Download Motion in a Circle

JJ 2014 H2 PHYSICS (9646)
Motion in a Circle
Motion in a Circle
Content
1.
Kinematics of uniform circular motion
2.
Centripetal acceleration
3.
Centripetal force
Learning Outcomes
Candidates should be able to:
(a)
express angular displacement in radians.
(b)
understand and use the concept of angular velocity to solve problems.
(c)
recall and use v = rω to solve problems.
(d)
describe qualitatively motion in a curved path due to a perpendicular force, and
understand the centripetal acceleration in the case of uniform motion in a circle.
(e)
recall and use centripetal acceleration a = rω2, a =
(f)
recall and use centripetal force F = mrω2, F = m
© Physics Dept
v2
to solve problems.
r
v2
to solve problems.
r
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JJ 2014 H2 PHYSICS (9646)
Motion in a Circle
Introduction
There are many types of motion in the world: Linear motion, projectile motion, circular motion and
simple harmonic motion, to name a few, that we would study. Besides travelling in a straight line
as studied in Kinematics earlier, we are interested in solving problems that involved motion in a
curved path. For example, a car going around a bend, a satellite orbiting Earth, an electron
orbiting a nucleus, a roller-coaster ride. In this chapter, we will study the motion of a body moving
round a circular path with a uniform (i.e. constant) speed which is described as a uniform
circular motion.
a)
Express angular displacement in radians
The diagram shows how angle θ is defined:
θ=
arc length
s
=
radius of arc r
where s is the curved distance along the arc of radius r
Æ θ is measured in radians.
Hence one radian (1 rad) is the angle when the arc length is the same as the radius of the arc.
The radian is the angle subtended by an arc length equal to the radius of the circle.
For one complete circle: arc length, s = circumference of circle = 2πr
Æ For one revolution, θ =
arc length
circumference of circle 2πr
=
=
= 2π rad
radius of arc
radius
r
To describe the angle turned during motion in a circle, we use the term angular displacement.
Angular displacement is the angle turned about the centre of the circle.
If a particle moves from A to B, the change in angular
displacement in this case is Δθ = (θ2 - θ1) rad.
Angular displacement is measured in radians.
Example 1
The laser on a CD player is 5.0 cm from the central hole of the disc. What length of the disc is
scanned by the laser when the disc turns through an angle of 0.45 radians?
Solution:
θ=
arc length
s
=
radius of arc r
© Physics Dept
Æ Length scanned, s = θ r = (0.45)(5) = 2.3 cm
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JJ 2014 H2 PHYSICS (9646)
b)
Motion in a Circle
Understand and use the concept of angular velocity to solve problems
If something is spinning, it is not possible to state a unique velocity with which it is
travelling, as different parts of it are travelling with different velocities (due to its changing
directions). To overcome this problem, the term ‘angular velocity’ is used.
Angular velocity ω is the rate of change of angular displacement.
Symbol
:
ω (small letter ‘omega’)
SI Unit of angular velocity : radians per second (rad s-1)
For constant angular velocity, ω =
θ
=
t
2π
T
The period T of circular motion is the time taken for one revolution.
Since 2π radians is turned in one revolution :
time for one revolution =
or
c)
T=
angle turned in one revolution
angular velocity
2π
ω
Recall and use v = rω to solve problems
Let a particle travel around a circle with uniform angular velocity, ω.
Let the length of the arc travelled be s.
Since s = rθ
Linear velocity, v =
dθ
ds d (rθ )
=
=r
= rω
dt
dt
dt
v = rω
Example 2
The minute hand of a clock is moving at constant angular velocity ω.
Which point on the hand, A, B or C is moving the fastest?
Solution:
As all points A, B and C lie on the same minute hand, they must all be
moving with the same angular velocity ω.
From v = rω, points further from the centre (larger r) will have greater
speeds.
Hence, point C is moving the fastest.
© Physics Dept
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JJ 2014 H2 PHYSICS (9646)
Motion in a Circle
Example 3
A train is travelling on a track which is part of a circle of radius 600 m
at a constant speed of 50 m s-1. What is its angular velocity?
Solution:
v = rω Æ ω =
v
50
=
= 0.083 rad s-1
r 600
Example 4
What is the angular velocity of the Earth about its axis?
Solution:
Angular velocity of the Earth’s rotation about its axis
T=
2π
ω
Æ ω=
2π
2π
=
= 0.26 rad hr-1
T
24 hr
However, we typically state ω in rad s-1.
