Angles in Physics

Angles in Physics

• In physics, the angle is measured in radian not in degree, so is your

calculator
𝐴𝐴𝐴𝐴𝐴𝐴 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑠𝑠
𝜃𝜃 ≡ =
𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑟𝑟
𝑠𝑠 = 𝑟𝑟𝜃𝜃
𝑟𝑟
• One round turn = 360 degree
2𝜋𝜋𝜋𝜋
• One round turn = = 2𝜋𝜋 = 6.283 rad
𝑟𝑟

360
1 rad = = 57.3 degree
6.283
Chapter 2 Motion

Part1: Motion Along a

Straight Line
Topics for Chapter 2 part 1

• What is motion? Why do we need to understand

motion?
• How do we describe motion? Position, velocity,
acceleration
• One dimensional motion in terms of velocity and
acceleration
• Instantaneous and average velocity (acceleration).
Differences.
Topics for Chapter 2 part 1

• What is motion? Why do we need to understand

motion?
• How do we describe motion? Position, velocity,
acceleration
• One dimensional motion in terms of velocity and
acceleration
• Instantaneous and average velocity (acceleration).
Differences.
Topics for Chapter 2 part 1

• What is motion? Why do we need to understand

motion?
• How do we describe motion? Position, velocity,
acceleration
• One dimensional motion in terms of velocity and
acceleration
• Instantaneous and average velocity (acceleration).
Differences.
Topics for Chapter 2 part 1

• What is motion? Why do we need to understand

motion?
• How do we describe motion? Position, velocity,
acceleration
• One dimensional motion in terms of velocity and
acceleration
• Instantaneous and average velocity (acceleration).
Differences.
What is motion?

• Motion is: change of position with time (position as

function of time)
• Motion of basket ball, missile & train : change of
the basket ball position as a function of time
• Change of missile and train position with time
Why do we need to study motion/mechanics?

• Design of engine, moving components,

car, trains,
• Understand human motion in sport
• Design robots, artificial limbs
• Describe electron motion in electronic
devices
Description of Motion by physical quantities

• Mechanics is the study of motion and the cause of motion

(change of position with time).
• The objective is to find the position as a function of time

• Velocity and acceleration are other physical quantities used

to describe motion apart from position and time
• The reason we study velocity and acceleration is because they
are directly related to the cause of motion – forces/energy
• Then, from velocity and acceleration, we can deduce the
position (coordinates) as a function of time.
How is position described?
• First thing to do is to set up a coordinate
system (origin and rulers).
• Position is described by coordinates or
position vector (a vector that takes you
from the origin to the position)

• One dimension: x coordinate on the x-axis

• Two dimension: two coordinates: x, y and

position vector 𝑟𝑟⃗ = 𝑥𝑥𝚤𝚤̂ + 𝑦𝑦𝚥𝚥̂
• Three dimension: three coordinates: x, y, z,
position vector 𝑟𝑟⃗ = 𝑥𝑥𝚤𝚤̂ + 𝑦𝑦𝚥𝚥̂ + 𝑧𝑧𝑘𝑘�
How do we describe motion?
• Motion is change of position with time,

• Write coordinates as a function of time describes the motion

1Dimension: x as a function of t, x = x(t)
2Dimensin: x(t) and y(t); 𝑟𝑟(𝑡𝑡)
⃗ = 𝑥𝑥(𝑡𝑡)𝚤𝚤̂ + 𝑦𝑦(𝑡𝑡)𝚥𝚥̂

3Dimension: x(t), y(t) and z(t); 𝑟𝑟(𝑡𝑡)

⃗ = 𝑥𝑥(𝑡𝑡)𝚤𝚤̂ + 𝑦𝑦(𝑡𝑡)𝚥𝚥̂ + 𝑧𝑧(𝑡𝑡)𝑘𝑘�

• Apart from position, we also want to know: how fast is the motion,
how fast motion is changed, so we need…

Velocity (how fast the motion is),

Acceleration (how fast the motion is changed)
 Displacement (vector)

• A particle moving along the x-axis, its

coordinate x(t) changes with time t.
Δx Δx

• The change in the particle’s position x2 x1 x1 x2

(coordinate) is Δx = x2 - x1 in time Δt.

