MOTION IN A PLANE –PROJECTILE & CIRCULAR MOTION
TOPICS COVERED
Position and Displacement Vectors
Velocity
Rectangular Components of Velocity
Acceleration
Rectangular Components of Acceleration
Displacement equations of Motion in a Plane with constant acceleration
Relative Velocity
Magnitude and direction of Relative Velocity
Projectile Motion
Displacement and Velocity Equations of Projectile Motion
Trajectory of Projectile
Time of Flight of Projectile
Maximum Height of Projectile
Range of Projectile
Uniform Circular Motion
Relation between linear velocity and angular velocity.
Centripetal Acceleration
Position and Displacement Vectors
X
Y
O
r
i
x
j
y
P
The position vector r of a particle P located in a plane with reference to the origin of an x-y reference frame is given by
i
= x
r
j
+ y
O
Y
X
r
P′
r′
P
Δr
Δy
Δx
Direction of vav
Let the particle move through the curve from P at time t to P′ at time t′. Then
Displacement vector is
r
r′
-
Δr =
It is directed from P to P′.
or
Δr
= (x′
i
j
+ y′ ) -
i
(x
j
+ y )
i
= Δx
j
+ Δy
where Δx = x′ - x and Δy = y′ - y
where x and y are components of r along x- and y- axes.
Velocity
The velocity v of a particle is the ratio of the displacement to the corresponding time interval.
v =
Δt
Δr
=
i
Δx
j
+ Δy
Δt
=
i
Δx
Δt
j
+
Δy
Δt
or
The direction of vav is same that of Δr.
The instantaneous velocity is given by the limiting value of the average velocity as the time interval approaches zero.
v =
Δr
Δt
lim
Δt→0
dr
dt
v =
or
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.
i
= vx
v
j
+ vy
The velocity is the first differential coefficient of displacement.
Velocity in terms of rectangular components
v =
Δr
Δt
lim
Δt→0
=
i
Δx
Δt
j
+
Δy
Δt
lim
Δt→0
=
i
Δx
Δt
+
lim
Δt→0
j
Δy
Δt
lim
Δt→0
=
i
dx
dt
j
+
dy
dt
or
dx
dt
vx =
dy
dt
vy =
where
and
i
= vx
v
j
+ vy
The direction of velocity v is
tan θ =
vy
vx
or
θ = tan-1
vy
vx
The magnitude of velocity v is
v =
vx2 + vy2
X
Y
O
i
vx
j
vy
v
P
θ
Acceleration
The average acceleration a of a particle is the ratio of the velocity to the corresponding time interval.
=
i
Δvx
Δt
j
+
Δvy
Δt
a =
Δt
Δv
=
Δt
i
Δ (vx
j
+ vy
)
i
= ax
a
j
+ ay
or
The instantaneous acceleration is given by the limiting value of the acceleration as the time interval approaches zero.
a =
Δv
Δt
lim
Δt→0
dv
dt
a =
or
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, they may have any angle between 0° and 180° between them.
Acceleration in terms of rectangular components
a =
Δv
Δt
lim
Δt→0
=
i
Δvx
Δt
j
+
Δvy
Δt
lim
Δt→0
=
i
Δvx
Δt
+
lim
Δt→0
j
Δvy
Δt
lim
Δt→0
=
i
dvx
dt
j
+
dvy
dt
or
i
= ax
a
j
+ ay
The direction of acceleration a is
tan θ =
ay
ax
or
θ = tan-1
ay
ax
The magnitude of acceleration a is
a =
ax2 + ay2
where
dvx
dt
ax =
d2x
dt2
=
and
dvy
dt
ay =
d2y
dt2
=
DISPLACEMENT EQUATIONS OF MOTION IN A PLANE WITH CONSTANT ACCELERATION
Let the velocity of the object be v0 at time t=0 and v at time t. Then
a =
v – v0
t – 0
=
v – v0
t
v = v0 + at
or
In terms of components,
vx = v0x + ax t
vy = v0y + ay t
Let the position vector of the object be r0 at time t=0 and r at time t.
