Important Point
Chapter 13: Kinematics of a particle
Kinematics
Curvilinear Motion
Angular Motion
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Important Point
Dynamics is concerned with bodies that have accelerated
motion.
Kinematics is a study of the geometry of the motion.
Kinetics is a study of the forces that cause the motion.
Rectilinear kinematics refers to straight-line motion.
Figure: During the time this rocket undergoes rectilinear motion, its
altitude as a function of time can be measured and expressed as s = s(t).
Its velocity can then be found using v = ds
dt and its acceleration can be
dv
determined from a = dt .
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Chapter 13: Kinematics of a particle
Kinematics-Motion without taking into account the forces
that cause them.
Particle- Object with mass but no physical extension.
Read through section 13.1
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Straight-line (Rectilinear) motion
Straight line path of a point is defined using a single coordinate
axis s.
Position of P at time t is s = s(t)
Displacement of P during time interval [t0 , t] is s(t) − s(t0 )
ds
Velocity of P at time t is v (t) =
dt
dv
d 2s
Acceleration of P at time t is a(t) =
= 2
dt
dt
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Straight-line (Rectilinear) motion
Straight line path of a point is defined using a single coordinate
axis s.
Position of P at time t is s = s(t)
Displacement of P during time interval [t0 , t] is s(t) − s(t0 )
ds
Velocity of P at time t is v (t) =
dt
dv
d 2s
Acceleration of P at time t is a(t) =
= 2
dt
dt
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Straight-line (Rectilinear) motion
Straight line path of a point is defined using a single coordinate
axis s.
Position of P at time t is s = s(t)
Displacement of P during time interval [t0 , t] is s(t) − s(t0 )
ds
Velocity of P at time t is v (t) =
dt
dv
d 2s
Acceleration of P at time t is a(t) =
= 2
dt
dt
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Straight-line (Rectilinear) motion
Straight line path of a point is defined using a single coordinate
axis s.
Position of P at time t is s = s(t)
Displacement of P during time interval [t0 , t] is s(t) − s(t0 )
ds
Velocity of P at time t is v (t) =
dt
dv
d 2s
Acceleration of P at time t is a(t) =
= 2
dt
dt
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Straight-line (Rectilinear) motion
Straight line path of a point is defined using a single coordinate
axis s.
Position of P at time t is s = s(t)
Displacement of P during time interval [t0 , t] is s(t) − s(t0 )
ds
Velocity of P at time t is v (t) =
dt
dv
d 2s
Acceleration of P at time t is a(t) =
= 2
dt
dt
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Example
The car moves in a straight line such that for a short time its
acceleration is
a(t) = 1.8t + 0.6 m/s 2
Given that s(1) = 2.4m and s(2) = 6m, determine its velocity and
position when t = 3s.
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Example
The airplane releases its drag parachute on landing at t = 0. Its
velocity is given by
v (t) =
80
m/s
1 + 0.32t
.
1
What is its acceleration at t = 3s.
2
How far does plane travel from time t = 0 to t = 10s.
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Example
Race car starts from rest and accelerates at a = 5 + 2t (m/s 2 ) for
10s. The brakes are then applied and the car has constant
acceleration a = −30m/s 2 until it come to rest. Determine
1
Maximum velocity of the car
2
Total distance traveled by the car.
3
Total time of travel.
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Straight-line motion: Acceleration as a function of velocity
or position
Example
When a train is traveling along a straight track at 2 m/s, it begin
to accelerate at a = (60v −4 ) m/s 2 , where v is in m/s. Determine
its velocity v and the position 3 s after the acceleration.
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Straight-line motion: Acceleration as a function of velocity
or position
Example
Velocity of rocket sled is 400m/s when its hits a water break. The
subsequent acceleration is a = −0.003v 2 m/s 2 .
1
Find the time taken for sled’s velocity to decrease to 100m/s.
2
Obtain the total distance covered in this time.
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Example
A steel ball is released from rest in a container of oil. Its downward
acceleration is a = 2.4 − 0.6v cm/s 2 , where v is the ball’s velocity
in cm/s.
1
What is the ball’s downward velocity 2s after it is released?
2
what distance does the ball fall in the first 2 s after its release?
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Example
A particle travels to the right along a straight line with a velocity
5
m/s, where s is in meters. Determine its deceleration
v = 4+s
when s = 2 m.
Example
Engineers analyzing the motion of a linkage determine that the
velocity of an attachment point is given by v = A + 4s 2 m/s,
where A is a constant. When s = 2 m, its acceleration is measured
and determined to be a = 320 m/s 2 . What is its velocity of the
point when s = 2 m?
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Example
A particle travels to the right along a straight line with a velocity
5
m/s, where s is in meters. Determine its deceleration
v = 4+s
when s = 2 m.
Example
Engineers analyzing the motion of a linkage determine that the
velocity of an attachment point is given by v = A + 4s 2 m/s,
where A is a constant. When s = 2 m, its acceleration is measured
and determined to be a = 320 m/s 2 . What is its velocity of the
point when s = 2 m?
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Curvilinear Motion
Occurs when particle moves along a curved path. Since path is
often described in three dimensions, vector analysis will be used to
formulate the particles position, velocity and acceleration.
Position:
r = x i + y j + zk
Velocity:
v = dr
dt
Acceleration:
a = dv
dt
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Example
During a test flight in which a helicopter starts from rest at t = 0
from the origin. Components of acceleration during the time
interval t ∈ [0, 10] sec are ax = 0.6t and ay = 1.8 − 0.36t m/s 2 .
