UNIT 4
Kinematics of Particle
We know velocity of a body is
V = (ds/dt) OR V = ( dx / dt)
&
Acceleration
A =dv / dt a = dx /dt
= d/dt ( ds/dt) | = d / dt (dx / dt)
A = d2s/ dt2 a = d2x/ dt2
= V. Dv/ds | = V.dv/dx
A body or particle sometimes moves along the straight line with variable acceleration .
This variable acceleration may be the function of time or position or velocity .
(1) When acceleration is function of time (t) [ a= f(t) ]
a= dv /dt) = f (t)
∴dv= f(t) dt
Integrating both sides, we got
= (t) dt
This will give us equation for velocity as a function of time
V = f (t)
Velocity,
V = ds /dt = f (t)
∴ds = f (t) dt .
Integrating above equation we get S ( displacement ) in terms of ‘t’.
While solving the problems on variable acceleration , following cases will arise :-
(1)Given equation of motion is in terms of displacement (s) & time (t)
S = f (t) or x = f (t)
Differentiating both sides will give (w.r.t. Time t)
V = ds / dt = f(t) or V = dx / dt = f (t)
Again differentiating above equation, w.r.t. (t)
A= dv /dt = d2s / dt2 = v.dv/ds = f (t) OR
∴ a= dv/dt = d2x/dt2 =v. Dv/dx = f (t)
(2) given equation is in terms of acceleration (a) & time (t) .
A = f (t)
Integrating once will give us velocity &Integrating again will give us the displacement .
(3) Given equation is in terms of acceleration (a) & displacement ( x or s)
A = f(s) or a = f(x)
Integrating once will give us velocity equation Integrating twice will give the equation of motion for displacement .
(4) Given equation is in terms of velocity (v) &time (t)
V = f(t)
Integrate above equation to get displacement. Differentiate above equation to get acceleration .
( in all above cases , use given condition to find the constants of integration .)
Examples based on variable acceleration -
(Equation in terms of s& t or x & t – given)
1 ) the position of a particle which moves along a straight line is defined by the relation S= t3 - 6t2 - 15 + 40 , where S is in meters and t in sec ;
Determine :- a) time at which velocity will be zero .
b) position & distance travelled by particle at that time .
c) Acceleration at that time.
d) Distance travelled by particle from t= 4 sec to t= 6sec
→ s = t3 - 6t2 -15t +40
Differentiating w.r.t. Time‘t’
Ds/dt = 3t2 -12t - 15
∴ v = ds/dt =3t2 -12 t -15
Again differentiating w.r.t. Time‘t’
a= dv/dt = d2s/ dt2= 6t -12
a) time at which velocity will be zero
For, V =0
∴ v = 3t2 – 12t -15
∴ 0 =3t2- 12t -15
Solving the equation
[t = 5 sec]
B) t= 5 sec. At this time Displacement will be,
∴Ss= t2-6t2 – 15t +40
∴Ss= (5)2 - (6 x 52) -(15 x 5 )+ 40
∴ Ss= -60 m.
At t = 0 sec; displacement will be ,
S0= t2- 6t2 – 15t + 40
∴S0 = 40 m.
2 ) the motion of particle is defined by x = t3 - 6t2 - 36 t - 40 in meter.
Determine (1) when the velocity is zero. (2) velocity , acceleration & total distance travelled when x = 0 .
→given,
X=t3 - 6t2 -36t- 40
Differentiating w.r.t. t we get
∴ v =dx /dt =3t2 – 12t – 36
Again differentiating w.r.t. ‘t’
∴a= dv/dt =d2x/ d2t= 6t – 12
(1) When the velocity is zero.
For, v= 0
V= 3t2-12t – 36
0 = 3t2 - 12t - 36
Solving above equation, we get
[t = 6 sec.]
( 2) velocity , acceleration & total distance travelled at x=0 .
For x = 0,
x =t3 - 6t2 – 36t – 40
O =t3- 6t2 – 36t -40
Solving above equation, we get
T = 10sec.
