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UNIT 5 KINEMATICS
Structure
5.1 Introduction Objectives 5.2 Kinematics of a Particle : Along a Straight Path 5.2.1 Linear Motion 5.2.2 Simple Harmonic Motion 5.3 Kinematics of a Particle : Along a Curved Path 5.3.1 Motion of Rotation 5.3.2 Curvilinear Motion 5.3.3 Projectile 5.4 Kinematics of a Rigid Body 5.4.1 Motion of Translation 5.4.2 Rotation of a Rigid Body about a Fixed Axis 5.4.3 Plane Motion of a Rigid Body
, 5.5 Relative Motion
5.5.1 Velocity in a Moving Frame of Reference 5.6 Summary 5.7 Key Words 5.8 Solutions/Answers to SAQs
This unit will introduce you the various types of particle and rigid body motions. The prerequisites for this is the concept of a particle, a rigid body, and the displacement of a particle. The basic knowledge of algebra co-ordinate geometry, trigonometry, differential and integral calculus is essential. The understanding of various types of motions, velocities and accelerations of particles and rigid bodies will be developed after study of this unit.
Objectives
After studying this unit, you should be able to :
know different types of linear particle motions, velocities and accelerations,
identify the simple harmonic motion,
(^) differentiate circular and curvilinear motion and projectile,
h o w rigid body motions, and various methods to calculate velocity and
acceleration in plane motion, and
- define a relative velocity of a body.
5.2 KINEMATICS OF A PARTICLE : ALONG A STRAIGHT PATH
We can define a particle as a material point without dimensions, but contains a defrnite
quantity of matter. In kinematics we are concerned only with the time displacement relationship of a given motion of a particle or a rigid body.
When the size of a particle is extremely small compared with its range of motion, in cenain cases we can consider it as a particle. Stars and planets, although quite large in diameter, can be considered as very small when compared with their range of motion and may be
considered as particles in a very large space. Similarly the dimensions of a bullet of a rifle
are very small compared with those of its trajectories, and the bullet may be considered as a
particle.
When a particle moves, it describes a curve that is called a path. If the path described by a particle is a straight line the particle is said to have rectilinear motion. The displacement is a vector quantity and therefore, all vector methods are applicable to the displacement, velocity and acceleration relationships.
Dyaukr
5.2.1 Linear Motion
Consider a particle moving along the x axis as shown in Figure 5.1, the displacement of this particle can be defined by x-coordinate measured from fixed reference point 0. When the particle is on the right hand side of 0, we can call this displacement positive and when the particle is on the left hand side of 0 we can call the displacement negative. $ I- (^7) As particle moves, the displacement varies with time. If the displacement is x at any time t,
o the motion of the particle can^ be^ completely defined^ by^ the relation^ x^ =f^ (t), where^ f^ (t)
stands for any function of time.
nrrr S.1 'Ihe^ relationship^ between^ x^ and t will depend upon how the particle moves along^ x^ axis.
tben constant x, represents an initial displacement at time t = 0, while the constant v
represents the rate of increase of displacement. Such motion of a particle is called a uniform rectilinearmotion. Consider a particle at A when time is t as shown in Figure 5.2. After a small interval of time At the particle moves further and occupies position B. Such that AB = hr. Then we can defme,
~ ~ 4 s = Average change of displacement, also called as average velocity during time At. At
A B *
(0 dx as At -t 0 is called the instantaneous velocity which can be expressed as - FiyrrSi2 At dt ?herefore, mathematically the differentiation of x with respect to t gives us the instantaneous velocity.
Example 5. The displacement of a particle is given by the equation x = l o t 3 - 7 t 2 + 6 t + 2.
- Calculate the displacement, velocity, at t = 4 seconds.
2) Also calculate the distance travelled in 4th second. Distances a ~ e measured in
centimeters. Solution
- ~ e h a v e x = 1 0 r ~ - 7 t ~ + 6 t + 2.
.= Displacement, at t = 4 secs, will be
x , = 1 0 ~ 4 ~ - 7 ~ 4 ~ + 6 x 4 + 2
= 640-112+24+ = 554 cm.
Velocity
:. at r = 4 secs, v 4 = 3 0 x 4 * - 1 4 x 4 + 6
= 430 cmfsec.
- Distance travelled in 4th second = x4- x,
we have x4 = 554 cm [calculated already]
a n d x , = 1 0 ~ 3 ~ - 7 ~ 3 * + 6 ~ 3 + 2
= 270 - 63 + 18 + 2
= 227 cm
= 16 m/sec Example 5.
The acceleration time diagram for a particle is as given below. Calculate velocity of
the particle at various time intervals, plot velocity time diagram and calculate the
displacement of the particle at various intervals. Assume initial velocity v o = 0 and
initial displacement so = 5.30 crns.
