Motion in two Dimension
Projectile Motion
A body moving horizontally as well as vertically under the action of gravity simultaneously is called a projectile. The motion of projectile is call
ed projectile motion. The path followed by a projectile is called its trajectory.
Examples of projectile motion are
1. Kicked or thrown balls
2. Jumping animals
3. A bomb released from a bomber plane
4. A shell of a gun.
Analysis of Projectile Motion
Let us consider a body of mass m, projected an angle θ with the horizontal with a velocity V0. We made the following three assumptions.
1. The value of g remains constant throughout the motion.
2. The effect of air resistance is negligible.
3. The rotation of earth does not affect the motion.
Horizontal Motion
Acceleration : ax = 0
Velocity : Vx = Vox
Displacement : X = Vox t
Vertical Motion
Acceleration : ay = – g
Velocity : Vy = Voy – gt
Displacement : Y = Voy t – 1/2 gt2
Initial Horizontal Velocity
Vox = Vo cos θ …………………. (1)
Initial Vertical Velocity
Voy = Vo sin θ …………………. (2)
Net force W is acting on the body in downward vertical direction, therefore, vertical velocity continuously changes due to the acceleration g produced by the weight W.
There is no net force acting on the projectile in horizontal direction, therefore, its horizontal velocity remains constant throughout the motion.
X – Component of Velocity at Time t (Vx)
Vx = Vox = Vo cos θ ……………….. (3)
Y – Component of Velocity at Time t (Vy)
Data for vertical motion
Vi = Voy = Vo sin θ
a = ay = – g
t = t
Vf = Vy = ?
Using Vf = Vi + at
Vy = Vo sin θ – gt ……………….. (4)
Range of the Projectile (R)
The total distance covered by the projectile in horizontal direction (X-axis) is called is range
Let T be the time of flight of the projectile.
Therefore,
R = Vox x T ………….. {since S = Vt}
T = 2 (time taken by the projectile to reach the highest point)
T = 2 Vo sin θ / g
Vox = Vo cos θ
Therefore,
R = Vo cos θ x 2 Vo sin θ / g
R = Vo2 (2 sin θ cos θ) / g
R = Vo2 sin 2 θ / g ……………… { since 2 sin θ cos θ = sin2 θ}
Thus the range of the projectile depends on
(a) The square of the initial velocity
(b) Sine of twice the projection angle θ.
The Maximum Range
For a given value of Vo, range will be maximum when sin2 θ in R = Vo2 sin2 θ / g has maximum value. Since
0 ≤ sin2 θ ≤ 1
Hence maximum value of sin2 θ is 1.
Sin2 θ = 1
2θ = sin(-1) (1)
2θ = 90º
θ = 45º
Therefore,
R(max) = Vo2 / g ; at θ = 45º
Hence the projectile must be launched at an angle of 45º with the horizontal to attain maximum range.
Projectile Trajectory
The path followed by a projectile is referred as its trajectory.
We known that
S = Vit + 1/2 at2
For vertical motion
S = Y
a = – g
Vi = Voy = Vo sin θ
Therefore,
Y = Vo sinθ t – 1/2 g t2 ………………….. (1)
Also
X = Vox t
X = Vo cosθ t ………… { since Vox = Vo cosθ}
t = X / Vo cos θ
(1) => Y = Vo sinθ (X / Vo cos θ) – 1/2 g (X / Vo cos θ)2
Y = X tan θ – gX2 / 2Vo2 cos2 θ
For a given value of Vo and θ, the quantities tanθ, cosθ, and g are constant, therefore, put
a = tan θ
b = g / Vo2 cos2θ
Therefore
Y = a X – 1/2 b X2
Which shows that trajectory is parabola.
Uniform Circular Motion
If an object moves along a circular path with uniform speed then its motion is said to be uniform circular motion.
Recitilinear Motion
Displacement → R
Velocity → V
Acceleration → a
Circular Motion
Angular Displacement → θ
Angular Velocity → ω
Angular Acceleration → α
Angular Displacement
The angle through which a body moves, while moving along a circular path is called its angular displacement.
The angular displacement is measured in degrees, revolutions and most commonly in radian.