∴velocity (for t =0)
V10 = 3t2 – 12 t – 36
=(3x102)- (12x10)-36
V10 = 144m/sec
∴acceleration (for t = 10)
A = 6t – 12
= (6x10)-12
A = 48m/sec2
Distance travelled =|x10 – x6| + |x6 –x0|
∴x10 = 103 – ( 6 x 102)- ( 36 x 10 ) – 40
= 1000 – 600 – 360- 40
X10 = 0m
∴x6 = 63 – (6x62) –(36x6) – 40
X6 = -256m
∴x0 = 03 – 6 x 02 -36x 0 -40
X0 =-40 m
∴distance travelled = |0 – (-256)| +|-256 –(-40)|
=256 + 216
= 472m
Distance travelled = |ss- s0|
= |-60-40|
=100m
(c) Acceleration at t = 5 sec
A = 6t-12
= (6 x 5) -12
A = 18 m/s2
(d) Distance travelled from 4 to 6 sec
As at t= 5sec, v = 0
Thus,
Distance travelled = distance travelled from 4 to 5 sec + distance travelled from 5 to 6 sec.
= |s5 – s4|+ |s6 – s5|
∴at t = 6,
s6 = 63 - (6 x 62) – (15 x 6) + 40
=-50m
At t =4,
s4 = 43 - (6 x 42) – (15 x 4 ) + 40
= -52m
Distance travelled = | -60 – (-52)|+|-50 – (-60)|
= 8+10
= 18m
When a particle moves along a curved path, then motion of the particle is said to be curvilinear.
Basic terminology used to describe curvilinear motion: -
- Position vector: - ()
- Consider that particle is moving along the curve as shown in figure.
b. Let ‘P’ is the position of particle at any time instant ‘t’.
c. Let we have fixed reference axes x,y,z as shown.
d. The line ‘OP’ represents the position of particle &it is known as position vector of particle at time ‘t’.
Position vector
e. = xi + yj + zk and
- Magnitude of position
2. Displacement and distance:
- Consider that particle is moving along the plane curve P-P’ as shown in figure.
- Let particle is located at point P at time instant‘t’.
Position of particle at point P is given by vector. Now after time (t + ∆t), let particle is moved to a new position P’. This position of particle is given by the vector ()
- The vector joining P & P’ is ∆. (dashed line)
∆= change of position of particle during the time interval ∆t.
∆ = displacement of particle.
- The distance travelled by the particle along the curve from point P to P’ is ∆s. This is measured along the curved path & is scalar quantity.
3. Velocity :
In above figure ∆ = displacement vector
∆r = displacement of particle (magnitude).
∆t = time taken by particle to move from P to P’.
Avg. Velocity = = vavg.
- When the time interval approaches to zero, ∆t → 0.
Instantaneous velocity at P will be,
V =
V =
Speed =
4. Acceleration: -
Avg. Acceleration a =
For very small interval of time ∆t → 0
Thus a = &
The acceleration at point P =
(Instantaneous acceleration)
- We know that, velocity of particle is always tangential to the path along which it is moving.
(consider a particle moving along curve & is located at point P at any instant ‘t’)
So the tangent drawn at that point (at point P) represents the tangential direction. &The direction which is perpendicular to the velocity vector (or tangent at P) & passing through center of curvature is known as normal direction.
- Let us consider the particle moving along a curve contained in the plane as shown in figure above.
Let P is the position of the particle at a given instant. Now, let us attach unit vector at point P. is along tangential direction is along normal direction.
= unit vector along tangential direction
=unit vector along normal direction.
- Unit vectors can be written as
& =
Differentiating above equation w.r.t. Ø
We get,
--------- (2)
- Velocity components
As the velocity is always tangential to the path along which partial moves, Then,
Tangential component of velocity is equal to velocity itself and normal component will be zero.
vT= v
vN= 0
The velocity vector of particle is tangent to path.
It will be
- Acceleration of particle
-----(3)
But by double chain rule
--(4)
- Now let us find the value of
Consider particle moving along curve is at Pat time t. Then after small time interval ∆t, particle attains new position at P’. During this it covers ‘ds’ distance along the curve.
Let C = center of radius of curvature.
= radius of curvature of curved path
- Then arc length = radius × angle subtended by arc
Ds = × dØ
Putting above value in equation (4), we get,
Putting this value in equation (3)
--------acceleration vector
Where,
& -----tangential component of acceleration
----------normal component of acceleration
Magnitude of acceleration (total)
Direction of total acceleration is,
Tanα = where α = angle made by acceleration with normal direction
- aN is always directed toward center of curvature.
- aT reflects the change in speed of particle.
- aNreflects the change in direction of motion.
- When become infinity at inflection point then aN = 0
(MOTION CURVES):-
It is a graphical representation of displacement velocity acceleration with time.