Then the displacement is the product of average velocity and time interval.
t
r – r0
2
v + v0
=
t
2
=
v0
v0 + at
+
r – r0 = v0t + ½ at2
r = r0 + v0t + ½ at2
Displacement equations in terms of components,
x = x0 + v0xt + ½ axt2
y = y0 + v0yt + ½ ayt2
When the velocity of an object is measured with respect to an object which is at rest or in motion, the velocity measured is known as relative velocity.
Relative Velocity - (By Vector Algebra)
Consider two objects A and B moving with velocities vA and vB respectively.
To find the relative velocity of the object, say A with respect to the object B, the velocity -vB is superimposed on the object B so as to bring it to rest.
To nullify this effect, velocity -vB is superimposed on the object A also.
The resultant of vA and -vB gives the relative velocity vAB of the object A with respect to the object B.
Mathematically, vAB = vA + (-vB)
vA
vB
P
vAB = vA - vB
R
O
-vB
vAB = vA - vB
or
When two objects are moving along the same straight line:
B
O
R
α
θ
180°- θ
vA
vB
-vB
A
vAB = vA - vB
vAB
The magnitude of vector vAB is
The direction of vector vAB is
B sin θ
tan α =
A - B cos θ
vAB =
vA2 + vB2 - 2 vA vB cos θ)
or
Magnitude and Direction of the Relative Velocity in terms of the Magnitudes and Angle θ between them
vAB =
vA2 + vB2 + 2 vA vB cos (180° - θ)
PROJECTILE MOTION
An object that is in flight after it being thrown or projected is called a projectile.
Projectile
Projectile Motion
The motion of a projectile which is in flight after it being thrown or projected is called projectile motion.
It can be understood as the result of two separate, simultaneously occurring components of motion (along x- and y- axes).
The component along the horizontal direction (x- axis) is without acceleration.
The component along the vertical direction (y- axis) is with constant acceleration under the influence of gravity.
In our study, the air resistance is negligible and the acceleration due to gravity is constant over the entire path of the projectile.
Displacement and Velocity Equations of a Projectile Motion
Suppose that the projectile is launched with velocity v0 that makes an angle θ0 with the x-axis.
X
Y
O
v0
θ0
v0 cos θ0
v0 sin θ0
j
a = -g
Acceleration acting on the projectile is due to gravity which is directed vertically downward:
j
a = -g
or
ax = 0, ay= -g
v0x = v0 cos θ0
v0y = v0 sin θ0
If the initial position is taken as the origin O, then
x0 = 0, y0= 0
becomes
x = x0 + v0xt + ½ axt2
x = v0xt = (v0 cos θ0)t
The components of initial velocity v0 are:
and
y = y0 + v0yt + ½ ayt2
becomes
y = (v0 sin θ0)t - ½ gt2
The components of velocity at time t are:
vx = v0x + ax t
becomes
vx = v0x = v0 cos θ0
vy = v0y + ay t
and
becomes
vy = v0 sin θ0 - gt
The magnitude of velocity of the projectile at an instant ‘t’ is given by
vt =
v02 cos2 θ0 + (v0 sin θ0 – gt)2
The direction of velocity of the projectile at that instant ‘t’ is given by
v0 sin θ0 - gt
tan β =
v0 cos θ0
Note:
The horizontal component of velocity remains constant throughout the motion.
But, the vertical component reduces to zero at its peak of the path and again increases in the opposite direction.
j
a = -g
X
Y
O
v0
θ0
v0 cos θ0
v0 sin θ0
v0 cos θ0
v0 cos θ0
v0 sin θ0 - gt
-(v0 sin θ0 – gt)
v0 cos θ0
v
-θ0
v0 cos θ0
-v0 sin θ0
vt
β
vt
Trajectory (Path) of a projectile
The shape of the path of a projectile can be found by mathematical equation.
x = v0xt = (v0 cos θ0)t
From
we get
v0 cos θ0
x
t =
y = (v0 sin θ0)
- ½ g
v0 cos θ0
x
v0 cos θ0
x
2
On simplification,
y = (tan θ0)
-
x
2 (v0 cos θ0)2
g
x2
Since g, θ0 and v0 are constants, the above equation is in the form of
y = ax - bx2
where a = tan θ0 and b =
2 (v0 cos θ0)2
g
The above equation is the equation of a parabola.