Solution We have
d v -
d t
= a , or d v = a dt.
;. v 1 - v o = area of a - t diagram between, t = 0 and t = 1 seconds
(^2) 1 = 10 cm/sec 0 1 2 3 4 ~
Rgure 5 5 v2^ =^ v,^ +^ area of^ a^ -^ r^ diagrambetween^ t^ =^0 and^ t^ =^ 2 seconds
= 20 cm/sec
and v3 = 30 cm/sec
And now we plot v - t diagram as shown in Figure 5.6.
We have
time ~~~
area of v - t diagram between t
v d t
= O and t = 1 seconds.
:. s, = so + -^1 x 1 x 10 2 = 5.30 + 5. = 10.30 c m s2 = so + area of v -3 diagram between t = 0 and t = 2 seconds.
= 25.3 cm.
s, = so + area of v - f diagram between f = 0 and f = 3 seconds.
= 50.3 cm. Vector Representation of Motion 3 +
If the motion of a particle is defmed by a position vector r ( t ) = x ( t ) ?t y (t) J + z ( t ) k
++ +
where i, j and k are unit vectors along x, y and z axes respectively.
The direction cosines of this vector are cos a , cos f! , cos y
We have 3 + r ( t ) = x ( t ) ?+ y (t) j + z ( t ) t
Then velocity
and c o s a = -,^ v x^ a s $ = v 1,^ c o s y = 3 v
v v v
Similarly acceleration of the particle will be obtained by
Example 5.
The position vector of a particle at any instant f with respect to the axes x, y and z is
given by
r ( t ) = ( 4? + 5 ) z ( 3? + 2 ) j 3 + 02
Determine nature of its path and its position, velocity and acceleration at I = 3
seconds. Distances are measured in centimeters.
Solution
As coefficient of k m the given equation is zero, there is no displacement along z
direction at any time f. This can be described as a motion of a particle in plane
(XOY). If there is any coefficient for the k vector of the motion, the motion of the
particle will be in space.
The given equation of motion can be written as
Eliminating t 2, from the above two relations we get
.. 3 ~ 1 5 = 4 y - 8- :. 3 x - 4 y - 7 = 0. This is the equation of a straight line. The path is shown in the Figure 5.8 which can be obtained by calculating values of x and y for various values of 1 as given in Table 5.1.
5.2.2 Simple Harmonic Motion ~ h c ~ ~ ~ t i e s Simple Harmonic Motion can be defined as a straight line motion of a particle whose acceleration is proportional to the displacement from a fixed point called origin and is always directed towards the origin. Let us consider a particle moving with constant speed along a circular path. The motion of
the projection of this moving particle upon the diameter of the circle can be considered as
Simple Harmonic Motion. (^) D A X
Consider a circle of radius r. A particle is moving along the circumference of this circle
with constant angular velocity win the anticlockwise direction. A particle starts its motion (^) -
from the initial position A as shown in Figure 5.9.
At any time r it occupies the position B and OB makes an angle 0 with x axis as shown. Let I
OC be the projection of OB along x axis. ~^ w^ 5.9m
Then we have
OC = x = OB cos 8 = r cos or
The motion of C along x axis can be defined as Simple Harmonic Motion.
We have x = r cos wr.
..^4 =^ - r w^ sinwt ... velocity of C dt
and
= - d x ... acceleration of C.
negative sign indicates that acceleration is oriented towards origin. From the figure it is clear that the diametrical projection of such a point oscillates between
AOD and back to A as the particle rotates around the circle. w is called the natural circular
frequency. The circle in this discussion is called auxilliary circle. The radius is called amplitude of the motion. The time required to complete one oscillation is called the period of the motion. I
periodic time T = -^ 2~. 0
Frequency is defined as the number of oscillations per second or f = 1 = *
T 2 % '
Example 5.
The displacement of a particle is defied by x = 5 + 3 cos 31
Find the position of the particle at t = 3 seconds. Calculate amplitude and frequency
of the particle. Solution
at t = 3 x^ =^ 5 + 3 c o s 9 = 2.27 cm. We have
Amplitude will be calculated as the displacement when t = 0 .. x = 5 + 3cos0° = 8. :. amplitude = 8 cm.
5.3 KINEMATICS OF A PARTICLE : ALONG A CURVED PATH
When the motion of a particle is along a path other than straight path, it is called curvilinear motion. The motion along a curved path may be only rotational motion or may be motion of rotation alongwith motion on a straight path. In case of motion along a straight path velocity and acceleration of the particle is oriented along the path itself but in case of curvilinear motion velocity and acceleration will not be along the path.