(1) Displacement - time curve : (x –t diagram)
This diagram represents position of particle w.r.t to time here time is taken on x axis & displacement (s or x) is taken on y axis
Slope x t diagram at any point represents the velocity at the instant
∴at any instant t (time)
Velocity is given by v =ds/dt = dx/dt
(2)Velocity timecurve: [v - t diagram ]
This diagram represents velocity of particle w.r.t. time here velocity is taken on y axis & time on x axis
(a) The slop of v – t diagram represents acceleration at that instant
∴a= dv/dt
(b)area under the curve (v –t curve)represent change in displacement (x or s)between two instant of time
∴let us select elementary trip betweent1& t2.
∴The area of strip da = v x dt
Area bounded between t1 &t2can be find out by integrating area of elementary strip.
∴da= v.dt
∴∫da =∫v.dt
∴ A =
But v=dx/dt b
∴v.dt = dx
∴ A = = [
∴A = area between (t1 &t2 ) = x2 - x1
(3) Acceleration - time diagram (a –t curves):
This representation acceleration of particle w.r.t. time‘t’ acceleration is plotted on y axis & time is plotted on x axis.
The slope of the curve representation jerks J = da/dt
The area under a- t diagram between two instant of time (t1 & t2) represent change in velocity
Let us consider an elementary strip
∴ Area of strip da= na.dt
∴ Total area a =
But a= dv/dt∴ a. Dt = dv
∴ a =
∴ Area a= v2 - v1
The position coordinate (x) is directly found out from the following moment equation (from a – t diagram)
X1 =x0 + v0 t + mt
Where,
Xt= position of particle at time‘t’
X0 =initial position of particle
V0 = initial velocity of particle
Mt=moment of area under a - t diagram about the instant t.
(4) Velocity - displacement diagram (v -x)
Here plot or graph of velocity (on y axis ) & displacement on (x axis) is drawn.
If a normal is drawn to the tangent on curve the subnormal on x axis represents the acceleration
Let pr is the normal drawn at p
The subnormal =pq.tanθ
=v.tanθ
= v.dv/dx
= a
∴ a = V
∴ Acceleration = velocity x slope of v- x diagram
Numerical on motion diagram
The acceleration versus time for a particle moving along x axis is given in the figure given below. The time interval is 0 to 40 sec for some time interval plot
(1) V-t diagram (2) x – t diagram (3) also find max speed attained & max distance covered
Diagram:
We know that,
Change in velocity= area under a-t diagram from the given a – t diagram
∴at T =20sec
V20 - v0 = 1/2 x 20x12
V20- v0 =120 m/sec
But at T=0 v0= 0
∴ 20= 120m/sec
Now,
At t =40 sec
∴ v40- v20= ½x 20 x12 = 120
∴ v40 = 120+v20
∴ v40 = 240m/sec
Thus v -t diagram will be as follows
Now from above v -t diagram
Change in displacement = area under v-t diagram
At t=20sec
X20 -X 0 =1/3 x 20x 120 =800m
As (X0 = 0 at t = 0)
∴X20 = 800m
At t =40sec
X40 - X20 = (20x 120) + (2/3 x20 x120)
X40 = 2400+1600+ X20
X40 = 24001600+800
X 40= 4800m
∴ x -t diagram will be as follows
Max speed attained = 240m/sec
Max distance travelled = 4800m
When a particle is freely thrown in the air along any direction other than vertical it follows it follows the parabolic path .The motion of a particle along this parabolic path is called as projectile motion.
i.e. when we project the particle in the space, its motion is a combination of horizontal & vertical motion. This motion is called as projectile Motion.
Wind Resistance, curvature & rotation of the earth affects the actual path.
But these parameters are neglected.
- The path Traced by projectile is called as Trajectory.”
- The motion of projectile in Horizontal direction is uniform motion.
- Ax = Horizontal component of acceleration = 0
- The acceleration in vertical direction is affected by gravity. Thus motion in y direction is considered as “Motion under gravity.”
- :. y = -
Basic Terms involved in the projectile Motion
1) Time of flight :- (t)
- The time by the projectile to move from point of projection to the point of target is called as “Time of flight.”
- It is the total time during which projectile remains in space.
2) “Horizontal Range” :- (R)
It is Horizontal distance from point of projection the point of target. OR It is Horizontal distance bet/n point of projection & point of landing.
3) Maximum Height :- (H) or (Hmax)
It is the vertical distance bet/n the point of projection and the point © where the vertical component of velocity is zero.
4) Angle of projection :( )
- It is the angle made by velocity with the Horizontal.
- If velocity is directed up the horizontal, then it is called as angle of elevation.
-If the velocity is directed down the Horizontal, then it is called as angle of depression.