• Δx is called the displacement. Δx can be positive, moving to the right

or negative, moving to the left. Thus displacement is a vector
 Displacement and Average Velocity (vector)

• x(t) is a function of time, as shown in the x-t graph.

• The average x-velocity of the particle in time Δt is

vav-x = Δx/Δt = displacement divided by time.
x2
• In the x-t graph, the average x-velocity vav-x is x1
the slope of the line connecting two positions

• It describes how fast (rate) the particle moves in time

Δt and direction Δx Δx vav-x > 0
vav-x< 0
x2 x1 x1 x2 A position-time graph (an x-t
graph) shows the particle’s
• Or, you can find displacement/position from average position x as a function of time t.
velocity using Δx= vav-x Δt or x2 = x1 − vav-x Δt
 Instantaneous velocity
• Average velocity is for a time interval Δ𝑡𝑡, so it is not accurate enough,
we want velocity at an instance (at a particular time t)

• The instantaneous velocity is the velocity at a specific instant of time or

specific point along the path and is given by vx = dx/dt. (differentiation
of x with respect to t)

Uniform velocity, Constant velocity, Uniform motion means the

instantaneous velocity is a constant. So, the average velocity equals the
instantaneous velocity and is also a constant.
Finding instantaneous velocity on an x-t graph
• Let us look at the average velocity in a small time interval Δ𝑡𝑡:
Slope of line PQ is the average velocity 𝑣𝑣𝑥𝑥 = Δ𝑥𝑥/Δ𝑡𝑡
• Now let Δ𝑡𝑡 → 0, or Q approach P. PQ becomes the tangent at P, and
Δ𝑥𝑥 𝑑𝑑𝑑𝑑
𝑣𝑣av−𝑥𝑥 = ⇒ ≡ 𝑣𝑣𝑥𝑥 when Δ𝑡𝑡 → 0
Δ𝑡𝑡 𝑑𝑑𝑑𝑑
Δ𝑥𝑥 𝑑𝑑𝑑𝑑
𝑣𝑣𝑥𝑥 = lim =
∆𝑡𝑡→0 Δ𝑡𝑡 𝑑𝑑𝑑𝑑
• The average velocity becomes the
instantaneous velocity, which is the
differentiation of x(t) with respect to t:
• Instantaneous velocity is the slope of
the tangent at P.
 Distance [average speed (scalar)] and Displacement [velocity(vector)]

• Displacement Δx can be positive or negative, depending

on the direction of travel, but distance D is the total length
of travel and is the sum of the lengths of each segment.

• Distance D is always positive and is a scalar.

Δx = 2 – 3 = −1 Δx

D=4+3=7

• Average Speed vav (scalar) = D/∆t ≠ | Δx |/Δt =|𝑣𝑣̅ x |

• However, instantaneous Speed v (scalar) = lim |∆x|/∆t = |vx| = magnitude of

instantaneous velocity.
Average acceleration and Instantaneous acceleration

• Average acceleration describes the average rate of change of velocity

with time in a time interval Δt, describes how fast the motion is changed
• The average x-acceleration is aav-x = Δvx/Δt.

