P.1
PHYSICS
CIRCULAR MOTION
1. Motion in a circle
- a body which travels equal distances in equal times along a circular path has constant
speed but not constant velocity.
- since the direction of the velocity is changed from time to time, the body has an
acceleration.
2. Angles in radians θ
Consider a body moves in a circular path with centre 0.
s = r θ
s = distance travelled
r = radius of the circle
 = angular displacement
3. Angular velocity w
- it is the angle swept out in unit time by the radius joining the body to
the centre of the circle. (Unit : rad s-1 )
- Consider a body moving uniformly from A to B in time t
w
=
=
=
v =

t
s rt
v r
( v = constant speed of the body )
rw
T
=
2 r / v
T
=
2
w
(T = period of the circular motion )
4. Centripetal acceleration in a circular motion
- consider a body moving with constant speed v in a circle of radius r and centre  .
travels from A to B in a short interval of time t .
It
magnitude of v A and v B = v
-
If t is very small,  will also be small and CD will have almost the same length as
arc CD.
v  v 
v

a 
 v
 vw
t
t
-
a = v2 /r = rw2
The direction of the acceleration is towards centre O of the circle as can be seen if t is
made so small that A and B all but coincide; vector CD is then perpendicular to VA (or
V
B)
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i.e. along AO (or BO).
-
We say the body has centripetal acceleration.
This centripetal acceleration is not constant since its direction is varied from time to
time.
5. Centripetal force
-
We know from Newton’s second law that whenever a body is accelerating, there must
be a resultant force. If a body of mass m follows a circular path of radius r with speed
v, its acceleration is v2/r and it must be subject to a centripetal force.
mv 2
 mrw 2
r
The acceleration is directed towards the centre of circle so the force, too, must act in
F
-
this direction. For this reasons the force is called a centripetal force.
e.g.1 A ball of mass m attached to a string is swung round in a HORIZONTAL CIRCLE
with radius r and uniform speed v.
T
-
mv
r
tension of the string T provides the necessary centripetal force so that the ball can be
maintained in a circular path.
e.g.2 A mass is swung in a VERTICAL CIRCLE with uniform speed v and radius r.
i. . 0°<θ<90°
(II)
T  mg cos 
mv 2
r
T  mg cos 
mv 2
r
90°<θ<180°
T  mg cos(180   ) 
mv 2
r
mv 2
T  mg cos  
r
2
mv
T
 mg cos 
r
T  mg cos  
** N.B. (1) For 90° < θ < 180°
T = 0 when - mg cos  
i.e. v2 = - gr cosθ
mv 2
r
mv 2
r
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(2) maximum tension T  mg 
minimum tension T 
e.g.3
mv
,
r
mv
 mg ,
r
when θ = 0°
when θ = 180°
i.e. at B
i.e. at A
Overturning of a car
R1 + R2 = mg ………..(1)
take moment about c.g.,
(R2 – R1) a = f h ………..(2)
from (1), (2)
R2 =
1
fh
( mg  )
2
a
R1 =
1
fh
( mg  )
2
a
** Overturning just begin when R1 = 0,
R1 = 0 i.e. mg 
since
fh
…………………(3)
a
mv 2
 f ………………(4)
r
sub. (4) into (3),
mv 2 h
r a
gar
v2 
(condition for overturning just begin)
h
mg 
Hence, the general condition for overturning is
v2 
Question:
gar
h
How can we design a car so that it can have a large velocity but without
overturning?
P.4
6. Other applications with centripetal forces
a) The conical pendulum – The figure below shows a conical pendulum in which a mass
m, supported by a string of length  , follows a horizontal
circular path of radius r with constant speed v.
T cos θ = mg………(1)
T sin θ =
( 2)
(1)
mv 2
……….(2)
r
v2
tan  
rg
** N.B. (I)
the horizontal component of the tension in string provides the required
centripetal force.
(II) tan θ = v2 / rg i.e. at a fixed angle θ, v2 is directly proprotional to the
radius of the circle r.
b) Rounding a bend
(i) without banking
- A car is travelling round a circular bend with uniform speed on a horizontal
road.
- Air resistance is always against the instantaneous motion of the car. The
other and more important horizontal force is the frictional force exerted
inwards by the ground on the tyres of the car.
-
The centripetal force, this case is due to the resultant interaction of the car
with air and the ground.
*N.B. (i) the circular motion can be maintained only if the frictional force can
provided the required centripetal force - otherwise skidding occurs,
radius of the circular motion will then to increased.
(ii) safe cornering that does not role on friction is achieved by ‘banking’ the
road.
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(ii)
With banking
a car is moving round a banked track in a circular path of horizontal radius r.
If the only forces at the wheels are the normal reaction N (no frictional
force are introduced), then
N cos  mg....................(1)
N sin  
mv 2
....................( 2)
r
v2
tan  
gr
**N.B. (i) the centripetal force is only due to the horizontal component of the normal
reaction on the wheels by the road.
(ii) In this case, a car can have speed v without skidding if the banked angle of
the road is θ
s.t. tanθ = v2/rg.
( 2)
(1)
Question: What happens when (1) tanθ > v2/rg , and
(2) tanθ < v2/rg ?
c. Aircraft turning in flight
- An aricraft in straight, level flight experiences a lifting force at right angles to the
surface of its wings which balances its weight.
-
To turn, the aircraft is banked and
(i) the horizontal component of the lifting force supplies the necessary centripetal
force.
(ii) Now only the vertical component of the lifting force support the aircraft’s weight
and height will be lost unless the lifting force if increased by, for example,
increasing the speed.
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d. Looping the loop
-
A bucket of water can be swung round in a vertical circle without spilling.
(i) at position A,
mv 2
 mg  N
r
** if mg is greater than
Hence,
mg 
mv 2
at A, the water will not stays in the bucket.
r
mv 2
is required in this case.
r
(ii) at position B,
mv 2
N
r
(iii) at position C,
mv 2
 N  mg
r
(iv) at position D,
mv 2
N
r
Question:
For a fixed speed v and fixed radius r,
Which position will have (1) greatest N, and
(2) smallest N
e. Centrifuges
(i) Introduction – they are used to separate solids suspended in liquids of different
densities.
- when the mixture in a tube is rotated at high speed in a horizontal
circle, the loss dense matter moves towards the centre of rotation.
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(ii) Explanation – P1>P2 its that its difference can provide the necessary centripetal
force acting radially inwards.
- Thus PB>PA and there is a pressure gradient along the tube. And
hence any part of the liquid experiences an inward force.
Consider two particles X,Y in the liquid so that Mx < My.
For a constant centripetal force and rotating speed.
we have
M x v2
F
........................(1)
rx
F
M yv2
ry
........................(2)
since (1) = (2),
-
Mx My

