Download Circular Motion

Unit 2
Circular Motion
Circular Motion
The theory of circular motion is concerned with concepts we are familiar with,linear
motion and linear acceleration, along with the new concepts of angular speed and
angular acceleration.
How do they relate to each other?
Linear Motion
r = position
v = 𝑑𝑟
= velocity
𝑑𝑡
= rate of change of position
a = 𝑑𝑣
= acceleration
𝑑𝑡
= rate of change of velocity
Circular Motion
𝜃
= angular position
𝑑𝜃
𝑑𝑡
𝜃
=
= angular speed
= rate of change of angle
𝜃
= 𝑑𝑣
= acceleration
𝑑𝑡
= rate of change of velocity
When a particle moves along a straight line, it’s speed is calculated as the rate of change of distance. For a particle moving in
a circle we find the speed by finding the rate at which the radius is turning.
Tangential speed: A particle moving at constant speed v round a circle with centre O and radius r will at any instant be
moving in the direction of the tangent to the circle, so its speed is referred to as the tangential speed.
Angular Speed:
This can be thought of as the rotational speed as the particle rotates around its axis. Denoted ω and it is
𝜃
equal to . Where 𝜃 is in radians and time is in seconds.
𝑡
ω
𝜃
=𝑡
= 2π x number of revolutons in 1 second
If the angular speed is not constant, the definition of angular speed can be generalised as the rate of increase of the angle ϴ
𝑑𝜃
with respect ot time i.e ω =
𝑑𝑡
Period of Rotation:
2π
T=
ω
Uniform Circular Motion
The general expressions for circular motion are quite complex the particular case of uniform circular motion is far simpler.
With uniform circular motion 𝜃 is constant and therefore 𝜃= 0
𝜃=
Hence if

𝑑𝜃
𝑑𝑡
= a constant = 𝜔
𝜃 = 𝜔t + c
(c is a constant of integration)
If 𝜃 = 0 when t = 0 then 𝜃= 𝜔t and then the equations become,
𝑟 = (𝑟𝑐𝑜𝑠𝜔t, rsin𝜔t) = rer
𝑑𝑟
v = 𝑑𝑡 = r𝜔(-sin𝜔t, cos𝜔t) = r𝜔e𝜃
𝑑𝑣
a = 𝑑𝑡 = -r𝜔2(cos𝜔t,sin𝜔t) = -r𝜔2e𝑟
The magnitiudes of velocity and acceleration are:
v = r𝜔
a=
-r𝜔2
=
v2
r
So despite having a constant speed given by v = r𝜔 circular motion gives rise to an acceleration which is directed towards
the origin. This acceleration occurs not because of a change in speed but because of a continuous change in direction
whilst going round in a circle. The velocity vector is continuously changing and this must constitue an acceleration.
If a body is accelerating while in circular motion there must be a force acting on it.
F = ma
= -mr𝜔2er
=-
mv2
e
r r
So circular motion requires a central force of magnitude mr𝜔2 or
F = mr𝜔2 =
𝑚𝑣2
𝑟
𝑚𝑣2
𝑟
, otherwise it will fly off in a straight line at constant speed.
Example
What is the speed of the top of a minute hand of a clock, where the hand is 7cm long?
Find angular speed
ω=
ω=
𝜃
𝑡
2π
60 x 60
Leave as a fraction ( final answer is more accurate)
v=rω
v=7x
2π
60 x 60
v = 0.00012 m/s
Example 2
A particle has position vector at time t, in seconds, given by
r = ( 2 + 3cos2t, 4 + 3sin2t)
(a) Show that the motion is circular and
(b) Find it’s velocity and acceleration.
We need to show that the coordinates of the particle form the equation of a circle  (x – a)2 + (y – b)2 = r2
So rearrange and equating gives:
 (x,y) = ( 2 + 3 cost2t, 4 + 3sin2t)
Consider x component
x = 2 + 3cos2t
x – 2 = 3cos2t
(x – 2)2 = 9cos22t
Consider y component
y = 4 + 3sin2t
y – 4 = 3sin2t
(y – 4)2 = 9sin22t
 (x – a)2 + (y – b)2 = r2
 (x – 2)2 + (y – 4)2 = 9cos22t + 9sin22t = 9(cos22t + cos22t)
 (x – 2)2 + (y – 4)2 = 9
So path is a circle of radius 3 centre (2,4)
(b)
To find v we differentiate r
r = ( 2 + 3cos2t, 4 + 3sin2t)
 v = 𝑟 = (-6sin2t, 6cos2t) = 6( -sin2t, cos2t) Speed = 𝑣 = 62 ( −𝑠𝑖𝑛2𝑡
Speed = 6m/s
2
+ (𝑐𝑜𝑠2𝑡)2)
=1
To find a we differentiate v
a = 𝑣 = 6( -2cos2t, -2sin2t) = 12 (-cos2t, -sin2t)
𝑎 = 122 ( −𝑐𝑜𝑠2𝑡
𝑎 = 12 ms-2
2
+ (−𝑠𝑖𝑛2𝑡)2
Expressions for a particles velocity and acceleration whilst executing circular motion can be established using vectors
y
r
𝜃
Taking r = a constant and a varying 𝜃
we get the following for the components of the position of r
x
r = rcos𝜃, rsin𝜃
We can then differentiate this to find equivalent expressions for v and a.