• The instantaneous acceleration is ax = dvx/dt or average

acceleration when Δt go to zero:
𝑑𝑑𝑣𝑣𝑥𝑥 Δ𝑣𝑣𝑥𝑥
𝑎𝑎𝑥𝑥 = = lim
𝑑𝑑𝑑𝑑 Δ𝑡𝑡→0 Δ𝑡𝑡

You can find the change in velocity from the average

acceleration: Δvx = aav-xΔt

Uniform acceleration, constant acceleration means the

instantaneous and average acceleration are constants.
Find acceleration from velocity-time graph

Δ𝑣𝑣𝑥𝑥 𝑑𝑑𝑣𝑣𝑥𝑥
𝑎𝑎𝑥𝑥 = lim =
∆𝑡𝑡→0 Δ𝑡𝑡 𝑑𝑑𝑑𝑑
Find x from vx: Area under the vx- t curve gives the displacement x
constant
• Δx = vxΔt, is the area of a rectangle vx
velocity
= the displacement in time Δt t
Δt
• What if the velocity is not constant? vx
Δx = vxΔt is the area of a rectangle
• Divide the area under the curve into
small slices, the area of a small slice is
approximately by the area of a rectangle
• Sum of the area of all the rectangles
= approximately the area under the
curve, when slices are thin. t
t0 Δt t1
• ∑ Δx = ∑ 𝑣𝑣x Δ𝑡𝑡 = total displacement 𝑡𝑡1
= area under the curve: ∆𝑡𝑡 → 0: 𝑥𝑥1 − 𝑥𝑥0 = � 𝑣𝑣𝑥𝑥 𝑑𝑑𝑑𝑑
𝑡𝑡0
Find vx from ax: Area under the ax- t curve gives the velocity vx

• Likewise, area under ax - t curve is

the change in velocity

𝑡𝑡1
𝑣𝑣𝑥𝑥𝑥 − 𝑣𝑣𝑥𝑥𝑥 = � 𝑎𝑎𝑥𝑥 𝑑𝑑𝑑𝑑
𝑡𝑡0
Special Case: constant acceleration motion
 Motion with constant acceleration

• For a particle with constant

acceleration, the velocity
changes at the same rate
throughout the motion:
Δ𝑣𝑣𝑥𝑥
= 𝑎𝑎𝑥𝑥 = const.
Δ𝑡𝑡
• Velocity-time graph is a
straight line
t
The equations of motion with constant acceleration

• There are four equations

shown to the right for any
straight-line motion with
constant acceleration ax.
• First equation gives the
velocity at time t
• Second equation gives the
position of the particle at
time t
• Third equation gives you the
velocity at t in terms of the initial
velocity, acceleration and the
distance travelled (displacement
between t=0 and t).
• Fourth equation gives the distance
travelled in terms of initial
velocity, final velocity and time
spent.
Derivation of the four equations

• The 1st , 2nd and 4th equations can be

derived using the velocity-time graph.
• 3rd can be derived from 1st and 2nd or 4th.
 1st eqt.
axt=vx - v0x
vx= v0x+ axt