2π
2π
=
Hence ω =
= 7.3 x 10-5 rad s-1
T
24 × 3600 s
d)
Describe qualitatively, motion in a curved path due to a perpendicular force,
and understand the centripetal acceleration in the case of uniform motion in
a circle
Uniform circular motion is the motion of an object moving in circular path at constant
speed.
According to Newton’s 1st Law, every object continues in its state of rest or uniform
motion in a straight line unless a resultant external force acts on it to change that state.
If an object is travelling with constant angular velocity in a circle, the direction of
motion is continually changing (i.e. not travelling in a straight line). This implies that :
•
the velocity is changing but the speed is constant.
•
there must be a resultant force acting on the object, but directed towards the
centre of the circle.
© Physics Dept
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JJ 2014 H2 PHYSICS (9646)
Motion in a Circle
Analysis of Uniform Circular Motion
Consider an object moving in uniform circular motion at
constant angular velocity, ω, from A to B. If no resultant
force acts on the object, by Newton’s 1st law, the object
will travel from A to C at constant velocity, v.
C
v
B
v
F
However, since the object travels along the curved path
from A to B, this implies that a resultant force F must
have acted on the object to change its velocity.
O
Instead of arriving at C, the object arrives at B. This
means that the object must have been “pulled” towards
the centre of the circle "by this resultant force" F.
A
F
ω
Points to note:
1. Since the object is moving at constant speed, this means that there is no component of
acceleration in the direction of motion, implying there is no component of resultant force
acting along the tangential path of the object.
[Conversely, if the speed is changing (not constant), there will be a component of resultant force acting
tangentially to the circle. This component causes a change in speed.]
2. However, since the velocity keeps changing (i.e. only the direction of motion is changing),
this must imply that there must be a resultant force acting on the object. The only direction
that this resultant force can act is towards the centre of the circle. This resultant force is
called centripetal force.
Centripetal force is the resultant force acting on an object in uniform circular motion
and is directed towards the centre of the circle.
3. Since the object experiences a resultant force towards the centre of the circle, by Newton’s
2nd law, there must be an acceleration towards the centre of the circle. This acceleration is
called centripetal acceleration, which is a vector with units m s-2.
In uniform circular motion, the speed of the object does not change, but the direction is
changing (i.e. velocity is changing). Hence, the centripetal acceleration does not
change the magnitude of the velocity, but only changes its direction.
4. The centripetal force is NOT an additional force acting on the object. It is the resultant of
all the forces acting on the object and acts towards the centre of the circle. It can be a
combination of forces like tension, friction, gravitational force, electric force or nuclear force. It
is called centripetal force because of what it is doing - moving object in a circular path. It
does not appear in any free body diagrams as a fundamental force, rather it is a resultant
force.
5. The centripetal force is always perpendicular to the direction of motion. Hence, this
force does not do any work on the object.
© Physics Dept
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JJ 2014 H2 PHYSICS (9646)
Motion in a Circle
As there is no displacement in the direction of the resultant force (resultant force is always
perpendicular to the velocity), there is no work done on the object moving in uniform circular
motion. Hence there is no change in the kinetic energy of the object.
Example 5 [N93/I/6]
Which of the following statements is correct for a particle moving in a horizontal circle with
constant angular velocity?
A
the linear momentum is constant but the kinetic energy varies
B
the kinetic energy is constant but the linear momentum varies
C
both kinetic energy and linear momentum are constant
D
neither the linear momentum nor the kinetic energy is constant
E
the speed and the linear velocity are both constant.
Solution:
Ans
A
B
C
e)
Remarks about statement
Incorrect.
Linear momentum is a vector and since the direction is changing, the linear momentum varies
(i.e. it is NOT constant).
Kinetic energy is a scalar. As the object is moving at constant angular velocity, the kinetic energy
is constant.
Correct statement.
Incorrect.
KE is constant, but linear momentum (vector) is always changing due to the changing direction of
velocity.
D
Incorrect. KE is constant.
E
Incorrect.
Speed is constant, but velocity (vector) is changing.
Recall and use centripetal acceleration a = rω2, a =
v2
to solve problems.
r
Centripetal acceleration is given by [the proof of this equation is not necessary].
v2
= rω 2
Centripetal acceleration, a =
r
1.
The direction of the centripetal acceleration at any instant is in a straight line
towards the centre of the circle (circular path).
2.
For circular motions at constant speed, the smaller the curvature (radius r) &
larger the speed v, the larger the centripetal acceleration, a.
© Physics Dept
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JJ 2014 H2 PHYSICS (9646)
Motion in a Circle
Example 6
The maximum speed of the blades on rotary lawn-mowers is restricted to avoid danger from
flying stones. If the rate of rotation of the blades in one particular model is 3500 revolutions per
minute and the blade has a radius of 0.23 m, find
a)
the angular velocity of the blade.
b)
the linear velocity of the tip of the blade.
c)
the centripetal acceleration of the tip of the blade.