NOTE:
(1)
When differentiating trig terms with respect to time when 𝜃 is itself a function of time we must use the chain
rule
𝑑 𝑐𝑜𝑠𝜃
𝑑(𝑐𝑜𝑠𝜃) 𝑑𝜃
=
. 𝑑𝑡 = −𝑠𝑖𝑛𝜃.𝜃
So
𝑑𝑥
𝑑𝜃
(2)
To then differentiate −𝑠𝑖𝑛𝜃.𝜃 we need to use the product rule 
𝑑(−𝑠𝑖𝑛𝜃.𝜃)
𝑑𝑥
=
𝑑 𝑠𝑖𝑛𝜃
𝑑𝜃
𝑑𝜃
. 𝜃 + 𝑠𝑖𝑛𝜃 𝑑𝑡
𝑑 𝑢𝑣
𝑑𝑥
=v
= 𝑐𝑜𝑠𝜃 𝜃𝜃 + sin𝜃 𝜃
= 𝑐𝑜𝑠𝜃 𝜃2 + sin𝜃 𝜃
𝑑𝑢
𝑑𝑥
𝑑𝑣
+ 𝑢 𝑑𝑥 = 𝑣𝑢′ + 𝑢𝑣′
Hence
𝑟 = (𝑟𝑐𝑜𝑠𝜃, rsin𝜃)
𝑑𝑟
v = 𝑑𝑡 = (-rsin𝜃 𝜃,rsin𝜃 𝜃 ) = r𝜃(−𝑠𝑖𝑛𝜃, 𝑐𝑜𝑠𝜃)
𝑑𝑣
a = 𝑑𝑡 = (-r𝑐𝑜𝑠𝜃 𝜃2 -rsin𝜃 𝜃, -r𝑠𝑖𝑛𝜃 𝜃2 + rcos𝜃 𝜃) = -r𝜃2(cos𝜃,sin 𝜃) + r 𝜃(-sin𝜃,cos 𝜃)
So that
r = rer
v = r𝜃e𝜃
a = -r𝜃 2er + r𝜃 e𝜃
Where
er = (cos𝜃,sin𝜃) is the unit vector in the radial direction and
e𝜃 = (-sin𝜃,cos𝜃)is the unit vector in the tangential direction
Since er . e𝜃 = 0 they are perpendicular
e𝜃
er
Banked Inclines
Consider the forces acting on a car which is completing a turn on a banked incline – in the
absence of friction.
Note:
The normal force is always
perpendicular to the road surface
When considering the net force
acting we need to resolve forces
vertically and horizontally
So for our car
N ϴ Ncosϴ
ϴ
mg
Resolving forces
Vertically :
No acceleration forces are balanced
Mg = Ncosϴ
Horizontally:
Centripetal force (acceleration) acting towards
centre of the turn
 Nsinϴ = ma
So Nsinϴ is the force causing the acceleration
Example
A bob-sleigh with its two-person team has a total mass of 200Kg. On one stretch of the course
the ream rounds a horizontal bend of radius 25 m at a speed of 35ms-1 . They bank the sleigh
so that it rounds the bend with no sideways frictional force.
(a)
(b)
(a)
Calculate the acceleration of the sleigh as a multiple of g
Find the angle to the horizontal at which the sleigh is banked.
M = 200kg
r = 25m
v = 35ms-1
a=?
a=
a=
𝑣2
𝑟
352
25
a = 49 ms-2
a = 5g ms-2
49 ÷ 9.8 = 5
(b) Draw a force diagram
ϴ
N
Ncosϴ
Nsinϴ
mg
ϴ
Resolving vertically
Ncosϴ = mg
Ncosϴ = 200g
Resolving horizontally
Nsinϴ = ma
Nsinϴ = 200 x 5g
= 1000g
 tan ϴ =
Nsinθ
Ncosθ
 tan ϴ =
1000g
200g
 tan ϴ = 5
 ϴ = 78.7o
The Conical Pendulum
One end of a string of length l, is tied to a hook, and a particle of mass m is attached to the
other end. With the string taut and making an angle ϴ with the downward vertical, the particle
is set in motion so that it rotates in a horizontal circle about the vertical line through the hook.
Find the period of one revolution of the particle around the circle.
