( sin )2

y v t gt

If andx yv v are the horizontal and vertical velocity components at any time t,

xv = v cos and yv = v sin gt

If andx ya a are the horizontal and vertical acceleration components at any time t,

0 &x ya a g

Since we know,21sin t

2y v gt

Substituting the value ofcos

xt

v ,

We get

21

sincos 2 cos

x xy v g

v v

Or

2

2 2sin

2 cos

gxy x

v

This is a first degree equation iny and a second degree equation inx. This is the equation of a parabola. So, the path

followed by the projectile, i.e., the t rajectory of the projectile is parabolic. It is clear that the t rajectory is completely

known ifv cos is known.

It should be kept in mind that the above equation is valid only if lies between 0 and2

.

* Maximum Height

It is denoted by axh or H. It is also known as vertical range. It is the maximum height to which a projectile rises abovethe horizontal plane of projection. In order to calculate the maximum height H, we

make use of the fact that the velocity ( )yv t of the projectile at the maximum

height is zero. If1t be the time taken by the projectile to reach maximum height,

then

10 sinv gt

Or 1 singt v

Or 1sinv

tg

So, when 1' ' , ( ) Ht t y t

H = 21 11

( sin )2

v t gt

Or H =

2 2 2 2 2sin 1 sin sin sinsin or H

2 2 2

v v v v v g

g g g g

Or

2 2sinH

2

v

g


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Time of Flight

It is the total time taken by the projectile to return to the same level from where it was thrown. Time of flight is equal to

twice the time taken by the projectile to reach the maximum height. This is because the time of ascent is equal to the time

of descent.Time of flight, T = 2t, wheretis the time taken by the projectile to reach maximum height.

Now, at the highest point ( ) 0, , (0) sin y y yv t a g v v

We know that ( ) (0)y y yv t v a t

Substituting values 0 sinv gt

Or singt v

Orsinv

tg

2 sinT v

g

* Horizontal Range

It is the total horizontal distance from the point of projection to the point where the projectile comes back to the plane

of projection. It is denoted by R.

To calculate horizontal range R, consider horizontal motion of the projectile. The horizontal motion is uniform with

constant velocity v cos .

R = v cos time of flight

R = v cos 2 sinv

g

Or

2 (2sin cos )R

v

g

Or

2 sin2R

v

g ( 2sin c os sin 2 )

* Maximum Horizontal Range

For a given velocity of projection and a given place, the value of R will be maximum when the value of sin 2 is maximumi.e., 1.

For R to be maximum, sin 2 = 1 (maximum value)Or sin 2 = sin 90 Or = 45

So, for a given velocity, the angle of projection for maximum range is 45, i.e.,4

.

Maximum horizontal range,

2

maxR

v

g .


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A boy and a girl are tossing an apple back and forth between them. The sketch at the right shows the path the apple

followed when watched by an observer looking on from the side. The apple is moving

from the left to the right. Five points are marked on the path. Ignore air resistance.

(a) Make a copy of this figure. At each of the marked points, draw an arrow that

indicates the magnitude and direction of the apples velocity when it passes

through that point.

ACTIVITY

12. UNIFORM CIRCULAR MOTION

When an object follows a circular path at a constant speed, the motion of the object is called uniform circular motion.

The word uniform refers to the speed, which is uniform (constant) throughout the motion i.e. the body travelsequal distance in equal interval of time.

Consider an object moving with uniform speed v in a circle of radiusR as shown in Figure.

Since the rate change in velocity is acceleration, in this case the body is moving with uniform speed but velocity

of the object is changing continuously in direction, thus object undergoes acceleration.

Let us find the Magnitude and the direction of this acceleration.

Let r

and

r be the position vectors and v

and v

the velocities of the object when it is at point P and P as shown

in Fig. (a)

By definition, velocity at a point is along the tangent at that point in the

direction of motion.

Since the path is circular, v

is perpendicular to r

and so is v

to r

.

Therefore, v is perpendicular to r.

Since average acceleration is along v v

at

, the average acceleration

a

is perpendicular to r

andit is directed towards the centre of the circle.

In Fig. (b), 0t and the average acceleration becomes the instantaneous

acceleration. It is directed towards the

centre. Thus, we find that the acceleration

of an object in uniform circular motion is

always directed towards the centre of the circle.

By definition the magnitude of a

is given by: 0lim

t

va

t

Let the angle between position vectors r

and r

be . Since the velocity vectors

v

and v

are always perpendicular to the position vectors, the angle between them

is also . Therefore, the triangle CPPformed by the position vectors and the triangle

GHI formed by the velocity vectors , and v v v

are similar (Fig. (a)). Therefore, the

ratio of the base-length to side-length for one of the triangles is equal to that of the

other triangle. That is:

Fig. (b)

Fig. (a)


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v r

v R

Orr

v vR

Therefore,0 0 0

lim lim limR Rt t t

v v v r va

t t t

If t is small, will also be small and then arc PP can be approximately taken to be | r|:

r v t

vv

t

0lim

t

rv

t

Therefore, the centripetal acceleration acis:

2

c

v va v

R R

Thus, the acceleration of an object moving with speed v in a circle of radius R has a magnitude

2v

R and is

always directed towards the centre.

This is why this acceleration is called centripetalacceleration (a term proposed by Newton).

Since v andR are constant, the magnitude of the centripetal acceleration is also constant.

However, the direction changes pointing always towards the centre.