(i.e. M  r  close to the centre of rotation)
rx
ry
Hence, suspended matter of smaller density will move inwards because
such inward force is too large to it comparatively.
f. Rotor
(i)
Introduction – it consists of an upright drum. The drum is spun at
increasing speed about its central vertical axis and at a
certain speed the floor is pulled downwards.
The
occupants do not fall but remain ‘pinned’ against the wall of
the rotor.
N = normal reaction on the occupant
by the wall
f = frictional force acting on the
occupant by the wall
(ii) Explanation –
balance his
if the frictional force acting on the man by the will can
weight, then he will not fall down.
mv 2
...............(1)
r
f  mg
N
N  mg...............(2)
( 2)
(1)

gr
v2
**N.B. (1) sufficient frictional force can be obtained if
  gr v
2
i.e. min   gr
v
2.
(2) the man’s weight has no effect to his result.
P.8
g. Motion of bicycle rider round circular track
(i) When a person on a bicycle rides round a circular racing racing track with radius r.,
the frictional force (provides the centripetal force) f has a moment about the c.g.
which tends to turn the rider outwards.
(ii) When the rider leans inwards, moment of f is counter-balanced by the moment of
N about c.g.
M = mass of bicycle and rider
f = frictional force acting on the wheel
by the ground
N = normal reaction on the wheel by
the ground
v = speed of the bicycle
take moment about c.g.,
f h = N a ………….(1)
mv 2
..................(2)
r
N  mg..................(3)
f
resolve horizontally,
resolve vertically,
mv 2
h  mg a
r
a v2

h rg
sub. (2), (3) into (1),
tan  
**N.B. (i)
v2
rg
θ = angle of inclination to the vertical
at a circular track of fixed r,
tan   v 2
(ii) to avoid skidding,
mv 2
r
mv 2
mg 
r
2
v

gr
f
  tan 
condition to avoid skidding.