6- Curvilinear Motion - Lecture notes 1 PDF

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CURVILINEAR MOTION A particle moving along a curve other than straight line is said to be in a curvilinear motion. Consider the motion of a particle along a plane curve as shown in Fig. 8.0 A’

y

A

x

Fig.8.0 At time t, the particle is at position A defined by the positive vector r measured from a fixed origin. 0. If both the magnitude and direction of vector r are known at time t, then the position of the particle is completely defined. At time t  t the position of the particle is at A which is defined by r   r  r (vector addition)

s is the scalar distance of the particle from A to A’  r is the vector displacement from A to A’. The average velocity of the particle between A and A’ is defined as

v 

r t

This is a vector quantity whose direction is same as r t  0 gives

Taken limit of V as

lim it

V  t  0 V 

r t

 dr  r dt

 The magnitude is called speed and is given by v  v  ds  S

dt

Again average acceleration of the particle between A and A is defined as v / t . The instantaneous acceleration of the particle is the limiting value of the average acceleration as the time interval approaches zero.

a  lim it  t 0

 dv

8.1

v t .

dt

 v

Rectangular coordinates system (x, y)

This system of coordinates is particularly useful describing motion where x and y components if acceleration are independently generated or determined. The resulting curvilinear motion is then, obtained (a vector combination if the x-and y – components of the position vector r the velocity v and the acceleration a ) If r  xi  yj as shown in Fig.8.1. y j A yj

x

i Fig.8.1 To obtain the velocity we differentiate r with respect to time r = xi + yj 

v= r v = xi + yj similarly a = xi + yj = v = r r = (x2 - x1)i + (y2 – y1 )j = xi + yj  r  x  y i j t t t

Note that the vector has no time derivatives since their magnitudes and direction remain constant. It then follows from the equation above that .

v x  x,

and

.

vy  y

.

..

ax  vx  x . .. ay  v y  y

 v  v x  v y and

a  ax  ay

vectorially,  v  v xi  v y j , a  ax i  a y j Their magnitudes are v  v  v x2  v 2y a  a  a x2  a 2y tan θ 

vy vx

If x and y are known independently as a function of time i.e. x  f1 t  and y  f2 t  .

.

..

..

We may combine their derivative x and y to obtain v and similarly combining x and y to obtain a . On the other hand, if the acceleration components a x and a y are given as a function of time, once we have obtained vx and v y and given

x  f1  t and y  f2 t  Elimination of equations gives the equations of curved path i.e. y  f  x

Example 8.1 2 t and , where v x 50 16 is in metre per second, y is in metres and t is in seconds. It is also known that x = 0 when t =

The curvilinear motion of a particle is defined by vx

0.determine it velocity and acceleration when the position y = 0 is reached.

Solution

Vx 



x

x t dx   dx   Vx dt 0 0 dt

dx 

0

t

 50 16t dt 0





 x  50t  8t 2 m 

ax  V x 

d 50  16 t  dt

2

ax = -16m/s

y – component . 

dy y = 100 – 4t2, v y = y = = -8m/s dt .. 

. 

a y  y  V  dv = 8m/s 2 dt

when y = 0 then,

y = 100 – 4t2 = 0  100 = t2 4

t = 25 =  5s = 5s  vx at t = 5s = 50 – 16 x 5 = -30m/s ax at t = 5s = -16m/s 2 V y at t = 5s = -8 x 5 = -40m/s ay at t = 5s = -8m/s2 In vector form v = vi + vj = - 30i + (-40)j = -30i – 40j a = ax i + ayj = -16i – 8j V  V  30 2  40 2  50m / s a  a 

16

2

 8

2

 17 . 9 m / s

Example 8.2 The acceleration of a particle falling through the atmosphere is define by the equation a= g(1K 2V2 ) where g is the acceleration due to gravity, V is the velocity and K is a constant. If the particle starts from rest (i) Show that the velocity at any time is given by V = 1 tanh(Kgt). (ii) K Write an equation defining the velocity of the particle at any position S. (Hint: sinh x =

e x  e  x , cosh = e x  e  x ). 2 2 Solution

Since the acceleration function has been given as

2 2

a = g(1 –K V )