Therefore, the path of the projectile is a parabola.
y = (v0 sin θ0)t - ½ gt2
becomes
Time of Flight of a Projectile
Let tm be the time taken for the projectile to reach its maximum height and Tf be the total time of flight of the projectile.
At the point of maximum height and at t = tm , vy = 0.
At t = Tf , y = 0.
vy = v0 sin θ0 - gt
becomes
0 = v0 sin θ0 - gtm
y = (v0 sin θ0)t - ½ gt2
becomes
0= (v0 sin θ0)Tf - ½ gTf2
Note that Tf = 2 tm because of the symmetric nature of the parabolic path.
or
v0 sin θ0
g
tm =
or
2 v0 sin θ0
g
Tf =
Maximum Height of a Projectile
Let hm be the maximum height of the projectile after time tm.
y = (v0 sin θ0)t - ½ gt2
becomes
hm = (v0 sin θ0)
- ½ g
2
v0 sin θ0
g
v0 sin θ0
g
or
v02 sin2 θ0
2g
hm =
Aliter:
At hm (the maximum height of the projectile), vy = 0.
vy2 = v02 sin2 θ0 – 2gy
becomes
02 = v02 sin2 θ0 – 2ghm
or
v02 sin2 θ0
2g
hm =
Range of a Projectile
Let R be the Range of the projectile after time Tf (Time of flight). It is the horizontal distance covered by the projectile from its initial position (0,0) to the position where it passes y = 0.
becomes
x = v0xt = (v0 cos θ0)t
R = (v0 cos θ0) Tf
or
R = (v0 cos θ0)
2 v0 sin θ0
g
or
v02 sin 2θ0
g
R =
Note that the range will be maximum for the maximum value of sin.
i.e. when sin 2θ0 = 1. This is possible when θ0 is 45°.
v02
g
Rm =
Therefore, the maximum horizontal range is
When θ0 is 45°,
v02
4g
hm,45° =
and
Rm = 4 hm,45°
X
Y
O
v0
Rmax
α
α
45°
Range of a Projectile is same for complement angles of projection
v02 sin 2θ0
g
R =
For angles, (45° + α) and (45° - α), 2θ0 is (90° + 2α) and (90° - 2α) respectively.
The values of sin (90° + 2α) and sin (90° - 2α) are the same and are equal to cos 2α. Therefore, ranges are equal for elevations which exceed or fall short of 45° by equal amounts of α.
In other words, for complement angles of elevation, the ranges will be the same.
i.e. for θ0 and (90° - θ0) the values of sin 2θ0 and sin (180° - 2θ0) are the same.
v0
v0
UNIFORM CIRCULAR MOTION
When a body moves with constant speed on a circular path, it is said to have uniform circular motion.
P’
v’
ω
r'
A particle P moves on a circle of radius vector r with uniform angular velocity ω. Δθ is the angular displacement of the particle
Linear velocity v is constant in magnitude but changes its direction continuously.
This acceleration is called centripetal acceleration and is always directed towards the center.
v
P
r
O
Δr
The angular velocity is the rate of change of angular displacement.
Δs
Δθ
Δθ
ω =
Δθ
Δt
lim
Δt→0
The velocity vector v turns through the same angle Δθ and becomes v’.
The linear displacement PP’ is Δr. The linear distance Δs is the arc PP’.
When the particle moves from P to P’ in time Δt = t’ – t, the line OP (radius vector) moves through an angle Δθ. Δθ is called ‘angular displacement’.
In case of non-uniform circular motion, the particle has acceleration due to change in both speed(aT) and direction(aC). The net acceleration is determined by the resultant of both accelerations.
|v| =
Δs
Δt
lim
Δt→0
But Δs = r Δθ
|v| =
r Δθ
Δt
lim
Δt→0
|v| =
Δθ
Δt
lim
Δt→0
r
|v| =
r
|ω|
The linear velocity is the rate of change of linear displacement.