Diagram Coming Soon
s = arc length
r = radius of the circular path
θ = amgular displacement
It is obvious,
s ∞ θ
s = r θ
θ = s / r = arc length / radius
Radian
It is the angle subtended at the centre of a circle by an arc equal in length to its radius.
Therefore,
When s = r
θ = 1 radian = 57.3º
Angular Velocity
When a body is moving along a circular path, then the angle traversed by it in a unit time is called its angular velocity.
Diagram Coming Soon
Suppose a particle P is moving anticlockwise in a circle of radius r, then its angular displacement at P(t1) is θ1 at time t1 and at P(t2) is θ2 at time t2.
Average angular velocity = change in angular displacement / time interval
Change in angular displacement = θ2 – θ1 = Δθ
Time interval = t2 – t1 = Δt
Therefore,
ω = Δθ / Δt
Angular velocity is usually measured in rad/sec.
Angular velocity is a vector quantity. Its direction can be determined by using right hand rule according to which if the axis of rotation is grasped in right hand with fingers curled in the direction of rotation then the thumb indicates the direction of angular velocity.
Angular Acceleration
It is defined as the rate of change of angular velocity with respect to time.
Thus, if ω1 and ω2 be the initial and final angular velocity of a rotating body, then average angular acceleration “αav” is defined as
αav = (ω2 – ω1) / (t2 – t1) = Δω / Δt
The units of angular acceleration are degrees/sec2, and radian/sec2.
Instantaneous angular acceleration at any instant for a rotating body is given by
Angular acceleration is a vector quantity. When ω is increasing, α has same direction as ω. When ω is decreasing, α has direction opposite to ω.
Relation Between Linear Velocity And Angular Velocity
Consider a particle P in an object in X-Y plane rotating along a circular path of radius r about an axis through O, perpendicular to the plane of figure as shown here (z-axis).
If the particle P rotates through an angle Δθ in time Δt,
Then according to the definition of angular displacement.
Δθ = Δs / r
Dividing both sides by Δt,
Δθ / Δt = (Δs / Δt) (1/r)
=> Δs / Δt = r Δθ / Δt
For a very small interval of time
Δt → 0
Alternate Method
We know that for linear motion
S = v t ………….. (1)
And for angular motion
S = r θ …………….. (2)
Comparing (1) & (2), we get
V t = r θ
v = r θ/t
V = r ω ……………………… {since θ/t = ω}
Relation Between Linear Acceleration And Angular Acceleration
Suppose an object rotating about a fixed axis, changes its angular velocity by Δω in Δt. Then the change in tangential velocity, ΔVt, at the end of this interval is
ΔVt = r Δω
Dividing both sides by Δt, we get
ΔVt / Δt = r Δω / Δt
If the time interval is very small i.e., Δt → 0 then
Alternate Method
Linear acceleration of a body is given by
a = (Vr – Vi) / t
But Vr = r ω r and Vi = r ω i
Therefore,
a = (r ω r – r ω i) / t
=> a = r (ωr- ωi) / t
a = r α ……………………………… {since (ωr = ωi) / t = ω}
Time Period
When an object is rotating in a circular path, the time taken by it to complete one revolution or cycle is called its time period, (T).
We know that
ω = Δθ / Δt OR Δt = Δθ / ω
For one complete rotation
Δθ = 2 π
Δt = T
Therefore,
T = 2 π / ω
If ω = 2πf …………………… {since f = frequency of revolution}
Therefore,
T = 2π / 2πf
=> T = 1 / f
Tangential Velocity
When a body is moving along a circle or circular path, the velocity of the body along the tangent of the circle is called its tangential velocity.
Vt = r ω
Tangential velocity is not same for every point on the circular path.
Centripetal Acceleration
A body moving along a circular path changes its direction at every instant. Due to this change, the velocity of the body ‘V’ is changing at every instant. Thus body has an acceleration which is called its centripetal acceleration. It is denoted by a(c) or a1 and always directed towards the centre of the circle. The magnitude of the centripetal acceleration a(c) is given as follows
a(c) = V2 / r, ……………………… r = radius of the circular path
Prove That a(c) = V2 / r
Proof
Consider a body moving along a circular path of radius of r with a constant speed V. Suppose the body moves from a point P to a point Q in a small time Δt. Let the velocity of the body at P is V1 and at Q is V2. Let the angular displacement made in this time be ΔO .