3rd eqt.: 𝑣𝑣𝑥𝑥2 = 𝑣𝑣0𝑥𝑥

2
+ 2𝑎𝑎𝑥𝑥 𝑥𝑥 − 𝑥𝑥0 Another way
1st: 𝑣𝑣𝑥𝑥 = 𝑣𝑣0𝑥𝑥 + 𝑎𝑎𝑥𝑥 𝑡𝑡 1st: 𝑣𝑣𝑥𝑥 = 𝑣𝑣0𝑥𝑥 + 𝑎𝑎𝑥𝑥 𝑡𝑡
⇒ 𝑡𝑡 = (𝑣𝑣𝑥𝑥 − 𝑣𝑣0𝑥𝑥 )/𝑎𝑎𝑥𝑥
⇒ 𝑎𝑎𝑥𝑥 t = 𝑣𝑣𝑥𝑥 −𝑣𝑣0𝑥𝑥
Substitute it into 4th : 2(𝑥𝑥 − 𝑥𝑥0 ) = (𝑣𝑣𝑥𝑥 +𝑣𝑣0𝑥𝑥 )𝑡𝑡
1
2nd : 𝑥𝑥 = 𝑥𝑥0 + 𝑣𝑣0𝑥𝑥 𝑡𝑡 + 𝑎𝑎𝑥𝑥 𝑡𝑡 2
2
Multiply both sides together
𝑣𝑣0𝑥𝑥 𝑣𝑣𝑥𝑥 − 𝑣𝑣0𝑥𝑥
⇒ 𝑥𝑥 − 𝑥𝑥0 = 2
2𝑎𝑎𝑥𝑥 𝑥𝑥 − 𝑥𝑥0 𝑡𝑡 = (𝑣𝑣𝑥𝑥2 −𝑣𝑣0𝑥𝑥 )𝑡𝑡
1
𝑎𝑎𝑥𝑥
4th & 2nd eqt. + (𝑣𝑣𝑥𝑥 − 𝑣𝑣0𝑥𝑥 )2 /𝑎𝑎𝑥𝑥
2 Cancel t :
displacement = x - x0= Area = t (vx+v0x)/2 Multiply both sides by 2𝑎𝑎𝑥𝑥 2
⇒ 𝑣𝑣𝑥𝑥2 = 𝑣𝑣0𝑥𝑥 + 2𝑎𝑎𝑥𝑥 𝑥𝑥 − 𝑥𝑥0
(use expression for area of a trapezoid)
2
⇒ 2𝑎𝑎𝑥𝑥 (𝑥𝑥 − 𝑥𝑥0 ) = 𝑣𝑣𝑥𝑥2 − 𝑣𝑣0𝑥𝑥
x - x0 =(v0x+v0x+axt)t/2 = v0xt + axt2/2 2
⇒ 𝑣𝑣𝑥𝑥2 = 𝑣𝑣0𝑥𝑥 + 2𝑎𝑎𝑥𝑥 𝑥𝑥 − 𝑥𝑥0
Example of using calculus in constant acceleration
The example below show how calculus is used to find
velocity of constant acceleration motion
Differentiation is used to find velocity from position,
integration is used to find position vs time from acceleration Differentiation is needed
in exam, integration not
needed

needed in exam,
Two bodies with different accelerations
• Follow Example 2.5 in which the police officer and motorist have different
accelerations.
𝑥𝑥M0 = 0
𝑥𝑥P0 = 0
𝑣𝑣P0 = 0
Chapter 2 part 2

Motion in Two or Three

Dimensions
Topics for chapter 2 part 2

• How do we describe motion in three or two

dimension space? Use of vectors.
• The velocity vector of an object
• The acceleration vector of an object
• An example of two dimensional motion:
the curved path of projectile
• Circular motion
Introduction

• In reality objects move in three dimension

• We need to extend our description of motion to two and three
dimensions.
 Position vector and coordinates

• Position of a point P (an object) is given

by the coordinates or the position vector.
Position vector tells you the position of the
particle relative to the origin O.
P
• Position vector from the origin to point P
has components x, y, and z. These
components x, y, z are the coordinates of P O

• Motion of P is just the position vector 𝑂𝑂𝑂𝑂 = 𝑟𝑟⃗

or coordinates as a function of time:
 Average velocity—Figure 3.2
z
• The average velocity (a vector) between two points is
the displacement (a vector) or difference of position
vector Δ𝑟𝑟⃗ divided by the time interval between the
two points, and it has the same direction as the
displacement:
∥ Δ𝑟𝑟⃗ = 𝑟𝑟2 − 𝑟𝑟1
y
• Write Δ𝑟𝑟⃗ in its components:
x Δ𝑟𝑟=𝑟𝑟
⃗ 2 − 𝑟𝑟1
Δ𝑟𝑟⃗ = Δ𝑥𝑥 𝚤𝚤̂ +Δ𝑦𝑦𝚥𝚥̂ + Δ𝑧𝑧𝑘𝑘�
Δ𝑥𝑥 = 𝑥𝑥2 − 𝑥𝑥1
Δ𝑦𝑦 = 𝑦𝑦2 − 𝑦𝑦1

Δ𝑧𝑧 = 𝑧𝑧2 − 𝑧𝑧1
 Instantaneous velocity
• The instantaneous velocity is the instantaneous rate of change of
position vector with respect to time. It is a vector.