∴
or
ω, r and v are mutually perpendicular to each other and ω is perpendicular to the plane containing r and v.
v = ω x r
Relation between Linear and Angular Velocity
P’
v’
ω
r'
v
P
r
O
Δr
Δs
Δθ
Δθ
Δv
Δθ
Δv
Δθ
Δv
Δθ
Δv
Δθ
Δv
Δθ
v
v’
V
V
V
V
V’
β
β
β
β
Acceleration in Uniform Circular Motion : Centripetal Acceleration
Direction of acceleration of a particle in a uniform circular motion
As Δt→0, Δθ→0° and β→90°. It means the angle between Δv and v, i.e. β increases and approaches 90°. i.e. Δv becomes perpendicular to v.
r is perpendicular to v. And Δv is also perpendicular to v.
∴ Δv is acting along -r. (Note the negative sign)
Since acceleration is the rate of change of velocity, therefore it acts in the direction of Δv. Or it acts in the direction along the radius and towards the centre O. Hence, the acceleration is called ‘centripetal acceleration’.
Magnitude of acceleration of a particle in a uniform circular motion
The two isosceles triangles OPP’ and PAB are similar triangles.
∴
=
AB
PP’
PA
OP
or
|v|
|Δv|
=
|Δr|
|r|
or
|v|
|Δv|
=
|Δr|
|r|
|Δv|
|a|=
Δt
lim
Δt→0
∴
|a|=
lim
Δt→0
Δt
|v|
|Δr|
|r|
x
|a|=
lim
Δt→0
Δt
|Δr|
|v|
|r|
or
or
|a|=
|v|
|r|
|v|
|a|=
|v|2
|r|
or
or
acp =
v2
r
P’
v’
ω
r'
v
P
r
O
Δr
Δs
Δθ
Δθ
Δv
Δθ
v
v’
P
A
B
Centripetal acceleration in terms of angular speed.
acp =
v2
r
But v = rω
∴
acp =
(rω)2
r
or
acp = ω2r
Centripetal acceleration in terms of frequency can be expressed as:
ω = 2πν
∴
acp = ω2r
becomes
acp = (2πν)2 r
or
acp = 4π 2 ν2 r
ω
P
r
acp
O
v
Directions of r, v, ω and acp
The relative directions of various quantities are shown in the figure.
THE END
QUESTIONS ON PROJECTILE MOTION
Q1. A hiker stands on the edge of a cliff 490 m above the ground and throws a stone horizontally with an initial speed of 15m/s. Neglecting air resistance, find the time taken by the stone to reach the ground, and the speed with which it hits the ground(take g = 9.8m/s2)
Soln .
Here y = ½ gt2
490 = ½ x 9.8 t2
t = √ 100 = 10 s ans.
The horizontal and vertical components
of speed v of the stone at point P are
vx = u = 15 m/s
vy = uy+gt = 0+ 9.8 x 10 = 98 m/s
v= √ vx2 + vy2 = √ 152 + 982 = 99.1 m/s ans.
A
O
Ground
x
y =490 m
v
vy
vx
u
P
QUESTIONS ON PROJECTILE MOTION
Q2. A Projectile is fired horizontally with a velocity of 98 m/s from the top of a hill 490 m high. Find (i) time of flight (ii) the distance of the target from the hill and (iii) the velocity with which the projectile hits the ground. (take g = 9.8m/s2)
Soln .
(i) Here y = ½ gt2
490 = ½ x 9.8 t2
t = √ 100 = 10 s ans.
(ii) The horizontal distance of the target from the
hill, AP = x = horizontal velocity x time
= 98 x 10 = 980 m ans.
vx = u = 98 m/s
vy = uy+gt = 0+ 9.8 x 10 = 98 m/s
v= √ vx2 + vy2 = √ 982 + 982 = 138.59 m/s ans.
A
O
Ground
x
y =490 m
v
vy
vx
u
P