Solution:
a) Blade moves 3500 revolutions per minute.
Hence, time for ONE revolution =
b) v = rω
v2
c) a =
r
f)
Recall and use centripetal force F = mrω2, F = m
or
a = rω2
v2
to solve problems.
r
From Newton’s 2nd Law, if the mass remains constant, then resultant force, F = ma.
v2
= rω 2
In circular motion, centripetal acceleration, a =
r
Æ Hence, resultant force towards the centre of circle (centripetal force) is given by:
mv 2
= mrω 2
Centripetal force, F =
r
Procedure to solve circular motion questions:
Step 1:
Identify the object and the centre of the circular motion
Step 2:
Draw the free body diagram of the object in circular motion
Step 3:
Resolve all forces and find the resultant force pointing towards the
centre of the circular motion
Step 4:
equate this resultant force provides the centripetal force.
© Physics Dept
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JJ 2014 H2 PHYSICS (9646)
Motion in a Circle
Example 7
For each object, draw all the relevant forces, and write down the expression for the centripetal
force (free body diagrams).
a) of mass m tied to a string, describing a circular path in a vertical plane
o
v
v
T
v
mg
Object at bottom of
circle
circle
Object at
at side
side of
of
Object
circle
circle
Objectat
attop
topof
of
Object
circle
circle
Resultant force to the centre,
F = T - mg
mv 2
∴ T – mg =
r
Note:
There are two types of circular motion possible in this vertical circle - uniform speed and non-uniform speed.
The speeds of objects in a vertical plane are generally not constant (and difficult to maintain at constant speed),
hence is not uniform circular motion. This implies the centripetal acceleration cannot be the same magnitude
throughout the motion.
b) A stunt driver driving his car inside a vertical circular track.
vv
r
r
rr
vv
© Physics Dept
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JJ 2014 H2 PHYSICS (9646)
c)
Motion in a Circle
A conical pendulum (mass m) describing a circle in a
horizontal plane.
Notice that the horizontal component of the tension will now provide the centripetal force
while the vertical component balances the weight of the pendulum.
Their vector sum is not zero since there must be a resultant force as the centripetal force.
Solution:
Forces towards the centre of circle:
mv 2
Tsinθ =
r
Forces in the vertical:
Tcosθ = mg
Example 8
The diagram represents a cyclist making a left turn on a rough surface at
constant speed, v, as viewed from behind. The total mass of the bicycle and
rider is m and their combined centre of gravity is at G.
If R is the resultant force of the normal reaction and the frictional force,
which vector diagram represents the directions of the forces acting on the
bicycle and its rider?
Example 9
A satellite of mass 800 kg is orbiting the Moon in a circular path of radius 1760 km. The pull of
the moon on the satellite is 1300 N. Find the speed of the satellite and its period of rotation.
Solution:
In this example, the only force acting on the satellite is the force by the moon on the satellite,
which is already directed towards the centre of the circle (i.e. moon).
F = 1300 =
(
)
1300r
1300 1760 × 10 3
mv 2
=
= 1690 m s-1
Æv =
800
m
r
Period of rotation, T =
© Physics Dept
2π
ω
=
2π
2π
=
⎛ v ⎞ ⎛ 1690
⎜ ⎟ ⎜
3
⎝ r ⎠ ⎝ 1760 × 10
⎞
⎟
⎠
= 6540 s
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JJ 2014 H2 PHYSICS (9646)
Motion in a Circle
Banking of roads
The part of the road that is curved is usually tilted at an angle (i.e.
banked). Banking is beneficial to the safety of a car moving over a
curved road. This is because on a banked curve, the normal force
exerted by the road on the car contributes to the centripetal force
required for the car to turn. If the tilt angle is just right, the normal
force will be able to provide the entire centripetal force. This will allow
the car to negotiate the curve even if there is no friction between its
tyres and the road.
Example 10
If a roadway is banked at the proper angle, a car can round a corner without any assistance from
friction between the tyres and the road. Find the appropriate banking angle for a 900 kg car
travelling at 20.5 m s-1 in a turn of radius 85.0 m
Solution:
Resolving forces vertically:
Resolving forces horizontally:
tan θ =
Example 11 (Airplane taking a bank)
An aircraft is travelling at a constant speed of 180 m s-1 in a horizontal circle of radius 20 km.
(a) Find the centripetal acceleration of the aircraft.
(b) Draw the free body diagram of the aircraft.