From a = dv , dt

dV

dv = adt = g(1 –K2V2 )dt

2 2

1 K V

dV

V

0

= gdt,

1 K V

1 1 V   0  2 ( 1  KV )  2 ( 1  KV )  

V

  dV

= gt,

1 In ( 1  KV )  1 In ( 1  KV ) 2K 2K

1 2K

 In

= g 0 dt

2 2

( 1  KV )  In ( 1  KV ) V 0

= gt

= gt,

V

In 1  KV  =2gKt, 1  KV  0 = e2gKt ,

1  KV 1  KV

1+ KV = e2gKt - e2gK KV ,

KV+ e2gKt KV = e2gKt - 1, KV(1+ e2gKt) = e 2gKt –1 KV =

e e

2 gKt 2 gKt

V= 1



e

K e

  1 1

gKt

gKt

 egKt  e

 gKt

  x

x x e e but sinh x = e  e , cosh = 2 2

x

and

gKt  gKt  tanhKgt = e e  gKt gKt

e

V=

1 K

 e

tanhKgt

from a vdv , vdv = ads = g(1 –K2V2)ds ds

vdv

= gds 2 2 g (1  K V ) Resolve this into partial fraction 1 V  1 1  dV ,  2 K 0 1  KV 1  KV 

= 0s gds

V 1  In (1  KV )  In (1  KV )  = gs s vdv 0 ds  2 K  K K 0



1 In (1  K 2 V 2 )V 0 2K

= gs

gs

In(1-K2V2) = -2K2 gs 1-K2V 2 = e 2K2gs v

2







1 1  e  2Kgs K 2

v  1 K

1  e  2 Kgs

Example 8.3 t4 3 2   2 3 The position vector of a point moves in the x-y is given by r =  t  t i j 2 12  3 

where r is in metres and t is in seconds. Determine the angle between the velocity and acceleration vectors when t=2seconds.

Solution 4 r =  2 t 3  3 t 2  i  t j

 3

v



2

12

t3 = r = (2t - 3t) + j, 3 2

where t = 2s,

v = (8 –6)i +



a = v = (4t –3)i + t 2 j t2 j, 3

a = (8 – 3)i + 4j = 5i + 4j

v  a = v  a cos  ,

where is the angle between v and a

 2 i  8 j  3  

.(5i + 4j) =  

10  32 3

=



   

100 9



4 

64  9

 41  cos θ 

20.67 = 10 6.4cos = 21.34cos 3 cos = 20 .67 = 0.9686 21 .34  = cos-1 (0.9686) = 14.40

 25  16  cos θ 

8.2

PROJECTILE

An important application of 2-dimensional kinetic theory is the problem of projectile motion. Now let a particle be projected from a point with velocity v at an angle α (upwards) to the horizontal as shown in Fig. 8.3. Assuming the air-resistance is neglected. y Vy 

Vx  x O

A

Fig. 8.3 The horizontal component Vx (t) and the vertical component Vy(t) of its velocity at time t after projection are given by

Vx t   V cosα t V y t   V sin α  gt In writing this we have neglected other resistance (e.g. air resistance) other than gravity. Direction of motion at time t makes angles θ with the horizontal where U

y

U

x



V sin α  gt  tan θ V cos α

At the greatest height reached, the vertical component of velocity is zero. Hence the timet1 , taken to reach this is given by

0  V sin α  gt1

or

t1  V sin α g

and the greatest height s, is given by 0 = (vsin)2 -2gs1

or

V 2 sin 2 α s1  2g

If A is at the same horizontal level with 0 the time t2 taken to reach A is given by

1 s2  0  V sinαt 2  gt22 2 2 sin α

t2  V

g...