5.3.1 Motion of Rotation
A- Now we consider a particle moving along a circular path. Let the radius of the circle be r. Initially the particle is at A. At time t the particle occupies the position B. Position vector OB makes an angle 8 with OA, then 0 can be termed as angular displacement. :. 0 = f (1) Wgure 5. (^10) After a small interval of time At, the particle occupies the position C such that
LBOC = A0 as shown in Figure 5.1 1.
Then (^) 8 = average angular velocity. At and a+&
- (^) as At + 0 = - de = instantaneous angular velocity = c>ht Al dt *t. Consider o, angular velocity of the particle at B and o + A o will be angular velocity at C. A Then
- (^) as A ~ + O = - =d o a At dt This is defined as instantaneous angular acceleration and it acts along the tangent to the
Figure5.11 circle at the point. As the particle moves along the circular path, let S be the distance travelled along the path.
Then s = r e.
Linear velocity of the particle at B can be obtained by
Figure 5.
D ~ ~ C S (^) Resolving along tangent and normal at P.
a = [ lim ( v + A v ) cosA@. )" v c o s 0 Al+o At At
. 1 '
but as At + 0, A$ + 0. and lim. sin $ = $ and lim .cos $ = 1. A :. a = lim [ ( v t A ~ ) t - e I. + ( &+O A 1 At
4 + %.$ + U.$ At At 1
Av. A$ is the product of two small quantities, we neglect it. Also we have
A$ =^ AS -^ and As -^ - - v
P Ar
If a, = tangential component of acceleration, and a,, = normal component of
acceleration. Then A A a = a,t + a n n
= -. t^ dv^ A^ + -. n v2^ A dl P
la1 =
I X
Elgun?5.
d-@fq
This resultant acceleration will make an angle P with the normal to the curve at p. Then
shown in Figure 5.15. Here, we must remember that normal acceleration is oriented towards the centre of curvature C.
In case of rectilinear motion p= m and a,, = 0. The acceleration a = a, = -^ dv and will be
dt always along the path of the particle. If equation of the path of the particle is given by v = f ( x ). Then the radius of curvature at a point whose coordinates are x and v is given by
-^ d2v
Example 5. The motion of a particle in xoy plane is defined by Ule equation
r ( f ) = 3 t 1^ f^ + ( 4 t - 3 t 2 ) j f The distances are in metres. Find its radius of curvature and its acceleration when i t crosses the x axis again.
Solution
We have
:. The equation is a second degree curve and if we equate it to zero, we will get two values of x.
The path ctosses the x axis at x = 0, y = 0 and t = 0 second, x = 4 , 'Y (m)
y = 0, t = - seconds as shown in Figure 5.16.
The radius of curvature is obtained as below.
and
Wehave, x = 3 t y = 4 t - 3? :. vx = 3 m 1sec (^) :. vy = 4 - 6 t m 1 sec. :. t = C , v x = 3 mlsec., v y = 4 rnlsec.
:. v = d32 + 42 = 5 m/sec
Differentiating further, we get
Figure 5.
We have.
2 vo sin 0 .. = t ?!
2 v 0 s i n 0 < s i n 2 0
= vo cos 0 x - -
R R.
2 vo sin 0
T = time of projectile =
R
Example 5.
A ball is thrown horizontally from the top of a building 60 m height, hits the
horizontal ground 25 m from the base of the building as shown in Figure 5.18. What is the initial velocity of the ball? Solution
Let the top of the building comer be origin. Then v, = v and v,, = 0
We have
:. t = 3.5 seconds X = V f
Example 5.
A gun fires a projectile with a muzzle velocity of 300 m 1 sec as shown in Figure 5.19. Find its angle of inclination so that it strikes a target situated at a horizontal distance of 4000 m and 200 m above it.
Solution Coordinates of the target will be ( 4000 ,200 ). Substituting these values in the equation of motion. We have I
2 0 0 = 4 0 0 0 t a n 0 - 9.81^ x^ 400d
2 x 3002 cos
:. t a d 0 - 4.587 tan 0 + 1.23 = 0 :. 0 = 15.96O, or 76.91'
Kinematics
SAQ 4
Efpre 5.
Find the velocity an4 acceleration after 0.4 seconds from the extreme
position of a body moving with simple harmonic motion with an amplitude of 0.8 m
and period of complete oscillation of 1.8 seconds. (^) 9
SAQ 5 A projectile is fixed from the edge of a 160 m high cliff with an initial velocity of 200 m/sec at an angle of elevation 35" with horizontal. Neglecting air resistance, find a) the horizontal distance from the gun to the point where the projectile strikes the ground, and b) the greatest elevation above the ground reached by the projectile.
SAQ 6
Determine path, velocity and acceleration of point 'M' in the middle of b e Y (^) connecting rod of the crank gear shown in Figure 5.20. If OA = OB = 2a and angle $
increases in proportion with time $ = of.