 Instantaneous velocity (Graph)

Instantaneous velocity is always

tangent to the path P
𝑣𝑣⃗
Consider a 2-D example of the ∆𝑟𝑟⃗ Q
blue path on the right: 𝑟𝑟⃗1
𝑟𝑟⃗2
Displacement: ∆𝑟𝑟⃗ = 𝑟𝑟⃗2 − 𝑟𝑟⃗1 = 𝑃𝑃𝑃𝑃

𝑣𝑣⃗𝑎𝑎𝑎𝑎 = ∆𝑟𝑟⃗ /∆𝑡𝑡 ∥ ∆𝑟𝑟⃗

As ∆𝑡𝑡 → 0, Q  P and 𝑣𝑣⃗𝑎𝑎𝑎𝑎 → 𝑣𝑣⃗

But as ∆𝑡𝑡 → 0, 𝑃𝑃𝑃𝑃 becomes

tangent to the path Instantaneous velocity is always tangent to the path
Difference between speed and velocity
• Speed is a scalar, which is distance travelled divided by time
• Velocity is a vector = displacement/time

• Distance travelled can be larger than the displacement length

• Average speed may not equal average velocity
• Instantaneous speed is the magnitude of the velocity,
i.e. the length of the velocity vector. Constant speed v

• Constant speed does not mean constant velocity

because velocity direction can change, while speed is
constant (circular motion, come back to this later) Radius r
Average acceleration
• The average acceleration during a time interval Δt is
defined as the velocity change during Δt divided by Δt.

Δ𝑣𝑣=𝑣𝑣
⃗ 2 − 𝑣𝑣1
Δ𝑣𝑣𝑥𝑥 = 𝑣𝑣𝑥𝑥𝑥 − 𝑣𝑣𝑥𝑥𝑥
Δ𝑣𝑣𝑦𝑦 = 𝑣𝑣𝑦𝑦𝑦 − 𝑣𝑣𝑦𝑦𝑦
Δ𝑣𝑣𝑧𝑧 = 𝑣𝑣𝑧𝑧𝑧 − 𝑣𝑣𝑧𝑧𝑧

It is a vector with
three components:
Instantaneous acceleration
• The instantaneous acceleration
is the instantaneous rate of
change of the velocity with
respect to time.
• The components of the
instantaneous acceleration are
• ax = dvx/dt,
• ay = dvy/dt,
• az = dvz/dt.

Any particle following a curved path is

accelerating, even if it has constant speed,
because the direction is changing
 Direction of the acceleration vector
• Velocity is always tangential to the path.
• But the direction of the acceleration vector is more
or less arbitrary, depending on whether the speed is
constant, increasing, or decreasing, as shown:

𝑎𝑎⃗∥

𝑎𝑎⃗∥

𝑎𝑎⃗⊥

𝑎𝑎⃗⊥  Velocity direction change. 𝑎𝑎⃗∥  Velocity magnitude change.

Working on motion in 2 and 3 dimensions.
• For 2-D and 3-D cases, we need to use vectors to describe motion
• It is convenient to express the vectors in terms of the unit vectors,
i.e. their components.
• Each component can be treated as one-dimension motion.

• When you calculate the velocity and acceleration using the

components, you are actually doing one dimensional
problems for each components independent of others.
𝑑𝑑𝑑𝑑 𝑑𝑑𝑦𝑦 𝑑𝑑𝑧𝑧 𝑑𝑑𝑣𝑣𝑥𝑥 𝑑𝑑𝑣𝑣𝑦𝑦 𝑑𝑑𝑣𝑣𝑧𝑧
𝑣𝑣𝑥𝑥 = , 𝑣𝑣𝑦𝑦 = , 𝑣𝑣𝑧𝑧 = 𝑎𝑎𝑥𝑥 = , 𝑎𝑎𝑦𝑦 = , 𝑎𝑎𝑧𝑧 =
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

Total:
Calculating average and instantaneous velocity

• A rover vehicle moves on the surface of Mars.