Using trig we can see that
Height
(h) ϴ
l
radius = lsinϴ
Radius
To answer questions like this we resolve our forces horizontally and vertically
Vertically
Horizontally
Tcosϴ = mg
Tsinϴ = ma
Tsinϴ = mrω2
The Tsinϴ force
causes the
centripetal
acceleration
Tsinϴ = mlsinϴω2
Tcosϴ ϴ Tension (T)
Tsinϴ
 T= mlω2
mg
Substitute this value of T into the vertical equation
mlω2cosϴ

= mg
ω=
g
lcosθ
2𝜋
Since the time (T) to complete one revolution would be
𝜔
2𝜋
 T=
𝑙𝑐𝑜𝑠𝜃
g
T = 2𝜋
lcosθ
𝑔
NOTE:
Since lcosϴ is just the depth h below the hook the means the period of revolution can be
written as 2π
ℎ
.
𝑔
This is independent of l, ϴ and m. Meaning that if you have a number of
particles (potentially of different mass) attached to strings of different lengths, all moving in
circular paths at the same depth below the hook, they will all take the same time to make a
complete revolution
Example
A stone of mass 0. 5kg is attached to one end of a light inextensible string of length 0.4
metres. The other end of the string is attached to a fixed point O. The stone is released
from rest with the string taut, and inclined at an angle of 40° below the horizontal
through O.
(a)
Calculate the speed of the Particle as it passes beneath O.
(b)
Calculate the speed of the stone when the string makes an angle of 20° with the
vertical through O.
(a)
Calculate the speed of the Particle as it passes beneath O.
Loss in height  Loss in EP and gain in EK
O
40°
rcos50 50°
Loss in EP = mgr – mgrcos50 = mg(r – rcos50)
Gain in EK =
1
2- 1 mu2
mv
2
2
r
 mg(r – rcos50) = 1 mv2- 1 mu2
2
2
1
IMPORTANT:
Always consider whether it’s v or u first in
this formula
Gain  v first (final speed)
Loss  u first (initial speed)
1
rg – rgcos50 = 2 v2- 2 02
0.8g – 0.8gcos50 = v2
v=
0.8 x 9.8 − 0.8 x 9.8 x cos 50
v = 1.67 ms-1
(b)
Calculate the speed of the stone when the string makes an angle of 20° with the vertical through O.
First watch this (again) Vertical circle and conservation of energy
So since the mass was released from rest at and angle of 40° below the horizontal the 20° referred to here must be to
the downward vertical (since 20° to the upward vertical would be higher than where it started)
Gain in height  Gain in EP and loss in EK
O
20°
Gain in EP = mgr – mgrcos20 = mg(r – rcos20)
Loss in EK =
rcos20
1
1
mu2- mv2
2
2
Notice it’s u first
r

1
1
mg(r – rcos20) = 2 mu2- 2 mv2
1
1
rg – rgcos20 = 2 (1.67)2- 2 v2
v2 = 1.672 – 2rg + 2rgcos20
v=
1.672 − 2 x 0.8 x 9.8 + 2 x 0.8 x 9.8 x cos 20
v = 1.53 ms-1
Complete – vertical circle worksheet
Gravitiation
gravitational Force - 3mins
It is only comparatively close to the surface of the Earth that the force of gravity can be considered to be constant.
As a body travels further from the Earth the force of gravity decreases. Newton’s inverse square law of gravitation states
that the magnitude of the force of attraction between two particles is inversely proportional to the square of the distance
between the two particles.
We will use Newton’s Universal Law of Gravitation
𝐺𝑀𝑚
F=
𝑟2
Where G is the universal constant of Gravitation G = 6.67 x 10-11 m3kg-1s-2
We can easily re-arrange this formula to find an expression for GM. This can help us when solving problems of
this type.
𝐺𝑀𝑚
F=
𝑟2
mg =
𝐺𝑀𝑚
𝑟2
mgr 2 = 𝐺𝑀𝑚
gr 2 = 𝐺𝑀
𝐺𝑀 = 𝑔𝑟 2
Example 1
A satellite moves with constant speed in a circular orbit around a planet, at the surface of which the
magnitude of the acceleration due to gravity is 7.8 ms-2. The radius of the planet is 7500 kilometres and the
satellite orbits at a height of h kilometres above the surface of the planet, in a plane through the centre of
the planet. At this height the magnitude of the acceleration due to gravity is 6.6 ms-2. Calculate h.
Example 2
A satellite moves in a circular orbit around the earth in the plane of the equator at a height of 920
kilometres above the surface of the Earth, of radius 6400 kilometres. Calculate (a) the magnitude of
the acceleration due to gravity at this height and (b) the time for a complete revolution.
Example 3
A satellite moves in a circular orbit around a planet of radius 3500 kilometres. The constant speed of the satellite is 3200
ms-1 and it takes 130 minutes to complete one orbit. Calculate the height of the satellite above the surface of the planet.
Given that the mass of the satellite is 120 kilograms, and that the magnitude of the acceleration due to gravity at the
surface of the planet is 4.2 ms-2, calculate the magnitude of the force due to gravity experienced by the satellite while in
orbit.