A
Figure 5.U)
5.4 KINEMATICS OF A RIGID BODY
The study of motion of a rigid body in space is a very complicated problem. In this study we will only consider 1 ) motion of translation,
- motion of rotation and plane motion.
5.4.1 Motion of Translation
When a rigid body moves in plane in such a manner that a straight line in the body remains parallel to itself. Such motion of the body is called the motion of translation.
Figure 5.
Line AB moves parallel to itself at all the times, such that at different instants it assumes thc positions A , B,, A, B, etc. The vector AB = r,, remains constant during the motion, since the magnitude and direction of this vector does not change. Consider two wheels A and B and a bar PQ hinged at both ends to these wheels. When the wheels rotate about their axes
Dynamics
We must note that angle $ is measured in radians and the units of w will be radiansJsecond also mentioned as sec- '. If the body undergoes N number of revolutions per minute then we have w = - x: radians~second. 60
Further we can define instantaneous angular acceleration a as below
a = lim = - = A Aw^ d o^ d Al-to At dt d? Example 5. .The angular displacement of a rotating body starting from rest is defined by @ ( t ) = 4t3 - 312. Find its angular displacement, angular velocity and angular acceleration at t = 2 sec. Solution
At t =. 2 sec , :. w = 48 - 12 = 36 rad/sec
a = 48 - 6 = 42 radlsec
Velocity and Acceleration of a Point in a Rotating Body The velocity and acceleration of any point P can be expressed in terms of angular velocity and angular acceleration. Please refer Figure 5.24 (a). The position vector of the point P be r ( t ). As the body rotates
P will trace out a circle of radius a in a plane parallel to XOY plane, where a = r sin 0. The
centre C of the circle will be along z axis. The path of P is shown separately in Figure 5.24.
(b) Po is the starting point and distance s = a $ is measured from Po.
Then linear velocity of P can be obtained by
---:--Fp - -/ J Starting point I *a*
This will be directed along the tangent to the circle at P. The magnitude of velocity v is proportional to the distance of P from the centre of its path.
Consider the Figure 5.25. The acceleration of point P can be obtained as shown. We know
that P moves along a circular path of radius a. The acceleration of P will have two
components, viz tangential component and nonnal component.
Therefore, we have
\ A x i s
\
but
We can write coordinates of B as xB = xA + a cos I$ and yB = yA + a sin $. The displacement of a body undergoing plane motion is equivalent to a translation of a pole and a rotation about an axis passing through this pole. The final position of AB defied as A'B' can be considered as below. Fist, let the body occupy the position A'B" such that all the particles have motion of translation i.e. they move parallel to AB. Then we imagine that the body rotates about an axis passing through A' in the anticlockwise
direction through an angle A$ such that point B" occupies the position B'.
Also we can consider that the body has occupied fist position of B' A" and then rotated about an axis through B' and occupied final position of A'. Velocity of Points in a Rigid Body
Figarc 5.27 : Velocity of Points in a Rigid Body
We consider the equation of pole A in the body as
TA ( 1 ) = XA (!) ?+ yA (1) At any instant position vector of any other point B is given by rB = rA + rAB
r, = A B is a vector of constant magnitude because distance AB is constant. But the
direction of this vector can change with respect to time so that -^ &AB exist. Then by dl differentiatingwe get
4 i. B
- is the differentiationof a vector of constant magnitude but changing in direction. If o dl is the angular velocity with which the body rotates about the axis then we can write
This is a vector VAB of magnitude a. rAB, acting at right angles to AB. Then V B = VA + VAB. Here we must note that velocity of B consists of two components.
- The component vA which represents the velocity of translation of pole A.
2) The component^ vAB of magnitude^ a^ rAB is caused by the rotation of point^ B
about an axis passing through A, perpendicular to the plane XOY. This component will be perpendicular to AB. V B = V A + 0 X rAB Example 5.1 1 A wheel of 0.5 m radius rolls over a horizontal surface without slipping as shown in the Figure 5.28(a). The velocity of its centre is 2 mlsec and is constant. Calculate the angular velocity of the wheel and velocities at C, D and E.
Solution
As the wheel rolls on the ground and does not slip or slide on the ground, the velocity of the point of contact of wheel and ground is same but the velocity of ground must be equal to zero. :. we get (^) v, = 0. Now consider the wheel as the rigid body and the body rotating about an axis through the pole A and B is the point on the rim of the wheel. Considering this motion as a plane motion we have
= - AB. ai ,acting at right angles to BA in the opposite direction of vA
= 4 rad 1 sec.
Now, refer Figure 5.28(b)
Similarly,refer Figure 5.28(c)