O
θ
(c) Calculate the angle of the bank, θ, of the aircraft.
Solution:
(a)
v2
Centripetal acceleration, a =
r
180 2
20x10 3
= 1.6 m s-2
=
© Physics Dept
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JJ 2014 H2 PHYSICS (9646)
Motion in a Circle
(b)
L
O
θ
mg
(c)
Since the plane is in equilibrium in the vertical direction,
Resultant force (vertically), Fy = may = 0
L cos θ = mg -------------------------------(1)
Since plane is performing circular motion, there must be a resultant force towards the
centre of the circle (i.e. centripetal force),
Centripetal force, F =
Lsinθ =
Equations (2) / (1):
tanθ =
mv 2
r
mv 2
----------------------------(2)
r
v2
gr
⎛v2
Æ θ = tan −1 ⎜⎜
⎝ gr
⎛
⎞
180 2
⎟ = tan −1 ⎜
⎜ (9.81) 20 × 10 3
⎟
⎝
⎠
(
)
⎞
⎟ = 9.4°
⎟
⎠
Example 12 (The rotating spaceman)
A spaceman in training is rotated in a seat at the end of a horizontal rotating arm of length 5 m. If
he can withstand accelerations up to 9g, what is the maximum number of revolutions per second
permissible? Calculate the period of the rotating arm in maintaining this no. of revolutions.
Solution:
Using a = rω2 Æ 9g = 5ω2 Æ ω =
9g
= 4.2 rad s-1
5
Maximum number of revolutions in 1 second =
Period of rotation, T =
© Physics Dept
2π
ω
=
ω
4.2
=
= 0.67
2π
2π
2π
= 1.5 s
4.2
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JJ 2014 H2 PHYSICS (9646)
Motion in a Circle
Things I have learnt today / Things I found interesting / Questions I still have
Reference Textbooks:
1.
2.
3.
4.
Comprehensive Physics for ‘A’ level (3rd edition), K.F. Chan, Charles Chew & S.H. Chan.
Federal Study Aids.
Physics (2nd edition), Robert Hutchings. Bath Advanced Science.
Calculations for A-level Physics (3rd edition), T.L. Lowe & J.F. Rounce. Stanley Thornes
(Publishers) Ltd.
Physics (2nd edition), James S. Walker
Extra Reading
What is the maximum G force a human can survive?
That depends on the duration thereof.
If subjected to constant acceleration, it would be doubtful if someone could withstand 9g for
more than a few minutes. After a few minutes at that rate, one faints from the lack of blood to
the brain, and presumably after that, some damage will occur, like heart failure, brain
hemorrhage and so on.
But for very short durations, very high accelerations can be supported, although some
damage can result. Colonel John Paul Stapp of the US Air Force did several experiments,
strapping himself to a rocket sled, and determined that 32g was an acceleration someone
could walk away from, which then became the acceleration used in the design of fighter jet
seat.
Stapp subjected himself to 45g and above, but since he did that so often, ended up with
damage to his eyes.
The British Formula One racer David Purley crashed in 1977, his car going from 173 km/h
(107 mph) to 0 in only 66 cm (two feet) (which means he hit a wall and the car structure
compressed to decelerate him). He broke many bones, but survived. This deceleration of
178g is believed to be the highest ever survived by a human being.
[Excerpt from http://answers.yahoo.com/question/index?qid=20070101085015AAok4JJ]
© Physics Dept
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JJ 2014 H2 PHYSICS (9646)
Motion in a Circle
Summary
1)
A body travelling at constant speed v in a circular path has acceleration towards the
centre of its path. This acceleration is called a centripetal acceleration.
2)
The centripetal force is the resultant force acting towards the centre of the circular
path. It is perpendicular to the direction of motion (tangential to the circular path), and
hence, no work (as defined) is done by the centripetal force.
3)
Procedure to solve circular motion questions:
Step 1:
Identify the object and the centre of the circular motion
Step 2:
Draw the free body diagram of the object in circular motion
Step 3:
Resolve all forces and find the resultant force pointing towards the
centre of the circular motion
Step 4:
This resultant force provides the centripetal force.
Examples of Applications:
1.
Washing Machine
In washing machine, the drums provide the necessary centripetal force (reaction of the
drum acting on the clothes) to keep the clothes in circular motion. In the drums there
are holes but they cannot provide the centripetal force. When the drum is rotated with
high speed, water particles acquire linear speed and the speed is high so that they move
in a spiral path. The radius of the circular path is increasing corresponding to the speed.
Thus, it is the lack of necessary centripetal force that causes the water particle to move
away from the centre through the holes along a tangential path and not along the radius.
© Physics Dept
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