• Follow Example 3.1.
Enlarge the figure for clarity
 North
West
Calculating average and instantaneous acceleration
• Return to the Mars rover.
• Follow Example 3.2.
Typical 2-dimension motion
-- projectile motion
Projectile motion

• The projectile motion is a typical 2 dimension

motion that happens near above the surface of earth
• We learn more how to handle 2 dimension motion
using projectile motion
• The trick: separate the motion into the motion along
x and y directions (the x and y components of the
position vector) and treat them individually as 1-D
What is a projectile motion?

• A projectile is any body given an

initial velocity that then follows a path 𝑔𝑔⃗
determined by the effects of gravity
and air resistance.

• We usually neglect air resistance and

the curvature and rotation of the earth.
We consider constant gravity effect
only. Gravity gives an acceleration − g g = 9.8m/s2
perpendicular to the surface and
downward.
 The x and y motion are separable

• In projectile motion, the horizontal motion is not affected by

vertical motion. The two motions can be separated and independent
• The horizontal position of the ball is the same as a ball moving with
a constant velocity with zero acceleration

The vertical position is

the same as that of one
thrown up with velocity 𝑎𝑎⃗𝑦𝑦 = 𝑔𝑔⃗
v0y and influenced by the
gravitational acceleration
− g = -9.8m/s2
𝑎𝑎⃗𝑥𝑥 = 0
Interesting Video quiz
If they are not shown in the lecture, you can watch the
videos on the web.
http://media.pearsoncmg.com/aw/aw_0media_physics/vt
d/video2.html
http://media.pearsoncmg.com/aw/aw_0media_physics/vt
d/video5.html
http://media.pearsoncmg.com/aw/aw_0media_physics/vt
d/video7.html#
http://media.pearsoncmg.com/aw/aw_0media_physics/vt
d/video9.html
Equations for projectile motion
We apply the four equations to the x direction motion
(constant velocity) and the y direction motion (constant
acceleration) independently

𝑎𝑎𝑦𝑦 = −𝑔𝑔 = −9.8𝑚𝑚/𝑠𝑠 2

+𝑦𝑦
𝑔𝑔⃗
+𝑥𝑥

but with a common thread: time t (both motions occur at the same time)
 The equations for projectile motion
• If we set x0 = y0 = 0, the equations 𝑥𝑥 = 𝑣𝑣0 cos𝜃𝜃0 𝑡𝑡
describing projectile motion are 1 2
𝑦𝑦 = 𝑣𝑣0 sin𝜃𝜃0 𝑡𝑡 − 𝑔𝑔𝑡𝑡
shown at the right. 𝑣𝑣0𝑥𝑥 = 𝑣𝑣0 cos 𝜃𝜃0 2
𝑣𝑣0𝑦𝑦 = 𝑣𝑣0 sin 𝜃𝜃0
𝑣𝑣𝑥𝑥 = 𝑣𝑣0 cos𝜃𝜃0
Eliminate parameter t = x/v0cosθ0 : 𝑣𝑣𝑦𝑦 = 𝑣𝑣0 sin𝜃𝜃0 − 𝑔𝑔𝑔𝑔

𝑔𝑔𝑥𝑥 2
𝑦𝑦 = 𝑥𝑥 tan𝜃𝜃0 − 2 2
2𝑣𝑣0 cos 𝜃𝜃0
• The trajectory is a parabola. 𝐻𝐻
𝑣𝑣02 sin2 𝜃𝜃0 𝑣𝑣02 sin 2𝜃𝜃0
𝐻𝐻 = 𝑦𝑦𝑀𝑀 = , 𝑅𝑅 =
2𝑔𝑔 𝑔𝑔
𝑅𝑅
Height and range of a projectile
• A baseball is batted at an angle.
• Follow Example 3.7.
Another example of two dimension motion
– Uniform Circular Motion
Uniform circular motion

