•
•
•
•
If the net force acting on the body is zero, it is said to be in equilibrium. The acceleration of such
objects will be zero according to Newton’s Law. Zero acceleration implies a constant velocity.
•
Equilibrium: An object is said to be in equilibrium when the net force acting on an object is zero and
if the object is not moving.
•
It is also in equilibrium when the net force acting on an object is zero but if the object is moving at a
constant non-zero velocity.
•
Dynamics is concerned with the motion of objects.
Only when: σ 𝐹 ≠ 0 and/or σ 𝑀 ≠ 0
•
Engineers use mathematical models to describe the dynamics of an object, for example, in the
design of aircraft, cars, linkages, piston-crankshaft mechanisms, etc.
Dynamics can be roughly divided into 2 main areas:
1.
Kinematics: describes the motion of an object.
2.
Kinetics: is concerned with the forces acting on an object and the motion of the object.
Weeks 5 and 7: Kinematics and Kinetics of Particles
•
Rectilinear motion
•
Curvilinear motion
•
Relative motion
•
Kinetics of particles
•
Work/energy methods
•
Impulse & momentum
Weeks 8 and 9: Rigid Body Mechanics
•
Angular Motion
•
Kinematics of Rigid Bodies
•
Kinetics of rigid bodies
•
Moment of inertia
•
Work-energy methods for rigid bodies
•
To gain knowledge of the fundamentals of engineering dynamics;
•
To become familiar with common mechanisms;
•
To learn techniques for analysing the performance of mechanisms;
•
To develop and practise tactics for problem identification, formulation and
solution.
Definition of Particles:
A particle is an approximate model of a body whose size is negligible compared with the other
dimensions surrounding it.
A particle:
- can be modelled as a point
- its size can be ignored
- all forces (kinetics) act through a single point (which corresponds to its centre of mass)
•
Earth orbiting the sun
•
A car travelling on a freeway
•
A flying airplane
Kinematics of particles examines the position (𝑠), velocity (𝑣) and acceleration (𝑎) of particles.
The motion of particles can be described by using coordinates measured from fixed reference
axes. Whilst the motion of a particle is three-dimensional, we initially will consider plane motion.
Plane motion may be motion along a straight line (rectilinear motion) or motion along a curved
path (curvilinear motion).
Rectilinear motion is the motion along a straight line.
•
position (𝑠)
•
velocity (𝑣)
•
acceleration (𝑎)
mainly as functions of time.
In order to define the motion of a particle, we need to define a co-ordinate system and an origin.
We define position relative to a reference point (origin).
𝑠1 = 3 m
𝑠2 = −2 m
Note: Sign is important (it signifies the direction)
The displacement of 𝑃 during an interval of time from 𝑡0 to 𝑡 is the change in position
𝑠(𝑡) – 𝑠(𝑡0), where 𝑠(𝑡) denotes the position at time 𝑡.
Average Velocity
The average velocity between two positions P and P’ is defined as:
𝑣ҧ =
change in position ∆𝑠
=
change in time
∆𝑡
Instantaneous Velocity
The instantaneous velocity at P can be determined by taking the limit as ∆𝑡 → 0
𝑣 = lim
∆𝑡→0
∆𝑠 𝑑𝑠
=
= 𝑠ሶ
∆𝑡 𝑑𝑡
Note the ‘dot’ in 𝑠ሶ indicates the derivative with respect to (w.r.t.) time.
Velocity (𝑣 = 𝑠 ሶ) is the rate of the change of position with respect to time (m/s).
Instantaneous Velocity (v )
v is the gradient of the displacement vs time at 𝑃
•
Positive gradient = positive v
•
Zero gradient = stationary point, may be a maxima or minima.
∆𝑠 𝑑𝑠
𝑣 = lim
=
= 𝑠ሶ
∆𝑡→0 ∆𝑡
𝑑𝑡
We can integrate the velocity to establish a displacement equation,
𝑑𝑠
𝑣 = 𝑑𝑡
rearranging gives,
𝑑𝑠 = 𝑣𝑑𝑡
Integrating both sides
𝑠2
𝑣2
න 𝑑𝑠 = න 𝑣 𝑑𝑡
𝑠1
𝑣1
𝑣2
⇒
𝑠2 − 𝑠1 = න 𝑣 𝑑𝑡
𝑣1
Area under the 𝑣 − 𝑡 curve
Velocity is often given as a function of time, eg.
𝑣 = 3𝑡 − 2𝑡 3
The integration can then be performed mathematically.
𝑠2
𝑣2
𝑡2
න 𝑑𝑠 = න 𝑣 𝑑𝑡 = න (3𝑡 − 2𝑡 3 ) 𝑑𝑡
𝑠1
⇒
𝑣1
𝑡1
3𝑡 2 2𝑡 4
𝑠2 − 𝑠1 =
−
2
4
𝑡2
𝑡1
• The average acceleration is the gradient of the line joining 𝑃 and 𝑃’.
