Motion in 2 dimensions Vidyamandir Classes

Vidyamandir Classes

Motion in Two Dimensions

Motion in Two Dimensions For a particle moving along a straight line, all the vector quantities - position, velocity, displacement and acceleration have only one non-zero component and hence can be treated as positive or negative numbers. When a particle is moving along a curve, each of these quantities can have two non-zero components. Hence the motion is said to be two dimensional because two numbers (components) are associated with a vector quantity. We will be studying two types of curvilinear motions : (i)

Projectile Motion :

Motion of a particle under the effect of earth’s gravity

(ii)

Circular Motion :

Motion of a particle along a circle

PROJECTILE MOTION

Section - 1

We have already seen that when a particle is given a vertical initial velocity in earth’s gravitational field, it moves along a vertical line. Imagine the motion of a particle when it is given an initial velocity u directed at an angle  with the horizontal and   90.

1

(a)

Such a particle will move horizontally as well as vertically i.e. along a curve.

(b)

For convenience, we take origin at the point from where the particle is thrown and X-axis, Y-axis along horizontal and vertical respectively.

(c)

The velocity of the particle at any instant is directed along the tangent to the path and can have horizontal and vertical components.

(d)

The only force acting on the particle is its weight (mg) directed downwards. Hence acceleration is g directed vertically downwards.

(e)

As acceleration does not change with time, the projectile motion is a uniformly accelerated motion. At all time instants, ax = 0 and ay = g.

(f)

The motion of a particle as projectile can be imagined as being made up of two parts : horizontal and vertical which are independent of each other. The equations of uniformly accelerated motion which we used in the study of straight line motion can be applied to horizontal and vertical components separately.

Section 1

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Analysis of the motion :

Consider the path of a particle thrown with an initial velocity u from origin O at an angle  with the horizontal. 

Horizontal component of u = u cos 



Vertical component of u = u sin 

As the particle goes up, vertical component decreases and becomes zero at the top. The horizontal velocity remains constant i.e. u cos , because there is no component of acceleration along the horizontal. When the particle comes down, the vertical component becomes ve and increases in magnitude. Let P be the position of particle after time t sec and v be the velocity at P then : vx = Xcomponent of final velocity at P vy = Ycomponent of final velocity at P 

The vector OP represents the displacement.



sx = X-component of displacement sy = Y-component of displacement

The equations v = u + at, s = ut + 1/2 at2, v2 = u2 + 2as will be applied separately to horizontal and vertical motions.

Analysis of Horizontal Motion : ax = 0 m/s2 vx = ux + 0(t) 

vx = u cos  at all time instants, i.e. the horizontal velocity is constant. sx = horizontal component of displacement in a time interval of t sec. sx = ux t + 1/2 ax t2 = u cos (t) (as ax is equal to zero)



Horizontal displacement = (horizontal velocity)  (time)

Analysis of vertical motion : ay = g vy = uy  gt 2

Section 1


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sy = uy t  1/2 gt2 vy2 = uy2  2gsy uy = the vertical component of initial velocity sy = vertical component of displacement during t sec. vy = vertical component of final velocity

Vertical component of velocity at any time : (i)

is zero, if the particle is moving horizontally (at the highest point)

(ii)

is +ve if it is going up.

(iii)

is –ve if it is coming down.

Projectile Thrown From Ground Level :

Consider a particle thrown with an initial velocity u at an angle  above the horizontal from the ground. It goes up and reaches the maximum height at A and then comes back to ground at B. 

h = maximum height attained



R = range OB = net horizontal displacement



t = time of flight = total time for which it was in air

Interval from O to B : ux = u cos  uy = u sin 

(i)

ax = 0

ay = g

sx = OB = R

sy = 0 (as the displacement vector is horizontal)

Time of Flight : sy = uy t + 1/2 ay t2 (in the interval from O to B) 

0 = (u sin) t + 1/2 (g) t2



t

2u sin  2u y  g g


Section 1

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Motion in Two Dimensions (ii)

Range : sx = ux t = Range 

R = (ucos) (2usin /g)



R

u 2 sin 2 2u xu y  g g

Interval from O to A : At the topmost point, the tangent to the path is horizontal and hence velocity vector is horizontal. 

vy = 0 at the topmost point.

(i)

Maximum Height above the Ground : vy2 = uy2 + 2 ay sy

(in the interval from O to A)

0 = u2 sin2   2gh  (iv)

2 u 2 sin 2  u y h  2g 2g

Equation of Trajectory : It is the equation of the curve along which the particle moves.

