MOTION IN TWO DIMENSIONS
Motion in two Dimension
Projectile Motion:::::A body moving horizontally as well as vertically
under the action of gravity simultaneously is called aprojectile.
Themotion of projectile is calledprojectile motion. The path followed
by a projectile is called its trajectory.Examples of projectile motion
are1. Kicked or thrown balls2. Jumping animals3. A bomb released from
a bomber plane4. A shell of a gun.Analysis of Projectile MotionLet 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 MotionAcceleration : ax = 0Velocity : Vx =
VoxDisplacement : X = Vox tVertical MotionAcceleration : ay = -
gVelocity : Vy = Voy - gtDisplacement : Y = Voy t - 1/2 gt2Initial
Horizontal VelocityVox = Vo cos θ ...................... (1)Initial
Vertical VelocityVoy = 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 motionVi = Voy = Vo sin θa
= ay = - gt = tVf = Vy = ?Using Vf = Vi + atVy = 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 rangeLet 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 / gVox = Vo cos
θTherefore,R = Vo cos θ x 2 Vo sin θ / gR = Vo2 (2 sin θ cos θ) / gR =
Vo2 sin 2 θ / g .................. { since 2 sinθ 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.@import
"/extensions/GoogleAdSense/GoogleAdSense.css";The Maximum RangeFor a
given value of Vo, range will be maximum when sin2 θ in R = Vo2 sin2
θ/ g has maximum value. Since0 ≤ sin2 θ ≤ 1Hence maximum value of sin2
θ is 1.Sin2 θ = 12θ = sin(-1) (1)2θ = 90ºθ = 45ºTherefore,R(max) = Vo2
/ g ; at θ = 45ºHence the projectile must be launched at an angle of
45º with the horizontal to attain maximum range.Projectile
TrajectoryThe path followed by a projectile is referred as its
trajectory.We known thatS = Vit + 1/2 at2For vertical motionS = Ya = -
gVi = Voy = Vo sin θTherefore,Y = Vo sinθ t - 1/2 g t2
....................... (1)AlsoX = Vox tX = 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)2Y = X tan θ - gX2 / 2Vo2 cos2 θFor a given
value of Vo and θ, the quantities tanθ, cosθ, and g are constant,
therefore, puta = tan θb = g / Vo2 cos2θThereforeY = a X - 1/2 b
X2Which shows that trajectory isparabola.Uniform Circular MotionIf an
object moves along a circular path with uniform speed then its motion
is said to be uniform circular motion.Recitilinear MotionDisplacement
→ RVelocity → VAcceleration → aCircular MotionAngular Displacement →
θAngular Velocity → ωAngular Acceleration → αAngular DisplacementThe
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 lengthr = radius of the circular pathθ = amgular
displacementIt is obvious,s ∞ θs = r θθ = s / r = arc length /
radiusRadianIt is the angle subtended at the centre of a circle by an
arc equal in length to its radius.Therefore,When s = rθ = 1 radian =
57.3ºAngular VelocityWhen 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
intervalChange in angular displacement = θ2 - θ1 = ΔθTime interval =
t2 - t1 = ΔtTherefore,ω = Δθ / ΔtAngular 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 AccelerationIt 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 asαav = (ω2 - ω1) / (t2 - t1) =
Δω / ΔtThe units of angular acceleration are degrees/sec2, and
radian/sec2.Instantaneous angular acceleration at any instant for a
rotating body is given byAngular 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 VelocityConsider 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 / rDividing both sides
by Δt,Δθ / Δt = (Δs / Δt) (1/r)=> Δs / Δt = r Δθ / ΔtFor a very small
interval of timeΔt → 0Alternate MethodWe know that for linear motionS
= v t .............. (1)And for angular motionS = r θ
................. (2)Comparing (1) & (2), we getV t = r θv = r θ/tV =
r ω ........................... {since θ/t = ω}Relation Between Linear
Acceleration And Angular AccelerationSuppose 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 isΔVt
= r ΔωDividing both sides by Δt, we getΔVt / Δt = r Δω / ΔtIf the time
interval is very small i.e., Δt → 0 thenAlternate MethodLinear
acceleration of a body is given bya = (Vr - Vi) / tBut Vr = r ω r and
Vi = r ω iTherefore,a = (r ω r - r ω i) / t=> a = r (ωr- ωi) / ta = r
α .................................... {since (ωr = ωi) / t = ω}Time
PeriodWhen 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 thatω = Δθ / Δt OR Δt = Δθ / ωFor one complete rotationΔθ
= 2 πΔt = TTherefore,T = 2 π / ωIf ω = 2πf ........................
{since f = frequency of revolution}Therefore,T = 2π / 2πf=> T = 1 /
fTangential VelocityWhen a body is moving along a circle orcircular
path, the velocity of the body along the tangent of the circle is
called its tangential velocity.Vt = r ωTangential velocity is not same
for every point on the circular path.Centripetal AccelerationA 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 followsa(c) = V2 / r,
........................... r = radius of the circular pathProve That
a(c) = V2 / rProofConsider a body moving along a circularpath 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 / rSince the body is
moving with constant speedTherefore,|V1| = |V2| = VTherefore,ΔV / V =
Vs / rΔV = (V / r) ΔsDividing both sides by ΔtTherefore,ΔV / Δt =
(V/r) (V/r) (Δs / Δt)taking limit Δt → 0.Proof That a(c) = 4π2r /
T2ProofWe know thata(c) = V2 / rBut V = r ωTherefore,a(c) = r2 ω2 /
ra(c) = r ω2 ...................... (1)But ω = Δθ / ΔtFor one complete
rotation Δθ = 2π, Δt = T (Time Period)Therefore,ω = 2π / T(1) => a(c)
= r (2π / T)2a(c) = 4 π2 r / T2 .................. Proved@import
"/extensions/GoogleAdSense/GoogleAdSense.css";Tangential
AccelerationThe 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 thespeed is uniform
(unchanging) the body do not passes tangential acceleration.Total Or
Resultant AccelerationThe resultant of centripetal accelerationa(c)
and tangential acceleration a(t) is called total or resultant
acceleration denoted by a.Centripetal ForceIf a body is moving along a
circular pathwith 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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