• The average acceleration between two positions 𝑃 and 𝑃’ is defined as:
𝑎ത =
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 ∆𝑣
=
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡𝑖𝑚𝑒
∆𝑡
The instantaneous acceleration at P can be determined by taking the limit as ∆𝑡 → 0
∆𝑣 𝑑𝑣
𝑎 = lim
=
= 𝑣ሶ
∆𝑡→0 ∆𝑡
𝑑𝑡
Acceleration 𝑎 = 𝑣ሶ is the rate of the change of velocity w.r.t. time (m/s2).
Since 𝑣 = 𝑠ሶ
then 𝑎 = 𝑣ሶ = 𝑠ሷ
That is
𝑑𝑣 𝑑 2 𝑠
𝑎=
=
𝑑𝑡 𝑑𝑡 2
Graphically, 𝑎 is the gradient (slope) of the graph at 𝑃
For discontinuous functions, a graphical method can be used.
•
In the first part of the graph, the velocity is increasing linearly → the acceleration is positive
and constant.
At max v, a = 0
•
In the last part, v is decreasing linearly and a is negative and constant.
A sprinter reaches his maximum speed 𝑣𝑚𝑎𝑥 in 2.5 seconds from rest with constant acceleration.
He then maintains that speed and finishes the 100 meters in the overall time of 10.40 seconds.
Determine his maximum speed 𝑣𝑚𝑎𝑥 .
W6 Example 1 (Web view)
If acceleration is a function of time, velocity and displacement can be found through time
integration
𝑑𝑣
𝑎=
𝑑𝑡
𝑑𝑣 = 𝑎 𝑑𝑡
𝑣2
𝑡2
න 𝑑𝑣 = න 𝑎 𝑑𝑡
𝑣1
𝑡1
𝑡2
⇒ 𝑣2 − 𝑣1 = න 𝑎 𝑑𝑡
𝑡1
= area under the 𝑎 − 𝑡 curve
Acceleration (𝑎) is often given as a function of time, eg.
𝑎 = 𝑡 2 − 7𝑡
The integration can then be performed mathematically.
𝑣2
𝑡2
𝑡2
න 𝑑𝑣 = න 𝑎𝑑𝑡 = න (𝑡 2 − 7𝑡)𝑑𝑡
𝑣1
𝑡1
⇒ 𝑣2 − 𝑣1 =
𝑡1
𝑡3
3
−
𝑡
7𝑡 2 2
2 𝑡1
𝑑𝑠
𝑣=
𝑑𝑡
v is the rate of change of position
𝑑𝑣
𝑎=
𝑑𝑡
a is the rate of change of velocity
We may eliminate 𝑑𝑡 from these equation
𝑑𝑠 𝑑𝑣
𝑑𝑡 =
=
𝑣
𝑎
⇒
𝑑𝑠 𝑑𝑣
=
𝑣
𝑎
Rearranging (cross-multiplying) gives:
𝑣𝑑𝑣 = 𝑎𝑑𝑠
Integrate both sides:
𝑣2
𝑠2
න 𝑣𝑑𝑣 = න 𝑎𝑑𝑠
𝑣1
𝑣2
⇒
2
𝑠1
𝑣2
𝑠2
= න 𝑎𝑑𝑠
𝑣1
𝑠1
We can use all the equations so far to derive (recall from Physics) the kinematic equations for
constant acceleration:
𝑣 = 𝑣0 + 𝑎0 (𝑡 − 𝑡0 )
1
𝑠 = 𝑠0 + 𝑣0 (𝑡 − 𝑡0 ) + 𝑎0 (𝑡 − 𝑡0 )2
2
𝑣 2 = 𝑣02 + 2𝑎0 (𝑠 − 𝑠0 )
𝑣22
2
−
𝑣12
2
𝑠
= ‫ 𝑠׬‬2 𝑎𝑑𝑠 = area under the 𝑎 − 𝑠 graph
1
s1
s2
𝑥2
න 𝑑𝑥 = 𝑥 ቚ
𝑥1
𝑥2
𝑥2
𝑥1
𝑥2
= 𝑥2 − 𝑥1
𝑥1
𝑥2
1
𝑥 −𝑛+1
−𝑛
න 𝑛 𝑑𝑥 = න 𝑥 𝑑𝑥 =
𝑥
−𝑛 + 1
𝑥1
𝑥 𝑛+1
𝑛
න 𝑥 𝑑𝑥 =
𝑛+1
𝑥1
Except when n = 1
𝑥1
1
𝑥2
න 𝑑𝑥 = ln = ln 𝑥2 − ln 𝑥1
𝑥
𝑥1
𝑥1
𝑥1
𝑥2
when 𝑛 = 1 → then we must use the natural logarithm, ln
𝑥2
𝑥2
Using the separable variables method
𝑑𝑣
𝑑𝑣
=𝑎 𝑣 ⇒
= 𝑑𝑡
𝑑𝑡
𝑎 𝑣
and integrating to obtain,
𝑣
𝑡
𝑑𝑣
න
= න 𝑑𝑡
𝑎
𝑣
𝑣0
𝑡0
Solve for 𝑣(𝑡) and calculate 𝑠(𝑡) and 𝑎(𝑡)
A small projectile is initially fired with an initial velocity 𝑣𝑖 of 70 m/s downwards in air with
negligible resistance 𝑎𝑎𝑖𝑟 = 0 m/s. It then enters into a fluid medium. Due to the resistance of
the fluid the projectile experiences a deceleration equal to:
𝑎 = (−0.35𝑣3) m/s2
where 𝑣 is in
Projectile
m/s2.