Let the particle move from O to an arbitrary point P on the curve in time t. If the coordinates of P are (x, y) : sx = x sy = y

(because origin is same as the point of projection)



x = (u cos  ) t

and

y = (u sin  ) t  1/2 g t2

Eliminating t from two equations ;

 x  1  x  y  u sin    g   u cos   2  u cos  



4

y  x tan  

Section 1

g x2 2 u 2 cos 2 

2



gx 2 1  tan 2  or

y  x tan  



2u 2


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This equation can also be expressed as : x  y  x tan  1    R gR   2  By substituting u   2 sin  cos   

This is the equation of the curve along which the particle moves. This is called as the equation of the trajectory of the projectile. As y is a quadratic polynomial in terms of x i.e. of the form y = ax  bx2, the curve followed by the projectile is a parabola.

Projectile thrown horizontally from a height : For a horizontal projection of a particle from height h : ux = u ;

uy = 0

Let t be the time taken from A to B. ay = g

;

sy = h

Along vertical : h = 0 (t) + 1/2 (g) t2



t

2h g

Distance of B from O = sx = ux t 

sx  u

2h g

Illustration - 1

A ball is thrown from a field with a speed of 10 m/s at an angle of 45° with the horizontal. At what distance will it hit the field again and the maximum height attained by ball? Take g= 10.0 m/s2

SOLUTION : The horizontal range 

u 2 sin 2θ g

(10) 2 sin(2  45) 10m/s

2



100  10m. 10


Section 1

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Maximum height 



u 2 sin 2 θ (10) 2 sin 2 (45)  2g 2  10

100 1   2.5m. 2  10 2

Thus, the ball hits the field at 10 from the point of projection and the maximum height is 2.5 m. Illustration - 2

A block slides off a horizontal table top 1 m high with a speed of 3 m/s. Find :

(a)

the horizontal distance from the edge of the table at which the block strikes the floor.

(b)

the horizontal and vertical components of its velocity when it reaches the floor. (Take g = 9.8 m/s2)

SOLUTION : (b)

In the interval from O to B

vy = uy + ay t = 0  9.8 (10/7)

ux = 3 m/s, uy = 0 m/s , sy = 1 m As the initial velocity is horizontal, vertical component = 0 m/s. sy = uy t + 1/2 ay t

2



1 = 0 (t) + 1/2 (g) t



t

(a)

sx  u x t  3

vx = ux = 3 m/s



vy = 1.4 10 m/s

Horizontal component = ux = 3 m/s = vx Vertical component = vy =  1.410 m/s

2

2 10  s g 7

10 m  AB 7

= horizontal distance

Illustration - 3

A projectile is given an initial velocity of 5 m/s at an angle 30 below horizontal from the top of a building 25 m high. Find : (i)

the time after which it hits the ground.

(ii)

the distance from the building where it strikes the ground. (Take g = 10 m/s2)

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Section 1


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SOLUTION : The projectile is thrown from O and lands at A on the ground. From O to A : sy = 25 m, uy = 5 sin 30 = 2.5 m/s. (ve because vertical component is downwards)

25 = 2.5 t  1/2 (10) t2 

10 t2 + 5 t  50 = 0

On solving, we get : t = 2s,  2.5s The relevant time = 2s. (ii) The distance of A from the building = sx sx = ux t = (5 cos 30) 2 = 53 m.

(i) ay =  g =  10 m/s2 sy = uy t + 1/2 ay t2 Note : (i)

If the angles of projection are complementary ( + = 90), then (for the same projection velocity) the range on the horizontal plane is same.

(ii)

The range on the horizontal plane is maximum for angle of projection  = 45. (Rmax = u2/g)

Illustration - 4

A packet is released from a plane flying horizontally at an altitude of 910 m with a velocity of 216 kmph. When and where will the packet reach the ground ?

SOLUTION : t

The packet has two motions:

910  12.73 s 5

(i) Vertical motion downward s starting from rest (ii) Horizontal motion with a uniform velocity 60 m/s (i)

Height = 910 m h

(ii)

Horizontal distance covered, s = vt = 60 m s–1 × 12.73 = 763.8 m

The packet reaches the ground at a distance of 763.8 m from the point of reaches after 12.73 s

1 2 1 gt 910 m  10m s2 t 2 2 2


Section 1

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CIRCULAR MOTION

Section - 2

Uniform Circular Motion : A particle moving along a circular path with a constant speed is said to be in uniform circular motion. Examples of uniform circular motion are : (i)

An athlete running on a circular track with constant speed.

(ii)

Motion of tips of the second hand, minute hand and hour hand of a wrist watch.