Another important example of two dimension motion is the

uniform circular motion
Examples: the orbit of satellite around the earth, a car goes
around a circular roundabout or corner, a merry-go-around
Uniform circular motion—Figure 3.27

• For uniform circular motion, the speed is

𝑎𝑎∥ = 0
constant but the direction of velocity Constant speed v
vector is constantly changing
 changing velocity  acceleration Radius r

• No change in speed  𝑎𝑎∥ = 0

• The acceleration is perpendicular to the
velocity and pointing to the center: It is
The period T is the time
called centripetal acceleration, ac for one revolution, so

𝑣𝑣 2 𝑣𝑣 is the constant speed v = 2πr/T, and ac = 4π2r/T2.

𝑎𝑎𝑐𝑐 = 𝑟𝑟 is the radius of circle
𝑟𝑟
 Derivation of centripetal acceleration
Assume the object rotates an angle ∆θ in Δ𝑡𝑡:
By geometry, the v-vector (tangent) also rotates an angle of ∆θ

The velocity changes from 𝑣𝑣⃗1 to 𝑣𝑣⃗2 is ∆𝑣𝑣⃗ = 𝑣𝑣⃗2 − 𝑣𝑣⃗1 𝛥𝛥θ

∆𝑣𝑣
As ∆𝑡𝑡 → 0, ∆𝜃𝜃 → 0, ∆𝑣𝑣⃗ ⊥ 𝑣𝑣⃗ ⇒ 𝑎𝑎⃗ ≡ ⊥ 𝑣𝑣,
⃗
∆𝑡𝑡
or 𝑎𝑎⃗𝑐𝑐 is perpendicular to 𝑣𝑣⃗
|∆𝑣𝑣|
Magnitude of |𝑎𝑎⃗𝑐𝑐 | ≡ , but
∆𝑡𝑡

Magnitude of 𝑎𝑎⃗𝑐𝑐 in uniform circular motion is a constant = 𝑎𝑎𝑐𝑐 = 𝑣𝑣 2 /r

 Uniform circular motion revisited: vector component calculus method
Two-dim motion: 𝑅𝑅 = 𝑥𝑥 𝑡𝑡 𝚤𝚤̂ + 𝑦𝑦(𝑡𝑡)𝚥𝚥̂ 𝑣𝑣 ≡ 𝜔𝜔𝜔𝜔
Uniform
𝑣𝑣⃗ 𝑣𝑣𝑦𝑦
𝑥𝑥 𝑡𝑡 = 𝑅𝑅cos𝜃𝜃 = 𝑅𝑅cos(𝜔𝜔𝑡𝑡) Motion 𝜃𝜃 𝑣𝑣⃗
𝑦𝑦 𝑡𝑡 = 𝑅𝑅sin𝜃𝜃 = 𝑅𝑅sin(𝜔𝜔𝜔𝜔) 𝜃𝜃 = 𝜔𝜔𝜔𝜔 𝜃𝜃
𝑣𝑣𝑥𝑥 (𝑥𝑥, 𝑦𝑦)
𝑑𝑑𝑑𝑑
𝑣𝑣𝑥𝑥 𝑡𝑡 = = 𝑥𝑥 ′ = −𝜔𝜔𝜔𝜔sin 𝜔𝜔𝑡𝑡 = −𝑣𝑣 sin𝜃𝜃 𝑎𝑎⃗ 𝑅𝑅
𝑑𝑑𝑑𝑑
𝑑𝑑𝑦𝑦
𝑣𝑣𝑦𝑦 𝑡𝑡 = = 𝑦𝑦 ′ = 𝜔𝜔𝜔𝜔cos(𝜔𝜔𝜔𝜔) = 𝑣𝑣 cos𝜃𝜃
𝑑𝑑𝑑𝑑