vi = 70 m/s
2
aair = 0 m/s
a)
Find a function of velocity with respect to time.
3
a = −0.35v m/s
2
s=0m
Fluid
b)
Determine the projectile’s velocity 4 seconds after it is fired.
c)
Determine the projectile’s position 4 seconds after it is fired
W6 Example 2 (Web view)
sf
Concepts:
•
Position and displacement
•
Velocity
•
Acceleration
•
Rectilinear motion with Constant acceleration
•
Acceleration specified as a function of time
•
Acceleration specified as a function of velocity
•
Acceleration specified as a function of position
•
Graphical method
•
Constant acceleration equations:
𝑣 = 𝑣0 + 𝑎0 (𝑡 − 𝑡0 )
•
𝑣 2 = 𝑣02 + 2𝑎0 𝑠 − 𝑠0
1
𝑠 = 𝑠0 + 𝑣0 (𝑡 − 𝑡0 ) + 𝑎0 (𝑡 − 𝑡0 )2
2
Acceleration as a function of time
𝑣=
•
𝑑𝑠
𝑑𝑡
𝑎=
𝑑𝑣
𝑑𝑡
As a function of displacement
𝑣𝑑𝑣 = 𝑎𝑑𝑠
•
As a function of velocity
𝑣
𝑡
𝑑𝑣
න
= න 𝑑𝑡
𝑎
𝑣
𝑣0
𝑡0
Curvilinear motion is the motion along a curved path. Although the path may be in 3-D, we are
only concerned with motion in a plane (2-D motion).
We want to determine position, velocity and acceleration at any instant on the curvilinear path.
•
Position
•
Velocity
•
Instantaneous velocity
•
Acceleration
•
Coordinate systems
•
Curvilinear x/y coordinate system
•
Projectile motion
•
Curvilinear n/t coordinate system
•
Circular motion
Plane Curvilinear Motion describes the motion of a particle along a curved path that lies in a
single plane.
Circular motion and projectiles:
Consider the continuous motion of a particle along a plane curve
P’
P
Path
s
v
Δs
Δr
P
P
r’
r
O
Position
(a)
r
s
O
Displacement
(b)
r
s
O
Velocity
(c)
s
Curvilinear motion
Occurs when the particle moves along a curved path
The position of the particle, measured from a fixed point O, is designated by the position vector
𝐫(𝑡) (a vector is written in boldface). The position vector is a function of time.
P
Path
s
r
O
Position
(a)
s
Displacement
Suppose during a small time interval (t, t+Δt) the particle moves a distance Δs along the curve to
a new position P’, defined by r’=r(t+Δt). The displacement Δr represents the change in the
particle’s position.
Δ𝑟 = 𝑟(𝑡 + Δ𝑡) − 𝑟(𝑡)
Velocity
Δ𝐫
The average velocity of the particle is defined as 𝐯𝑎𝑣𝑔 = Δ𝑡
The instantaneous velocity is determined from this equation by letting Δt → 0,
𝑑𝐫
𝐯=
= 𝑟ሶ
𝑑𝑡
• Direction of v is tangent to the curve
• Magnitude of v is the speed, which may be obtained
by noting the magnitude of the displacement Δr is
the length of the straight line segment from P to P’.
𝑣=
𝑑𝑠
𝑑𝑡
Acceleration
If the particle has a velocity v at time t and a velocity v’ = v(t + Δt) at time t’ = t + Δt.
The increment of velocity is Δv = v(t + Δt) - v(t)
The average acceleration
𝐚𝑎𝑣𝑔 =
Δ𝐯
Δ𝑡
Instantaneous acceleration
𝑑𝐯
𝑑2𝐫
𝐚=
= 𝑣ሶ = 2 = 𝑟ሷ
𝑑𝑡
𝑑𝑡
•
Three coordinate systems are used in problem solving in curvilinear motion. Each system
has its advantages. When problem solving, it is best to select the most suitable coordinate
system for the problem.
•
Using a different coordinate system does NOT change the position, velocity and
acceleration.
Three coordinate systems are used in problem solving in curvilinear motion. Each system
•
has its advantages. When problem solving, it is best to select the most suitable coordinate
system for the problem.
Using a different coordinate system does NOT change the position, velocity and acceleration.
•
The 3 coordinate systems are:
1.