Radian : It is a convenient unit for measuring angles in physics. The arc AB of the circle, has length  and subtends an angle  at the centre C. If ACB = radians Then

=

arc radius

  = r

when  = r, then  = 1 radian. One radian is defined as the angle subtended at the centre of the circle by an arc which is equal in length to its radius. Angle subtended by the circumference at the centre. 2r = = 2 radians r

2 radians = 360°

360o 1 radian = 2

1radian  57.3o

Angular Velocity () When a particle moves along a circle, it covers some arc length (= s) along the circumference in time t sec. The angle subtended by this arc at the centre or the angle rotated by the radius vector is known as angular displacement in time t sec. If the particle goes from A to B, Angular displacement = AOB = 

8

Section 2


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Distance covered = arc AB = s By definition of  in radians,  

s r

The angular displacement per unit time is known as angular velocity.



angular displacement time



 s / r 1s     t t rt

s Now    v = speed of the particle t





v  r

v  r

Time period of revolution (T) : If the particle completes one revolution,  = 2. Let the time for one revolution = T 



2 T

Let n = frequency of revolution = number of revolutions completed in one second. 

n

1   T 2





2  2 n T

and

v  r 

2 r  T

v  2 rn

Velocity and Acceleration in Uniform Circular Motion : The magnitude of velocity does not change with time but the direction of the velocity (along tangent) keeps on changing from moment to moment. As the velocity vector changes direction with time, the acceleration is non-zero in uniform circular motion. Let v = speed of moving particle (same at all points) and r = radius of the circle. Let the particle start from A at t = 0. After time t, it reaches point P where the position vector is : r  t   OP  r cos t î  r sin t ˆj Self Study Course for IITJEE with Online Support

Section 2

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

v t  

dr   r sin t î  r cos t ˆj dt

Acceleration = a  t  

dv   r 2 cos t î  r 2 sin t ˆj dt



=  2 r cos  t î  r sin t ˆj 



a  ω 2 r

The magnitude of the acceleration is 2 r and it is directed opposite to r and hence towards the centre O. The figure shows the direction of velocity and acceleration for different positions of a moving particle on the circle. As the acceleration is directed towards the centre, it is known as centripetal acceleration or radial acceleration (along the radius). 

2 Centripetal acceleration =  r 

v2 r

Note :So far we have observed that in uniform circular motion, the magnitude of velocity (v) and magnitude of  v2  acceleration   are constant, while the direction of the velocity (along the tangent) and the direction of  r  acceleration (along the radius) keep on changing with time.

Non-Uniform Circular Motion : If the speed of the particle rotating in the circle changes with time, it is said to be in non-uniform circular motion. The acceleration of the particle in that case has two components: 1.

A radial (or centripetal) component which causes the changes in the direction of velocity. It is directed v2 . As the radial component is per-r v2 pendicular to velocity, it is also called normal component of acceleration denoted as an  . r

towards the centre and has a magnitude ar , given as :

ar 

Note that this component is also present in uniform circular motion. 2.

10

A tangential component which causes the change in magnitude of velocity. It is directed along the tangent and its magnitude is decided by the net tangential force acting on the particle. Its magnitude is given by at as :

Section 2


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Motion in Two Dimensions Δv Δt

where v is the speed of the particle.

The tangential acceleration is in the direction of motion if the particle speeds up and opposite to the direction of motion if the particle slows down. Note : (a)

In uniform circular motion, tangential component at = 0 m/s2 because speed does not change.

(b)

The concept of radial (normal) and tangential acceleration can be applied to motion in curves other than circles. 3.

The net acceleration is : a  ar2  at2 and it makes an angle  with tangent : a   tan 1 r at

Illustration - 5

A racer bike negotiates a curve of radius 50 m on a flat road. What is its angular displacement when it covers a linear distance of 78.5 m ?

SOLUTION : Radius = r = 50 m



78.5  1.54 radian 50



360  1.54  90 2π

Linear distance = s = 78.5 m s Angular displacement  θ  r Illustration - 6

A particle travels in a circle of radius 20 cm at a speed that uniformly increases. If the speed changes from 10.0 m/s to 12.0 m/s in 2.0s, find the tangential acceleration.

SOLUTION : at 

Δv v2  v1 12  10    1 m/s2 Δt t2  t1 2


Section 2

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Motion in Two Dimensions SUBJECTIVE SOLVED EXAMPLES

Example - 1 A stone is thrown with a velocity of 19.6 m/s at an angle of 30 above horizontal from the top of a building 14.7 m high. Find : (i)

the time after which the stone strikes the ground.

(ii)

the distance of the landing point of the stone from the building.

(iii)

the velocity with which the stone hits the ground.

(iv)

the maximum height attained by the stone above the ground.