Since V1 and V2 are perpendicular to the radial lines at P and Q, therefore, the angle between V1 and V2 is also Δ0, Triangles OPQ and ABC are similar.
Therefore,
|ΔV| / |V1| = Δs / r
Since the body is moving with constant speed
Therefore,
|V1| = |V2| = V
Therefore,
ΔV / V = Vs / r
ΔV = (V / r) Δs
Dividing both sides by Δt
Therefore,
ΔV / Δt = (V/r) (V/r) (Δs / Δt)
taking limit Δt → 0.
Proof That a(c) = 4π2r / T2
Proof
We know that
a(c) = V2 / r
But V = r ω
Therefore,
a(c) = r2 ω2 / r
a(c) = r ω2 …………………. (1)
But ω = Δθ / Δt
For one complete rotation Δθ = 2π, Δt = T (Time Period)
Therefore,
ω = 2π / T
(1) => a(c) = r (2π / T)2
a(c) = 4 π2 r / T2 ……………… Proved
Tangential Acceleration
The acceleration possessed by a body moving along a circular path due to its changing speed during its motion is called tangential acceleration. Its direction is along the tangent of the circular path. It is denoted by a(t). If the speed is uniform (unchanging) the body do not passes tangential acceleration.
Total Or Resultant Acceleration
The resultant of centripetal acceleration a(c) and tangential acceleration a(t) is called total or resultant acceleration denoted by a.
Centripetal Force
If a body is moving along a circular path with a constant speed, a force must be acting upon it. Direction of the force is along the radius towards the centre. This force is called the centripetal force by F(c).
F(c) = m a(c)
F(c) = m v(2) / r ………………… {since a(c) = v2 / r}
F(c) = mr2 ω2 r ………………….. {since v = r ω}
F(c) = mrω2
Projectile Motion
A body moving horizontally as well as vertically under the action of gravity simultaneously is called a projectile. The motion of projectile is call
ed projectile motion. The path followed by a projectile is called its trajectory.
Examples of projectile motion are
1. Kicked or thrown balls
2. Jumping animals
3. A bomb released from a bomber plane
4. A shell of a gun.
Analysis of Projectile Motion
Let us consider a body of mass m, projected an angle θ with the horizontal with a velocity V0. We made the following three assumptions.
1. The value of g remains constant throughout the motion.
2. The effect of air resistance is negligible.
3. The rotation of earth does not affect the motion.
Horizontal Motion
Acceleration : ax = 0
Velocity : Vx = Vox
Displacement : X = Vox t
Vertical Motion
Acceleration : ay = – g
Velocity : Vy = Voy – gt
Displacement : Y = Voy t – 1/2 gt2
Initial Horizontal Velocity
Vox = Vo cos θ …………………. (1)
Initial Vertical Velocity
Voy = Vo sin θ …………………. (2)
Net force W is acting on the body in downward vertical direction, therefore, vertical velocity continuously changes due to the acceleration g produced by the weight W.
There is no net force acting on the projectile in horizontal direction, therefore, its horizontal velocity remains constant throughout the motion.
X – Component of Velocity at Time t (Vx)
Vx = Vox = Vo cos θ ……………….. (3)
Y – Component of Velocity at Time t (Vy)
Data for vertical motion
Vi = Voy = Vo sin θ
a = ay = – g
t = t
Vf = Vy = ?
Using Vf = Vi + at
Vy = Vo sin θ – gt ……………….. (4)
Range of the Projectile (R)
The total distance covered by the projectile in horizontal direction (X-axis) is called is range
Let T be the time of flight of the projectile.