Rectangular coordinates (𝑥/𝑦)
2.
Normal and tangential coordinates (𝑛/𝑡)
3.
Polar (Cylindrical) coordinates (𝑟/𝜃)
x-y

Position
Position vector is defined by r = xi + yj + zk
i, j, k are fixed Eulerian unit vectors
The magnitude of r is always positive and defined as
𝑟=
𝑥2 + 𝑦2 + 𝑧2
The direction of r is specified by the components of the unit vector er = r/r
Velocity
𝒗=
𝑑𝒓
= 𝑣𝑥 𝒊 + 𝑣𝑦 𝒋 + 𝑣𝑧 𝒌
𝑑𝑡
𝑣𝑥 = 𝑥ሶ 𝑣𝑦 = 𝑦ሶ 𝑣𝑧 = 𝑧ሶ
The velocity has a magnitude defined as the
positive value of
𝑣=
𝑣𝑥2 + 𝑣𝑦2 + 𝑣𝑧2
and a direction that is specified by the components of the unit vector ev = v/v and is always
tangent to the path.
Acceleration
𝐚=
𝑑𝐯
= 𝑎𝑥 𝑖Ԧ + 𝑎𝑦 𝑗Ԧ + 𝑎𝑧 𝑘
𝑑𝑡
𝑎𝑥 = 𝑣ሶ𝑥 = 𝑥ሷ
𝑎𝑦 = 𝑣ሶ𝑦 = 𝑦ሷ
𝑎𝑧 = 𝑣ሶ𝑧 = 𝑧ሷ
Acceleration has a magnitude
𝑎=
𝑎𝑥2 + 𝑎𝑦2 + 𝑎𝑧2
Since a represents the time rate of change in velocity, a will not be tangent to the path.
The motion of a particle is defined by
𝑥 = 2𝑡 2 − 4𝑡 − 4
𝑦 = 𝑥2 + 3
Where x and y are in metres and t is in seconds.
Find |v| and |a| of the particle when t = 4.0 s.
W6 Example 3 (Web view)
Rectangular coordinates are particularly useful for describing projectile motion. In projectile
motion, we neglect aerodynamic drag and the curvature and rotation of the earth. We also assume
that the altitude change is small enough so that the acceleration due to gravity can be considered
constant.
Normal–Tangential Coordinates
For the 𝑛 − 𝑡 coordinate system:
•
𝐞𝒏 is a unit vector in the normal direction;
•
𝐞𝒕 is a unit vector in the tangential direction
The tangential axis is defined to be in the direction tangential to the path, that is, in the direction
of velocity. The direction of this axis thus changes with time, and
𝒗 = 𝑣𝒆𝑡
where v is the speed and 𝐞𝑡 is a unit vector in the tangential direction.
The normal axis is perpendicular to the tangential axis and is towards the centre of curvature of
the path.
The direction of the normal axis also changes
with time. 𝐞𝒏 is a unit vector in the normal
direction.
n/t coordinates are not normally used to define position, but are commonly used to define
acceleration.
We know
𝒗 = 𝑣𝒆𝑡
Acceleration becomes
𝒂=
𝑑𝒗 𝑑 𝑣𝒆𝑡
=
= 𝑣𝒆
ሶ 𝑡 + 𝑣𝒆ሶ 𝑡
𝑑𝑡
𝑑𝑡
Since both 𝑣 and 𝒆𝑡 vary with time, it can be shown that
𝒆ሶ 𝑡 =
𝑣
𝒆
𝜌 𝑛
where  is the instantaneous radius of curvature.
Thus:
𝑣2
𝒂 = 𝑣𝒆
ሶ 𝑡 + 𝒆𝑛
𝜌
𝑣2
𝒂 = 𝑎𝑡 𝒆𝑡 + 𝑎𝑛 𝒆𝑛 = 𝑣𝒆
ሶ 𝑡 + 𝒆𝑛
𝜌
𝑎𝑡 = 𝑣ሶ
𝑎𝑛 =
𝑣2
𝜌
is the acceleration in the tangential direction
is the acceleration in the normal direction.
The magnitude of acceleration is
𝒂 =𝑎=
Note:
𝒂 =𝑎=
𝑎𝑥2 + 𝑎𝑦2 =
𝑎𝑛2 + 𝑎𝑡2
𝑎𝑛2 + 𝑎𝑡2
The acceleration does NOT change using a different coordinate system.
A baseball player releases a ball with the initial conditions shown. Determine the radius of curvature
of the trajectory (a) just after release and (b) at the highest point of the ball. For each case, determine
the time rate of change of the speed.
W6 Example 4 (Web view)
Polar Coordinates
Specify the location of P using
1.
Radial coordinate r, which extends outward
from the fixed origin O to the particle and the
2.
Circumferential coordinate θ, which is the
counterclockwise angle between a fixed
reference line and the r axis.