Therefore, a centripetal acceleration is not a constant vector.We have another way of describing the velocity and the acceleration of an object in uniform circular motion. As the

object moves from P to P in time t(= tt), the line CP turns through an angle called angular distance.

We define the angular speed (Greek letter omega) as the time rate of change of angular displacement:

=

t

Now, if the distance traveled by the object during the time tis s, i.e. PP is s, then:

sv

t

But s =R Therefore:

v R R

t

v R

We can express centripetal acceleration acin terms of angular speed:

2 2 22

2

c

c

v Ra R

R R

a R


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The time taken by an object to make one revolution is known as its time period T

The number of revolution made in one second is called its frequency (= 1/T).

However, during this time the distance moved by the object is s 2R

Therefore,2R

2RT

In terms of frequency , we have

2 2

2

2R

4 Rc

v

a

Whirling Bucket: Put a little water in a bucket - tie a string firmly to the bucket handle and then swing the bucket in a

vertical circle. As long as the rate of rotation is great enough the water stays in the bucket! Slowing the rate of rotation

can get the water to almost fall out at the top of the path and you can usually hear it slopping around at this critical

point.

Make sure that the handles of the bucket do not come off and that the bucket does not hit the floor at the lowest point

of the circle, or the ceiling. This is best done outside; in any case avoid the temptation to stand on anything.

The force with which you are pulling on the string provides the centripetal force needed to keep the bucket orbiting.

Check that you understand the direction in which this force acts towards your hand.

At the same time, you can feel a force pulling on your hand. This force is the equal and opposite reaction to your pull.

Since it acts outwards from the centre of the circle, it can be described as centrifugal, or centre-fleeing.

Now can you identify the centripetal force and its direction?

ACTIVITY


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1. A person stands at 80 m from a building and throws a ball which just enters a window 36 m above the ground. Calculate

the velocity of projection of the ball. [Ans. 40.1 ms1]

2. A particle is projected with a velocity v so that its horizontal range is twice the greatest height attained. Find its horizontal

range. [Ans.

24

5

v

g]

3. Find a vector A

and its magnitude with initial point P (2, 2, 1) and terminal point Q (3, 1, 2). [Ans. A = 11 ]

4. What is the magnitude and direction of i j . [Ans. = 45]

5. What is the property of two vectors A

and B

, if A B A B

. [Ans. = 90]

6. Find the unit vector perpendicular to each of the vectors 3 2i j k and 2 2 4i j k . [Ans.8 3 ]

7. Calculate the area of the parallelogram whose sides are represented by 3 i j k and i j k . [Ans.4.89]

8. A particle moves from position 2 3 6i j k to position 12 13 9i j k in metres when a uniform force 4 2i j knewton acts on it. Calculate the work done by the force. [Ans. W = 30 J]

9. For what value of, A 2i j k

is perpendicular to B 4 2 2i j k

. [Ans. = 3]

10. At which point of projectile motion (i) potential energy is maximum (ii) kinetic energy is maximum (iii) total mechanical

energy is maximum. [Ans. (i)1

2m u2 sin2 , (ii)

1

2m u2 cos2 (iii)

1

2mu2]

11. A particle is projected at an angle from the horizontal with kinetic energy K. What is the K.E. of the particle at thehighest point. [Ans. K Cos2]

12. Pick out the two scalar quantities in the following list: force, angular momentum, work, current, linear momentum,

electric field, average velocity, magnetic moment, relative velocity.13. What will be the effect on the horizontal range of a projectile when its initial speed is doubled keeping its angle of

projection same.

14. A unit vector is represented by ai bj ck . If the values ofa and b are 0.6 and 0.8 respectively, find the value ofc.

[Ans. Zero]

15. What is the velocity and momentum of the body at highest point of projectile motion.

16. Two equal forces have their resultant equal to either. What is the inclination between them.

17. A cyclist has to bend a little inwards from his vertical position while turning. Why.

18. Prove the following statement, For Elevation which exceed or fall short of 45 by equal amount, the range is equal.19. A lady walking due east on a road with velocity of 10 m/s encounters rain falling vertically with a velocity of 30 m/s. At

what angle she should hold her umbrella to protect herself from the rain.

20. A swimmer can swim with velocity of 10 km/h. w.r.t. the water flowing in a river with velocity of

5 km/h. (i) In what direction should he swim to reach the point on the other bank just opposite to his starting point.


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Physics/Class XI 20 2012 Vidyamandir Classes Pvt Ltd


***

21. The angle between vector A

and B

is 60. What is the ratio of A.B and AB

. [Ans.1

3]

22. A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from thestation. A dishonest cabman takes him along a circuitious path 23 km long and reaches the hotel in 28 min. What is (a) theaverage speed of the taxi, (b) the magnitude of average velocity. Are the two equal.

[Ans. 49.3 / , 21.4 /a km h b km h ]

23. In long jump, does it matter how high you jump. What factors determine the span of the jump.

24. Rain is falling vertically with a speed of 30 ms1. A woman rides a bicycle with a speed of 10 ms1 in the north to south

direction. What is the direction in which she should hold her umbrella. [Ans. 11

tan3

]

25. The position of a particle is2

3.0 2.0 4.0r t i t j k

given by where tis in seconds and the coefficients have the properunits forrto be in metres.

(a) Find v

and a

of the particle. [Ans. v = 3 4 /i j m s , 2a = 4 j /m s ]

(b) What is the magnitude and direction of velocity of the particle at t= 2s. [Ans. 8.54 /m s , 1tan 2.66 ]

motion in plane

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