So, the v-vector (tangent) also rotates an angle of θ

𝑑𝑑𝑣𝑣𝑥𝑥 𝑣𝑣 2
𝑎𝑎𝑥𝑥 𝑡𝑡 = = 𝑥𝑥 ′′ = −𝜔𝜔2 𝑅𝑅cos 𝜔𝜔𝑡𝑡 = − cos𝜃𝜃
𝑑𝑑𝑑𝑑 𝑅𝑅
𝑑𝑑𝑣𝑣𝑦𝑦 𝑣𝑣 2
𝑎𝑎𝑦𝑦 𝑡𝑡 = = 𝑦𝑦 ′′ = −𝜔𝜔2 𝑅𝑅sin(𝜔𝜔𝜔𝜔) = − sin𝜃𝜃
𝑑𝑑𝑑𝑑 𝑅𝑅

Magnitude of 𝑎𝑎⃗ = 𝑎𝑎𝑐𝑐 = 𝑎𝑎𝑥𝑥2 + 𝑎𝑎𝑦𝑦2 = 𝑣𝑣 2 /R cos 2 𝜃𝜃 + sin2 𝜃𝜃 = 1

Centripetal acceleration on a curved road

• A sports car has a lateral acceleration as its rounds a

curve in the road.
• Follow Example 3.11.
Thinking questions
Thinking question TQ2.1

This is the x-t graph

of the motion of a
particle. Of the four
points P, Q, R, and
S, the acceleration
ax is greatest (most
positive) at

A. point P. B. point Q. C. point R. D. point S.

E. not enough information in the graph to decide
TQ2.2 This is a motion diagram of an object moving along the x-
direction with constant acceleration. The dots 1, 2, 3, … show
the position of the object at equal time intervals ∆t.

5 4 3 2 1
x
x=0
Which of the following vx-t graphs best matches the motion
shown in the motion diagram?

vx vx vx vx vx

0 t 0 t 0 t 0 t 0 t
A. B. C. D. E.
 TQ 2.3

The motion diagram shows an object moving along a curved

path at constant speed. At which of the points A, C, and E does
the object have zero acceleration?

A. point A only
B. point C only
C. point E only
D. points A and C only
E. points A, C, and E
TQ2.4

An object moves at a constant speed in a clockwise direction

around an oval track. The geometrical center of the track is at
point O. When the object is at point P, which arrow shows the
direction of the object’s acceleration vector?

#2
#1 #3
P A. #1 (directly away from O)

#4 B. #2 (perpendicular to the track)

#5 C. #3 (in the direction of motion)
O
D. #4 (directly toward O)
E. #5 (perpendicular to the track)

Oval track
 TQ2.5

A zookeeper fires a tranquilizer dart directly at a monkey. The

monkey lets go at the same instant that the dart leaves the gun
barrel. The dart reaches a maximum height P before striking the
monkey. Ignore air resistance.
When the dart is at P, the monkey

A. is at A (higher than P).

B. is at B (at the same height
as P).
C. is at C (lower than P).
D. not enough information
given to decide

Hint: what would happen if there were

no gravity? What does gravity do?
TQ2.6

You drive a race car around a circular track of radius 100 m at a

constant speed of 100 km/h. If you then drive the same car around
a different circular track of radius 200 m at a constant speed of 200
km/h, your acceleration will be
A. 8 times greater.
B. 4 times greater.
C. twice as great.
D. the same.
E. half as great.
(d) Caution It’s a common misconception that at the highest point of free-
fall motion, where the velocity is zero, the acceleration is also zero. Wrong!
The acceleration in free fall is always ay = - g = - 9.80 m/s2. ❙