𝑟=
𝑥2 + 𝑦2
𝜃 = tan−1 ( 𝑦/𝑥)
Basis Vectors
Vectors in polar coordinates are defined by using the unit vectors er and eθ extending from P
•
er : along increasing r, when θ is fixed
•
eθ : along increasing θ when r is fixed
Position
Position vector:
𝐫 = 𝑟𝐞𝑟
(can be: rറ = 𝑟e𝑟 )
Note er and eθ change directions as P moves and are perpendicular to each other
Lagrangian Basis Vectors
We need to introduce Lagrangian unit vectors ei that move with the particle. These
will replace the Eulerian basis vectors i, j, k
er and eθ change direction w.r.t. time
• To evaluate 𝐞ሶ 𝑟 note that er changes its
direction w.r.t. time although its
magnitude = 1 (rotating of a unit vector)
ሶ 𝜃 𝐞ሶ 𝜃 = −𝜃𝐞
ሶ 𝑟
𝐞ሶ 𝑟 = 𝜃𝐞
Velocity
• Instantaneous velocity v is obtained by the time derivative of r (using product rule)
𝐯 = 𝐫ሶ =
• To evaluate 𝒆ሶ 𝑟 recall
ሶ 𝜃
𝐞ሶ 𝑟 = 𝜃𝐞
∴
ሶ 𝜃
𝐯 = 𝑟𝐞
ሶ 𝑟 + 𝑟𝜃𝐞
𝑑(𝑟𝐞𝑟 )
= 𝑟𝐞
ሶ 𝑟 + 𝑟𝐞ሶ 𝑟
𝑑𝑡
Instantaneous velocity v
𝐯 = 𝑣𝑟 𝐞𝑟 + 𝑣𝜃 𝐞𝜃
𝑣𝑟 = 𝑟ሶ
𝑣𝜃 = 𝑟𝜃ሶ
• Radial component vr is a measure of the rate of increase or decrease in the length of the
radial coordinate
• Circumferential component vθ is the rate of motion along the circumference of a circle having
a radius r
Speed
• Since vr and vθ are mutually perpendicular, the
magnitude of the velocity or speed is:
𝑣=
𝑟ሶ
2
+ 𝑟𝜃ሶ
2
• Direction of v is always tangent to the path at P
Acceleration
Let’s take the rate derivative of our expression for v
𝐚 = 𝐯ሶ =
𝑑
ሶ 𝜃)
(𝑟𝐞
ሶ 𝑟 + 𝑟𝜃𝐞
𝑑𝑡
ሶ 𝜃 + 𝑟𝜃𝐞
ሷ 𝜃 + 𝑟𝜃ሶ 𝐞ሶ 𝜃
= 𝑟𝐞
ሷ 𝑟 + 𝑟ሶ 𝐞ሶ 𝑟 + 𝑟ሶ 𝜃𝐞
Recall our expressions for the rate derivative of er and eθ
ሶ 𝜃
𝐞ሶ 𝑟 = 𝜃𝐞
∴
ሶ 𝑟
𝐞ሶ 𝜃 = −𝜃𝐞
ሶ 𝜃 + 𝑟ሶ 𝜃𝐞
ሶ 𝜃 + 𝑟𝜃𝐞
ሷ 𝜃 − 𝑟𝜃ሶ 2 𝐞𝑟
𝐚 = 𝑟𝐞
ሷ 𝑟 + 𝑟ሶ 𝜃𝐞
Generalised acceleration
Instant acceleration has two components:
𝐚 = 𝑎 𝑟 𝐞𝑟 + 𝑎 𝜃 𝐞𝜃
𝑎𝑟 = 𝑟ሷ − 𝑟𝜃ሶ 2
𝑎𝜃 = 𝑟𝜃ሷ + 2𝑟ሶ 𝜃ሶ
•
The term 𝜃ሷ = 𝑑 2 𝜃/𝑑𝑡 2 is called the angular acceleration since it measures the
change made in the angular velocity during an instant of time, use unit rad/s2
•
Since ar and aθ are always perpendicular, the magnitude of the
acceleration is:
𝑎=
𝑟ሷ − 𝑟𝜃ሶ 2
2
+ 𝑟𝜃ሷ + 2𝑟ሶ 𝜃ሶ
2
•
Direction is determined the vector components
•
Acceleration may not be tangent to the path!
Circular motion is an important special case of plane curvilinear motion where the radius of
curvature 𝜌 becomes a constant radius of the circle.
Circular motion is commonly associated with 𝑛/𝑡 coordinates. Circular motion is the motion of a
point moving around in a circular path, eg. linkages, tip of a turbine blade.
For circular motion, radius r is constant for all θ.
Coordinate System. Polar coordinates
Velocity and Acceleration. Since r is constant,
𝑟=𝑟
𝑟ሶ = 0
𝑟ሷ = 0
Velocity and Acceleration
If we take the generalised formulation and reduce it to circular motion, do we get our classical
equation?