5) Trajectory:-
It is the path traced by a projectile during its motion. It is parabolic in nature.
Projectile on Horizontal plane
Consider a projectile projected from point A with
u= initial velocity of projection &
Angle of projection.
Let t = total time of flight.
Thus projectile will land at point b after time‘t’ Both point A& B are Qn H.P
Diagram:
As the air resistance is neglected, the motion in X-direction is uniform motion & y dirn motion is Motion under Gravity”.
a) Time of flight (t)
t 2 u sin/
b) Horizontal Range (R)
R = u2. Sin2/
c) Maximum Range (R max)
d) For maximum Range angle of projection must be 45
R max = u2/
e) Maximum Height
H = u2. Sin2/2
Derivation of path Equation
[Eqn of Trajectory]
Vx= u cos = constant.
Vy = u sin
Consider a particle projected from A with initial velocity ‘u’ & angle of projection ‘’.
Let after time ‘t’ , the particle has reached at point p (x,y).
Consider the motion of projectile in X dirn (VM) :- [ Ap]
S= velocity * time
X = u cos. t1
:. t1 = X/ u cos . t1
:. t1 = X/ u cos. --------- (1)
Consider the motion of projectile in y dir/n (m. U.G) [ A p]
:. Sy= uyt1- ½ t2
Y = u sin .t1 – 1/2 t12
From eq/n (1), put the value of time t1
:. y = usin .(x/ucos.) – ½ (x/ u-cos.)2
:. y = X. Tan - gx2/u2 cos2
:. y X. Tan - gX2/2u2 cos2
Eqn of Trajectory
Projectile on Inclined plane
Let projectile is projected from point A.
Let angle projection with (inclined) plane.
= Angle of inclined plane with Horizontal.
Now let us select X axis along the inclined plane and y –axis perpendicular to the inclined plane.
:. X component of velocity = u cos
:. Y - ----- -------------------- u sin
Similarly for gravitation Acc/n ‘g’
X component = g sin
y component = g cos
a) Time of flight (t)
t= 2u. Sin /g cos
b) Range along the plane (R)
R= 2 u2 sin/ g cos2 . Cos (+)
c) Maximum Range (Rmax)
Rmax = u2/g (1+sin)
d)Max. Height (lar to plane)
H = u2 sin2/ 2g. Cos
*Special cases of projectile*
*projectile projected with Horizontal velocity:-*
X motion
Consider motion A B
(V.M)
X = u*t
Consider Motion (y- motion)
From AB (M.V.G)
S = ut + ½ gt2
h= 0 + ½gt2
:. t= 2h/g
:. Horizontal distance, = X = u
Y = x tanx - gx2/u2 cos 2 -eqn of trnjectory
But = 0 At point A.
:. –h = -y = 0 - gx2/2u2
:. h = gx2 / 2u2
*for given values of u, two angle gives us the same Range.
1 =
2 = π/2 -
Numerical on projectile Motion. (Projectile on Horizontal plane)
Question) A projectile is fired with a velocity of 60 m/s on Horizontal plane. Find its time of flight in the following 3 cases.
a) is Range is 4 times the max . Height
b) Its max height is 4 times Horizontal range.
c) Its max. Height & Horizontal range are equal.
Answer:
u = 60m/s
a) When R = 4 H
u2sin 2 /g=4 [ 4 2.sin2/2g]
:. u2/g 2sincos = 242/g sin2
:. Cos = sin
:. Cos - sin =o:. = 45
Time of flight t= 2usin /g = 2* 60* sin 45/9081 = 8.65 sec.
b) When H = 4R
:. u2sin2/2g = 4[ 42 sin 2/g]
:. Sin2 = 8 sin 2
:. Sin =(2*8) cos
:. Sin = 16 cos
:. Tan= 16 & = 86.42
t= 2usin/g = 2*60*sin86.42/9.81 = 12.21 sec
c) When H = R
u2sin2/2g = u2 sin2g :. Sin2 /2 = sin2
:.Sin2 = 2*2 sin cos
:. Sin = 4 cos
:.tan =4 :.= 75.96
t= 2usin /g
= 2*60*sin75.96/9.81
t = 11.87 sec.
Question)A projectile is aimed at an object Qn a H.p through the point of projection and tall 8 M 8 short when the angle of projection is 15, while it overshoots the the object by 18 m when the angle of projection is 45 Determine the angle of projection to Hit the object exactly.