Therefore,
R = Vox x T ………….. {since S = Vt}
T = 2 (time taken by the projectile to reach the highest point)
T = 2 Vo sin θ / g
Vox = Vo cos θ
Therefore,
R = Vo cos θ x 2 Vo sin θ / g
R = Vo2 (2 sin θ cos θ) / g
R = Vo2 sin 2 θ / g ……………… { since 2 sin θ cos θ = sin2 θ}
Thus the range of the projectile depends on
(a) The square of the initial velocity
(b) Sine of twice the projection angle θ.
The Maximum Range
For a given value of Vo, range will be maximum when sin2 θ in R = Vo2 sin2 θ / g has maximum value. Since
0 ≤ sin2 θ ≤ 1
Hence maximum value of sin2 θ is 1.
Sin2 θ = 1
2θ = sin(-1) (1)
2θ = 90º
θ = 45º
Therefore,
R(max) = Vo2 / g ; at θ = 45º
Hence the projectile must be launched at an angle of 45º with the horizontal to attain maximum range.
Projectile Trajectory
The path followed by a projectile is referred as its trajectory.
We known that
S = Vit + 1/2 at2
For vertical motion
S = Y
a = – g
Vi = Voy = Vo sin θ
Therefore,
Y = Vo sinθ t – 1/2 g t2 ………………….. (1)
Also
X = Vox t
X = Vo cosθ t ………… { since Vox = Vo cosθ}
t = X / Vo cos θ
(1) => Y = Vo sinθ (X / Vo cos θ) – 1/2 g (X / Vo cos θ)2
Y = X tan θ – gX2 / 2Vo2 cos2 θ
For a given value of Vo and θ, the quantities tanθ, cosθ, and g are constant, therefore, put
a = tan θ
b = g / Vo2 cos2θ
Therefore
Y = a X – 1/2 b X2
Which shows that trajectory is parabola.
Uniform Circular Motion
If an object moves along a circular path with uniform speed then its motion is said to be uniform circular motion.
Recitilinear Motion
Displacement → R
Velocity → V
Acceleration → a
Circular Motion
Angular Displacement → θ
Angular Velocity → ω
Angular Acceleration → α
Angular Displacement
The angle through which a body moves, while moving along a circular path is called its angular displacement.
The angular displacement is measured in degrees, revolutions and most commonly in radian.
Diagram Coming Soon
s = arc length
r = radius of the circular path
θ = amgular displacement
It is obvious,
s ∞ θ
s = r θ
θ = s / r = arc length / radius
Radian
It is the angle subtended at the centre of a circle by an arc equal in length to its radius.
Therefore,
When s = r
θ = 1 radian = 57.3º
Angular Velocity
When a body is moving along a circular path, then the angle traversed by it in a unit time is called its angular velocity.
Diagram Coming Soon
Suppose a particle P is moving anticlockwise in a circle of radius r, then its angular displacement at P(t1) is θ1 at time t1 and at P(t2) is θ2 at time t2.
Average angular velocity = change in angular displacement / time interval
Change in angular displacement = θ2 – θ1 = Δθ
Time interval = t2 – t1 = Δt
Therefore,
ω = Δθ / Δt
Angular velocity is usually measured in rad/sec.
Angular velocity is a vector quantity. Its direction can be determined by using right hand rule according to which if the axis of rotation is grasped in right hand with fingers curled in the direction of rotation then the thumb indicates the direction of angular velocity.
Angular Acceleration
It is defined as the rate of change of angular velocity with respect to time.
Thus, if ω1 and ω2 be the initial and final angular velocity of a rotating body, then average angular acceleration “αav” is defined as
αav = (ω2 – ω1) / (t2 – t1) = Δω / Δt
The units of angular acceleration are degrees/sec2, and radian/sec2.
Instantaneous angular acceleration at any instant for a rotating body is given by
Angular acceleration is a vector quantity. When ω is increasing, α has same direction as ω. When ω is decreasing, α has direction opposite to ω.
Relation Between Linear Velocity And Angular Velocity
Consider a particle P in an object in X-Y plane rotating along a circular path of radius r about an axis through O, perpendicular to the plane of figure as shown here (z-axis).
If the particle P rotates through an angle Δθ in time Δt,
Then according to the definition of angular displacement.