If 𝑟 is constant, then
𝑟ሶ = 0 𝑟ሷ = 0
𝑣𝑟 = 𝑟ሶ = 0
𝑣𝜃 = 𝑟𝜃ሶ
𝑎𝑟 = 𝑟ሷ − 𝑟𝜃ሶ 2 = −𝑟𝜃ሶ 2
𝑎𝜃 = 𝑟𝜃ሷ + 2𝑟ሶ 𝜃ሶ = 𝑟𝜃ሷ
If angular velocity 𝜃 is also constant (uniform circular motion) then 𝜃ሷ = 0
𝑣𝑟 = 𝑟ሶ = 0
𝑣𝜃 = 𝑟𝜃ሶ
2
𝑣
𝜃
𝑎𝑟 = −𝑟𝜃ሶ 2 = −
𝑟
𝑎𝜃 = 0
Acceleration is towards the centre of the circle
•
Curvilinear x-y including projectile motion
𝑣𝑥 = (𝑣𝑥 )𝑜 = 𝑣𝑜 cos 𝜃
𝑣𝑦 = (𝑣𝑦 )𝑜 + 𝑎𝑦 𝑡 = 𝑣𝑜 sin 𝜃 − 𝑔𝑡
•
Polar coordinates for circular motion
𝐫 = 𝑟𝐞𝑟
ሶ 𝜃
𝐯 = 𝑟𝐞
ሶ 𝑟 + 𝑟𝜃𝐞
ሶ 𝜃 + 𝑟ሶ 𝜃𝐞
ሶ 𝜃 + 𝑟𝜃𝐞
ሷ 𝜃 − 𝑟𝜃ሶ 2 𝐞𝑟
𝐚 = 𝑟𝐞
ሷ 𝑟 + 𝑟ሶ 𝜃𝐞
•
Curvilinear n-t coordinates for paths
𝒗 = 𝑣𝒆𝑡
𝑣2
𝒂 = 𝑣𝒆
ሶ 𝑡 + 𝒆𝑛 = 𝑎𝑡 𝒆𝑡 + 𝑎𝑛 𝒆𝑛
𝜌
•
•
•
•
•
Kinetics is concerned with the relationship between the forces acting on an object and the motion
of the object
• At this stage, we are only interested in particles
• Therefore, we are only concerned with the translational motion of the centre of mass of the
particle (i.e., rectilinear translation and curvilinear translation)
• We are NOT yet interested in the rotation of the object about its centre of mass
To work in kinetics, we need both forces and kinematics (motion)
•
Kinetics looks at the forces needed to maintain a particular motion or the motion caused by
particular forces
•
We can look at Statics as a special case of Dynamics where the acceleration is zero
1.
How do we apply Newton’s 2nd Law along with Free Body Diagrams to particles?
2.
How can we model forces from springs, strings, friction, gravity, etc. in a Dynamics setting?
3.
How do Work/Energy and Impulse/Momentum principles apply in Dynamics?
4.
How do we know when to apply each of the 3 approaches (Newton’s laws, Work/Energy &
Impulse/Momentum)?
We are probably due a reminder on Newton’s 2nd Law:
“The change of motion is proportional to the motive force
impressed; and is made in the direction of the right line in
which that force is impressed.” – Sir Isaac Newton
Isaac Newton
Newton’s 2nd Law is also known as “The Balance of Linear Momentum”:
• Newton’s statement of the law is actually in a form that we will call Impulse/Momentum
• Leonhard Euler wrote the law in its differential form and referred to it as the Balance of Linear
Momentum
• Where 𝐺 = 𝑚𝑣 is the linear momentum
𝐅=
𝑑𝐆
𝑑𝑡
or
𝐅 = 𝐆ሶ
Leohnard Euler
For a particle, the mass 𝑚 is constant (𝑑𝑚/𝑑𝑡 = 0)
Then the balance of linear momentum reduces to
𝐅 = 𝑚𝐚
• Note that 𝐹 and 𝑎 are vectors
• This is called an “equation of motion”
• Also note that 𝐹 is the sum of the forces (net force) if 𝑎 is the total acceleration
• The acceleration and the sum of the forces are in the same direction
• So if we know the motion of the particle from kinematics, we can find the net force
• Likewise, if we can find the net force, we can use kinematics to find the path
A block of mass m = 5 kg is acted on by a force of 20 N. Neglecting friction, find the acceleration
of the block in each case.
W6 Example 5 (Web view)
When we draw a FBD, we draw the body free of any attachments. At the same time, we draw ALL
the forces that are acting on the free body. There forces will include:
•
force due to gravity
•
forces from any attachments or contacts
•
externally applied forces
•
Indicate velocity or acceleration
Consider an elevator without a passenger. The elevator is
accelerating upwards.
1.
Draw the FBD of the elevator.
2.
Find the tension in the elevator cable when the
acceleration of the elevator 𝑎 = 0.6 m/s 2 , upwards. The
mass of the elevator 𝑚 = 400 kg.
3.