1
Determine the helicopter’s speed at t = 6s.
2
Obtain the position vector of the helicopter at t = 6s.
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Projectile Problem
Object projected through the air.
Air resistance is negligible.
The only force affecting the object’s motion is its weight.
Angular Motion
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Example
The skier leaves the 20◦ surface at 10m/s.
1
Determine the distance d to the point where he lands.
2
What are the magnitudes of his components of velocity
parallel and perpendicular to the 45◦ surface just before he
lands?
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Example
Golfer pitches ball with initial angle θ0 = 50◦ . What range of
velocities v0 will cause ball to land within 2m of the hole?
Angular Motion
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Example
Athlete releases the shot 1.82m above the ground. Initial velocity
is v0 = 13.6m/s at an angle of 30◦ to the horizontal.
1
Find the horizontal distance the shot travels from point of
release to point where it hits the ground.
2
Determine the maximum height the shot attains.
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Example
A car travels at a constant speed of 100km/h on a straight road of
increasing grade whose vertical profile can be approximated by the
equation shown. When the car’s horizontal coordinate is
x = 400m, what is the car’s acceleration?
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Angular Motion
In the previous section we analysed curvilinear motion using
Cartesian coordinates. It is often simpler to use other coordinate
systems.
First introduce
Angular motion in a plane.
Derivative with respect to time of a unit vector rotating in a
plane.
Angular motion of a line in a plane
θ- angular position of line L relative to line L0 (rad)
Angular velocity of L relative to L0 is ω = dθ
dt
d 2θ
Angular acceleration of L relative to L0 is α = dω
dt = dt 2
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Example
Rotor of a jet engine is rotating at 10000rpm when the fuel is shut
off. The subsequent angular acceleration is
α = −0.00002ω 2 (rad/s 2 .
1 How long does it take the rotor to slow to 1000 rpm?
2
Determine the number of revolutions the rotor turns as it
decelerates from 10000 rpm to 1000 rpm.
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Example
The needle of a measuring instrument is connected to a torsional
spring that gives it an angular acceleration α = −4θrad/s 2 , where
θ is the needle’s angular position in radians relative to a reference
direction. The needle is given an angular velocity ω = 2rad/s in
the position θ = 0.
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Derivative of a rotating vector w.r.t. time
Angular Motion
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Curvilinear Motion-Normal and Tangential Components
We identify the position of a particle using a coordinate measured
along its path. Velocity and acceleration are given in terms of
components normal and tangential to the path.
et - unit vector tangential to the path
en - unit vector normal to the path
These unit vectors rotate as the particle moves.
Important Point
Chapter 13: Kinematics of a particle
Planar Motion
Curvilinear Motion
Angular Motion
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Example
Car starts from rest at A and the tangential component of its
acceleration is
at = 0.6 − 0.002v 2 (m/s 2 )
where v is the velocity in m/s. Determine the cars velocity and
acceleration in terms of normal and tangential components when it
reaches B.
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Expressing acceleration in terms of the radius of curvature
ρ- radius of curvature
ds- is approximate length of the corresponding arc of the
circle of radius ρ.
If the path is circular, ρ is just the radius of path.
We have
a = at et + an en
dθ
dv
et + v en
=
dt
dt
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Summary: Normal and tangential components of velocity
and acceleration
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Example
Helicopter starts from rest at t = 0. Cartesian components of
acceleration are
ax = 0.6t
and
ay = 1.8 − 0.36t (m/s 2 )
Obtain the tangential and normal components of acceleration and
the radius of curvature of the path at t = 4 s.
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Circular Motion
Particle moving in a circular path of radius R.
Angular Motion
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Example
The distance from the center of the medical centrifuge to its
samples is 300 mm. When the centrifuge is turned on its angular
acceleration is given by
α = 12 − 0.02ω (rad/s 2 )
(ω is the centrifuge’s angular velocity). Determine the tangential
and normal components of the velocity and acceleration when
t = 0.2 s.
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Curvilinear Motion- Polar Coordinates
We use coordinates (r , θ).
r - distance from O to P
θ- anticlockwise angle measured from + ve x axis.
In order to express vectors in polar coordinate we introduce unit
vectors er and eθ .
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
er - points in direction of radial line from O to P.
eθ - normal to er and points in direction of increasing θ.
Position: r = r er
Velocity: v = dr
dt
Angular Motion
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Example
If arm OA rotates counterclockwise with a constant angular
velocity of θ̇ = 2 rad/s , determine the magnitudes of the velocity
and acceleration of peg P at θ = 30◦ . The peg moves in the fixed
groove defined by the lemniscate, and along the slot in the arm.
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Example
The driver of the car maintains a constant speed of 40m/s.
Determine the angular velocity of the camera tracking the car
when θ = 15◦ .
Angular Motion
Important Point
Chapter 13: Kinematics of a particle
Curvilinear Motion
Angular Motion
Example
6) The slotted arm AB drives pin C through the spiral groove
described by the equation r = 1.5θ m , where θ is in radians. If the
arm starts from rest when θ = 60◦ and is driven at an angular
velocity of θ̇ = (4t) rad/s , where t is in seconds, determine the
radial and transverse components of velocity and acceleration of
the pin C when t = 1s.