Δθ = Δs / r
Dividing both sides by Δt,
Δθ / Δt = (Δs / Δt) (1/r)
=> Δs / Δt = r Δθ / Δt
For a very small interval of time
Δt → 0
Alternate Method
We know that for linear motion
S = v t ………….. (1)
And for angular motion
S = r θ …………….. (2)
Comparing (1) & (2), we get
V t = r θ
v = r θ/t
V = r ω ……………………… {since θ/t = ω}
Relation Between Linear Acceleration And Angular Acceleration
Suppose an object rotating about a fixed axis, changes its angular velocity by Δω in Δt. Then the change in tangential velocity, ΔVt, at the end of this interval is
ΔVt = r Δω
Dividing both sides by Δt, we get
ΔVt / Δt = r Δω / Δt
If the time interval is very small i.e., Δt → 0 then
Alternate Method
Linear acceleration of a body is given by
a = (Vr – Vi) / t
But Vr = r ω r and Vi = r ω i
Therefore,
a = (r ω r – r ω i) / t
=> a = r (ωr- ωi) / t
a = r α ……………………………… {since (ωr = ωi) / t = ω}
Time Period
When an object is rotating in a circular path, the time taken by it to complete one revolution or cycle is called its time period, (T).
We know that
ω = Δθ / Δt OR Δt = Δθ / ω
For one complete rotation
Δθ = 2 π
Δt = T
Therefore,
T = 2 π / ω
If ω = 2πf …………………… {since f = frequency of revolution}
Therefore,
T = 2π / 2πf
=> T = 1 / f
Tangential Velocity
When a body is moving along a circle or circular path, the velocity of the body along the tangent of the circle is called its tangential velocity.
Vt = r ω
Tangential velocity is not same for every point on the circular path.
Centripetal Acceleration
A body moving along a circular path changes its direction at every instant. Due to this change, the velocity of the body ‘V’ is changing at every instant. Thus body has an acceleration which is called its centripetal acceleration. It is denoted by a(c) or a1 and always directed towards the centre of the circle. The magnitude of the centripetal acceleration a(c) is given as follows
a(c) = V2 / r, ……………………… r = radius of the circular path
Prove That a(c) = V2 / r
Proof
Consider a body moving along a circular path of radius of r with a constant speed V. Suppose the body moves from a point P to a point Q in a small time Δt. Let the velocity of the body at P is V1 and at Q is V2. Let the angular displacement made in this time be ΔO .
Since V1 and V2 are perpendicular to the radial lines at P and Q, therefore, the angle between V1 and V2 is also Δ0, Triangles OPQ and ABC are similar.
Therefore,
|ΔV| / |V1| = Δs / r
Since the body is moving with constant speed
Therefore,
|V1| = |V2| = V
Therefore,
ΔV / V = Vs / r
ΔV = (V / r) Δs
Dividing both sides by Δt
Therefore,
ΔV / Δt = (V/r) (V/r) (Δs / Δt)
taking limit Δt → 0.
Proof That a(c) = 4π2r / T2
Proof
We know that
a(c) = V2 / r
But V = r ω
Therefore,
a(c) = r2 ω2 / r
a(c) = r ω2 …………………. (1)
But ω = Δθ / Δt
For one complete rotation Δθ = 2π, Δt = T (Time Period)
Therefore,
ω = 2π / T
(1) => a(c) = r (2π / T)2
a(c) = 4 π2 r / T2 ……………… Proved
Tangential Acceleration
The acceleration possessed by a body moving along a circular path due to its changing speed during its motion is called tangential acceleration. Its direction is along the tangent of the circular path. It is denoted by a(t). If the speed is uniform (unchanging) the body do not passes tangential acceleration.
Total Or Resultant Acceleration
The resultant of centripetal acceleration a(c) and tangential acceleration a(t) is called total or resultant acceleration denoted by a.
Centripetal Force
If a body is moving along a circular path with a constant speed, a force must be acting upon it. Direction of the force is along the radius towards the centre. This force is called the centripetal force by F(c).
F(c) = m a(c)
F(c) = m v(2) / r ………………… {since a(c) = v2 / r}
F(c) = mr2 ω2 r ………………….. {since v = r ω}
F(c) = mrω2
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