If a passenger with a mass of 80 kg stands on a set of
scales on the floor, what mass will the scales read the
elevator moves up with 𝑎 = 0.6
m
s2
↑?
• T is the tension in the elevator cable
• mg is the force due to gravity (N)
• Using Newton’s 2nd law
• We have to sum the forces in the same direction as the motion.
𝑇 − 𝑚𝑔 = 𝑚𝑎
𝑇 = 𝑚𝑔 + 𝑚𝑎 = 450(0.6) + 450(9.81) = 4684.5𝑁 ≠ 𝑚𝑔
If a passenger with a mass of 80 kg stands on the floor of the elevator in Example 1. Suppose the
elevator moves up with a = 0.6 m/s2 ↑.
Find T.
Now include a passenger:
෍ 𝐹 = 𝑚𝑎 ⇒
𝑇 − (𝑚𝑒 + 𝑚𝑝 )𝑔 = (𝑚𝑒 + 𝑚𝑝 )𝑎 = 5517.3 N
Alternatively, we can analyse the elevator and passenger separately. We need to include the
reaction force R between the person and the floor of the elevator, that is, there is a reaction force R
acting from the floor of the elevator on the feet of the person, and, from Newton’s 3rd law, there is
an equal and opposite reaction force from the feet of the person acting on the elevator floor. We
now have 2 FBDs (one for the elevator and the other for the person).
• Also, using the kinematic relationship between the elevator and the person we have:
෍ 𝐹 = 𝑚𝑎
⇒
𝑇 − 𝑅 − 𝑚𝑒 𝑔 = 𝑚𝑒 𝑎 𝑒
(1)
𝑅 − 𝑚𝑝 𝑔 = 𝑚𝑝 𝑎 𝑝
(2)
𝑎𝑒 = 𝑎𝑝
• That is, both the person and the elevator have the same acceleration upwards.
• Adding equations (1) and (2) together we get:
⇒ 𝑇 − 𝑅 − 𝑚 𝑒 𝑔 + 𝑅 − 𝑚𝑝 𝑔 = 𝑚𝑒 𝑎 𝑒 + 𝑚𝑝 𝑎 𝑝
𝑇 − (𝑚𝑒 ⥂ +𝑚𝑝 )𝑔 = (𝑚𝑒 ⥂ +𝑚𝑝 )𝑎
same as before!
•
We denote the constant of proportionality by 𝜇𝑠
•
Hence, 𝐹 ≤ 𝜇𝑆 𝑅
•
When the block starts moving, the friction opposes
motion
•
The direction is opposite
•
As noted above, its magnitude is proportional to the
magnitude of 𝑅, the reaction (normal) force
•
Let’s call the constant of proportionality 𝜇𝑘
•
Now Newton’s 2nd Law gives us 𝐹 = 𝜇𝑘
Ff = μsN
•
The motion is known (i.e. there is no relative
motion between the contacting objects)
•
So we can use 𝐅 = 𝑚𝐚 to find 𝐹𝑓 even if 𝐚 = 0
•
The equation 𝐹𝑓 = 𝜇𝑠 𝑁 is only valid at impending
slip
Consider a block of mass 𝑚 on a fixed inclined surface. The block has an external force 𝑃 applied to
it. Assuming the friction coefficient between the block and the incline is µd = 0.3, find the
acceleration of the block.
Consider a block of mass 𝑚 on a fixed inclined surface. The block has an external force 𝑃 applied to
it. Assuming the friction coefficient between the block and the incline is µd = 0.3, find the
acceleration of the block.
Strings, ropes and cables are often used to connect particles, and often the connection is via a pulley.
Strings and cables are considered inextensible (ie. not stretching), massless, and can only transit
forces while in tension.
A FBD of a short length of string is:
The tension force T must be equal and opposite.
Simplifying assumptions are also made about pulleys. (Unless otherwise stated), we assume
pulleys are massless and frictionless. A FBD of a pulley shown in the figure below is shown. Since
the pulley is massless and frictionless, there can be no resultant motion. Hence, the magnitude
and direction of the reaction force on the pulley from its bearing is such that the sum of the forces
on the pulley is zero.
FBD of Pulley
A 10 kg mass and a 5 kg mass are connected by a cable which runs over a massless pulley. The
system is released. Determine:
(a) the acceleration of each mass,
(b) the tension in the cable.
W6 Example 6 (Web view)
• Acceleration in polar coordinates
𝐚 = 𝑎𝑟 𝐞𝑟 + 𝑎𝜃 𝐞𝜃
𝑎𝑟 = 𝑟ሷ − 𝑟𝜃ሶ 2
𝑎𝜃 = 𝑟𝜃ሷ + 2𝑟ሶ 𝜃ሶ
• Acceleration in curvilinear n-t
coordinates
𝑣2
𝒂 = 𝑣𝒆
ሶ 𝑡 + 𝒆𝑛 = 𝑎𝑡 𝒆𝑡 + 𝑎𝑛 𝒆𝑛
𝜌
The mass 𝑚 (2.3 kg) has a velocity of 5.4 m/s.