Answer:
let R = actual Range Required to hit the object.
ax I - =15
Range = R -8
:. u2* sin(2*15)/g = (R-8)
:. Multiply both sides by 2
:. 2/g = 2R-16----------(1)
Cos (2) =45
Range = R +18
u2sin2/g = R+18
u2/g. Sin 90 = R +18
:. 42/g = R =18------------ (2)
From (1) & (2) R +18 = 2 R- 16
:. 2R – R = 18+16 = 34.
:. 2R = 34m ---- Actual Range to hit the object
Actual Range
R (42/g) sin 2
34 = (R +18) sin2
34 = (34+18) sin2
34 = 52 sin 2
:. Sin 2= 0.653
= 20.38 - Angle of projection to hit the object
Question) A shot is fired from the gun .After 2 sec. The velocity of shot is inclined at 30 up the horizontal After 1 more second. It attains max height. Determine the initial velocity and angle of projection.
Answer:
Let, u = initial velocity = angle of projection. Let, after. 2 second, the shot fired from gun reaches at point D Here Vo makes 30 angle with Horizontal.
&Let after one more second , shot attains max , Height at point C as shown in figure.
At point D, V0 makes 30 with Horizontal.
:. X component of velocity at ‘D’ = VD cos 30
But we know that velocity in X dirn is constant (U.m.)
:. VD cos 30 = u cos
:. 0.87 VD cos 30 = ucos----(1)
Consider y-Motion from A to D.
By substituting the value of VD ineqn (1) & (2)
This is Motion under gravity 0.87 VD = u cos
:. V = u +at : U cos = 0.87*19.62
:. VD sin 30 = usin- gtAD : ucos = 0.87* 19.62
:. 0.5 VD = usin- 9.81*2 : ucos=17.069 m/s – (3)
:.0.5 VD = usin- 19.62 – (2)
Now
Consider y motion from D-c ,(M.V.G) Also, usin= 0.5VD + 19.62
V = u +at = 0.5*19.62+ 19.62
0 = VD sin30 – g*tDC =u sin= 29.43 m/s & -------- (4)
0= 0.5 VD – 9.81 *1 from eqn (3) & (4)
:. 0.5 VD = 9.81 usin/ucos = 29.43/17.069
:. VD = 19.62 m/s : tan = 1.724
:. = 59.886& u = 29.43/sin54.886 =34.02 m/s
Question) A projectile is fired from the edge of 150 m cliff an initial velocity of 180 m/s at 30 angle with Horizontal. Find 1) The Horizontal distance from the gun to the point where the projectile strikes the ground 2) The greatest elevation above the ground reached by projectile 3) striking velocity. Refer the given figure.
Answer:
Let
X= Horizontal distance between A& B
A= point of projection
B= point of striking.
We can see from the fig that A& B are not on same level. TAB = time Req = tAB
Consider the Horizontal motion from A to B (U.M)
:. Distance = velocity * time
X = 180 cos 30 * tAB
X= 155.88 tAB ------ (1)
Consider vertical motion from A to C, H+ 150+ ½*9.81* t2CB
: V = u + at 412.84+ 150 = 4.905 +tCB
Vcy = 180sin30- g*tAC : t2CB = 562.84/4.905
:0 = 90 – 9.81 tAC :. t2CB = 114.748
:tAC = 90/9.81 :. tCB = 10.71 sec
:tAC = 9.17 sec. tAB = 9.17 + 10.71 = 19.88 sec
Consider vertical motion from A to E from eqn (1)
H = u2sinsin2 /2g = 1802*sin2 30/2*9.81 X = 155.88 tAB = 155.88*19.88
H = 412.84 m. X = 3098.9 m
:. Now using. Eqn of motion
S = ut + ½ gt2
Horizontal distance from the to the point of striking is
X = 3098.9 m
Time req. From A to B = tAB = 19.88 sec.
Greatest Height Reached by projectile above the ground is
Hmax H + 150 = 412.84+150
Hmax = 562.84 m
Now,
Consider that VB = striking velocity. &
Ø = angle made by striking velocity with Horizontal. As shown.
Let VBX = X component of VB.
VBy = y component of VB.
But we know that, in X dirn, motion is uniform.
Thus VBX = u cos = 180 cos 30 = 155.88 m/sec.
To find VBy consider the motion from C to B.
:. V = u + gt
VBy = Vcy + g * tCB
VBy = 0 + 9.81 * 10.71
VBy = 105.06 m/sec
:. VB = VBX2 +V2BY = 155.882 + 105.062
VB = 187.9 m/sec
Tanø = VBy/ VBX = 105.06/155.88
:. Ø = 33.97