(Take g = 9.8 m/s2)

SOLUTION :

vy = 19.6  sin30  9.8  3

Consider the interval from O to C

vy = 19.6 m/s

ux = 19.6 cos30 = 9.8 3 m / s ax = 0 m/s2 uy = 19.6  sin30 = 9.8 m/s ; ay =  9.8 m/s2 Along vertical direction :

sy = 14.7 m

(i) sy = uy t + 1/2 ay t2

vy is ve because the stone is moving down when it hits the ground. Resultant velocity = vx2  v 2y

14.7 = 9.8 t + 1/2 (9.8) t2 =

(ii) From O to C, the horizontal displacement = sx

9.8 3 

2

2

 19.6   9.8 7 m / s



4.9 t2  9.8 t 14.7 = 0

Distance of C from the building = AC



t 2  2t  3  0

= 29.4 3 m



t = 1, 3 s



Stone lands at C after 3 seconds.

sx = uxt = (19.6 cos30)  3 = 29.4 3 m

(iii) The horizontal velocity remains constant. Hence at C,

vx = ux = 9.83 m/s.

Velocity is directed at an angle  given by :

vy = uy + ay t

12

Subjective Solved Examples


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 vy   tan 1   vx

 1  19.6    tan    9. 8 3  

(iv) Maximum height attained above ground = height of B above point of projection + height of building

 2  below horizontal.  tan 1    3

= h + 14.7 = (u2/2g) sin2 30 + 14.7 = 19.6 m

The trajectory of a projectile in a vertical plane is y  ax  bx 2 , where a and b are constants and x and y are respectively horizontal and vertical distance of the projectile from the point of projection. The maximum height attained by the particle and the angle of projection from the horizontal are : Example - 2

(A)

b2 , tan 1  b  2a

(B)

a2 , tan1  2a  b

(C)

a2 , tan 1  2a  b

SOLUTION :

(D)



g 1  tan 2 

y  ax  bx 2

(ii) b 

2a 2 , tan 1  a  b



2u 2

Comparing with

y   tan   x 



2



g 1  tan  x

2

2u 2

Hence hmax 

we get : (i) a  tan 

   tan

1



sin  

a

and

1  a2

4b



u 



2b

u 2 sin 2  2g

1  a2   

a



g 1  a2

2

a2 1  a2



a2 4b

C is correct.

Example - 3

A ball is thrown with a velocity of 72 m/s at an angle of 45 with the horizontal. It just clears two vertical poles of height 90 cm each. Find the separation between the poles. (Take g = 9.8 m/s2)



13

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SOLUTION : Alternative Method : yA = yB = 0.9 m Using equation of trajectory for y = 0.9 m, we should get the values of xA and xB. Let us first calculate the time t after which the ball is at the top of the poles. During this time interval : sy = 0.9 m



2 7 2

uy = u sin 45 = 7 m/s. sy = uy t + 1/2 ay t2



0.9 = 7 t  1/2 (9.8) t2

2

cos 2 45

has roots xA and xB. On simplification, the equation reduces to ;



Hence the ball is at A after after



x2  10x + 9 = 0

1 9 t or sec 7 7



g x2

0.9  x tan 45 

1 seconds and at B 7

9 seconds. 7

xA = 1m

and xB = 9 m

PQ = xB  xA 

PQ = 8 m

9 =9m 7 1 OP = sx = (u cos 45) = 1 m 7 OP = 8 m

OQ = sx = (u cos45)



Example - 3

A stone of mass 0.04 kg is whirled by a string in a horizontal circle of radius 0.5 m at 60 rpm. Find the linear speed of the stone and the centripetal acceleration.

SOLUTION : Frequency, f = 60 rpm  

60 rps = 1 rps 60

Angular velocity ω  2πf  2  3.14  1  6.28 rad s 1

14


Linear speed, v  rω  0.5  6.28  3.14 m s–1 centripetal acceleration 

v 2 (3.14)2  r 0.5

 19.72 m s–


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THINGS TO REMEMBER 2u sin  2u y  g g

(i)

Time of Flight :

t

(ii)

Range :

R

(iii)

Maximum Height above the Ground :

(iv)

Equation of trajectory :

(v)

On an inclined plane :

u 2 sin 2 2u xu y  g g

h

2 u 2 sin 2  u y  2g 2g

y  x tan  

Range =

g x2 2 u 2 cos 2 

2u 2 sin     cos  g cos 2 

or

x  y  x tan  1    R

; Time of flight =

2u sin     g cos 

Non-Uniform Circular Motion 1.

Radial (or centripetal) component : as an 

2.

ar 

v2 is also called normal component of acceleration denoted r

v2 . r

Tangential component : A t 

dv dt

where v is the speed of the particle.

The tangential acceleration is in the direction of motion if the particle speeds up and opposite to the direction of motion if the particle slows down.


Things to Remember

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My Chapter Notes


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Illustration - 1


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Motion in 2 dimensions Vidyamandir Classes

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