The coefficient of kinetic friction for the mass 𝑚 is 0.5.
The pulley is massless and frictionless.
The curved surface is circular with a radius 𝑅 = 0.9 m.
Find the acceleration of both masses. Mass 𝑀 = 5 kg.
W6 Example 7 (Web view)
What will it be?
What will it be?
W6 Live Experiment (Web view)
• Newton’s statement of the law is actually in a
form that we will call Impulse/Momentum
𝐅 = 𝑚𝐚
• Leonhard Euler wrote the law in its differential
form and referred to it as the Balance of Linear
Momentum, where 𝐺 = 𝑚𝑣 is the linear
momentum
𝐅=
𝑑𝐆
𝑑𝑡
or
𝐅 = 𝐆ሶ
• For particle kinetics, we must be competent with
cables/pully systems, springs, friction, gravity
and curvilinear motion
Contents
•
Relative motion of two particles along a straight line (1-D)
•
Relative motion in 2-D
•
Relative motion of rigid links
We have been describing particle motion using coordinates referred to fixed reference axes.
The particle displacements, velocities and accelerations so far determined have been absolute.
Examples of relative motion: Two cars
•
fluid particles and a point on a turbine blade – the analysis of the relative velocities of fluids
and mechanical systems are useful for the design of turbines, pumps, etc.
•
two parts of a mechanical linkage, eg. piston and crankshaft in a car engine – the analysis of
the relative velocities and accelerations are useful for linkage design (see Kinematics of rigid
bodies).
Relative motion of two particles along a straight line (1-D)
Consider two particles 𝐴 and 𝐵: for example 2 cars on a road
𝑥𝐴 is the position of particle/car 𝐴.
𝑥𝐵 is the position of particle/car 𝐵.
𝑥𝐵/𝐴 is the position of particle/car 𝐵 relative to particle/car 𝐴.
Note: 𝑥𝐴 and 𝑥𝐵 are with respect to fixed (or absolute) axes. 𝑥𝐵/𝐴 is with respect to a moving axis.
𝑥𝐵 = 𝑥𝐴 + 𝑥𝐵/𝐴
𝑥𝐵/𝐴 = 𝑥𝐵 − 𝑥𝐴
Differentiating with respect to time
𝑣𝐵 = 𝑣𝐴 + 𝑣𝐵/𝐴
𝑎𝐵 = 𝑎𝐴 + 𝑎𝐵/𝐴
In 2-D and 3-D motion, the positions, velocities and accelerations of the particles need to be
described in terms of vectors.
𝐫𝐵 = 𝐫𝐴 + 𝐫𝐵/𝐴
• 𝐫𝐴 is the position of particle 𝐴
• 𝐫𝐵 is the position of particle 𝐵
• 𝐫𝐵/𝐴 is the position of particle 𝐵 relative to particle 𝐴
• 𝐫𝐴/𝐵 is the position of particle 𝐴 relative to particle 𝐵
𝐫𝐴 = 𝐫𝐵 + 𝐫𝐴/𝐵
If 𝐫𝐵 − 𝐫𝐴 = 𝐫𝐵/𝐴 and 𝐫𝐴 − 𝐫𝐵 = 𝐫𝐴/𝐵
then 𝐫𝐴/𝐵 = −𝐫𝐵/𝐴
Differentiating with respect to time:
𝐯𝐵 = 𝐯𝐴 + 𝐯𝐵/𝐴
→
𝐯𝐴/𝐵 = −𝐯𝐵/𝐴
𝐚𝐵 = 𝐚𝐴 + 𝐚𝐵/𝐴
→
𝐚𝐴/𝐵 = −𝐚𝐵/𝐴
In relative motion, it is often convenient to draw a velocity vector diagram.
For the instant shown below, car 𝐴 is rounding the circular curve at a constant speed of
50 km/h, while car 𝐵 with an instantaneous speed of 60 km/h is slowing down at the rate of
8 km/h per second (i.e., 2.22 m/s2).
Determine:
a)
the velocity of car 𝐴 relative to car 𝐵,
b)
the acceleration that car 𝐴 appears to have to an observer in car 𝐵.
W6 Example 8 (Web view)
𝐫𝐴 and 𝐫𝐵 are the positions of particle 𝐴 and B
𝐫𝐵/𝐴 is the position of particle 𝐵 relative to 𝐴
𝐫𝐴/𝐵 is the position of particle 𝐴 relative to 𝐵
𝐫𝐴 = 𝐫𝐵 + 𝐫𝐴/𝐵
𝐫𝐵 = 𝐫𝐴 + 𝐫𝐵/𝐴
Differentiating with respect to time:
𝐯𝐵 = 𝐯𝐴 + 𝐯𝐵/𝐴
𝐚𝐵 = 𝐚𝐴 + 𝐚𝐵/𝐴