Download 7TH CLASSES PHYSICS DAILY PLAN

9TH CLASSES
PHYSICS
DAILY PLAN
SUBJECT: CIRCULAR MOTION
GOALS:
DURATION:
IN PRACTICE:
 a=V/t=V.V /r  a=
 a=(w.r)2/r  a= w2 .r

V2
r
The acceleration vector in this case perpendicular to the path
and always points toward the center of the circle.
a
PRESENTATION:
a
a
UNIFORM CIRCULAR MOTION :
The motion of an object that moves on a circle with constant
speed is called uniform circular motion.
V
a

An acceleration of this nature is called a centripetal
acceleration or radial acceleration.
Centripetal Force :
V

In uniform circular motion the object is under the action of a
force which is always directed toward the center of the
circle.

This force keeps the object moving in the circle with a
constant speed.
Period (T) : The time to complete one turn is called the period
of the motion.

These forces ara called centripetal forces. By appliying
Newton’s 2nd Law
Frequency (f) : The number of turns in one second is called the
frequency of the motion and it is equal to ;
f=1/T or T.f=1
PERIOD (T)
FREQUENCY (f)
SECOND (s)
1/s or s-1
Fc = m.a  Fc = m.
O
V

r
V
V
Linear Speed (V) : The distance travelled by the object during
the period (T) is S=2r then;
V= S/T  V=2r/T  V= 2fr
Angular Speed (w) : The angle swept by the radius line in unit
time is called the angular speed (w). If the angle swept is  in
time t, then ;
w=/t
We know that the radius sweeps 2 radians during one complete
turn. Thus ;
w=2/T  w=2f
V2
 Fc = m.w2.r
r
V
V
F
F
F
F
V
V
TYPICAL PROBLEMS :

An object moving around a horizantal circle ;
V
T
V=2r/T
w=2/T
r
V= w.r
Centripetal Accleration : There are two ways in which an
acceleration can be produced; by a change in the magnitude of
the velocity & by a change in the direction of the velocity.
V1
V1
A
r

O
B
A
B
r
r
V2
V1 D

O
r

L
Fc = m.a
V
V2 E
If  is small, the length of the arc AB will approximately
be equal to line AB. Thus;
AB = V.t  V/V=(V.V)/r
An object moving around a verticle circle ;
V
AOBDBE  V/V = AB /r

Ex: A 0,5kg mass is attached to a string 80cm long and it swings
in a horizantal circle with a speed of 4m/s.
a) Find the centripetal acceleration of the mass.
b) Find the tension in the string.
(a:20m/s , b: 10N )

V
m
Fc = m.a
V2
T= m.
r
T
T
M
T
W
W
T
W
K
N
W




V2
At K T+W.cos= m.
r
2
V
At L  T+W= m.
r
2
V
At M  T= m.
r
V2
At N  T-W= m.
r
Ex: A 0,2kg stone is attached to a stribg 50cm long. It swings in
a vertical circle at an angular frequency of f= 4rev/s
a) Find the centripetal acc.
b) Find the tension in the string when the stone is at the
highest point.
c) At the highest point, when it is 180cm above the ground the
string breakes and the stone flies off. Find the horizantal
distance that the stone travels before it strikes the ground.
(=3,g=10m/s2)
 A car turning an unbanked curved road ;
V2

r
V2
.m.g  m.

r
N
Ff  m.
W
V  .gr
Ff
r
Ex: A 600kg car travels around a circular path of radius r=100m
at uniform speed in 20s.
a) What must be the minimum coefficient of friction between
the tires and the road for the car to remain on the circular
path?
b) Find the centripetal force acting on the car? (=3)

A car turning a banked curved road ;
1.
If there is no friction
y
N

x
r
 Ny
N
x
W

W
V
V2
Nx = m.
 N.sin = m.
r
r
2
V= r.g.tan
 N.cos = m.g
Ny = m.g
2.
N
If there is a friction ;
N
N 
x Ff
 Ny
inclined
plane
Nx + Ffx = m.
V2
r
N.sin+ Ff.cos= m.
W
V2
r
Ex: A 1000kg car rounds a curved banked road of radius 75m at
a speed 15m/s. Find the minimum angle of banking for the car
to remain on the road.
( Assume that there is no friction and =3 )
HOMEWORK:
MULTIMEDIA:
DEMONSTRATION:
EXPERIMENT:
TEACHER:
DIRECTOR:
V
