Download Centripetal Force powerpoint

3. Centripetal Forces
Rotational Motion and
Astrophysics
Advanced Higher
Recap
Angular
Equations of
Motion
   0  t
1 2
   0 t  t
2
Motion in a
Circle
s  r
v  r
d

dt
at  r
d d 

 2
dt
dt
2
  0  2
2
2
Central (Radial) Acceleration
Consider a particle undergoing circular motion
The particle travels from A to B in
time Δt and with speed v. The change
in speed is given as Δv
arcAB 2r
t 

v
v
Average Acceleration:
v 2v sin  2v sin  v 2 sin 
aav 



2 r
t
t
r
v
When Θ is small and is
measured in radians this
means that
sin Θ = Θ
v
v
a
a
r
r
2
2
Since v  r
r
2
a
 a  r
r
2
2
Centripetal Force
The centripetal force is the one felt when
you move round in a circle.
This is due to you accelerating into the
centre of a circle.
You feel like you are being pulled
outwards…but you are in fact still moving
in a straight line.
Magnitude of the Force
The same rule still applies for working out
the force being applied to an accelerating
object
F  ma
2
and
v
2
a
 r
r
Therefore
2
mv
2
F
 mr
r
Car on a banked track
Same rules of “box on a slope” still apply
R is the “normal reaction
force of the slope, and Fr
is the frictional force of
the slope. We must find
the horizontal and vertical
component of each force.
For this example we
assume that friction is
reduced to zero.
Vertically (Equation 1)
R cos   mg
This motion has zero
acceleration
Horizontally (Equation 2)
mv 2
R sin  
r
This motion has constant
central acceleration
Divide Equation 2 by
Equation 1 to get:
2
v
tan  
gr
Vertical Circular Motion
Similar ideas have to be considered when
discussing vertical, but circular motion
Highest
Point
mv
Ttop  mg 
r
Lowest Point
mv 2
Tbottom  mg 
r
2
2
Ttop
mv

 mg
r
Tbottom
Thoriz
mv

r
mv 2

 mg
r
2
As there is no vertical component of
tension at this point
Example
1. A piece of string has a breaking force
of 56N. This string is used to whirl a
mass of 150g in a horizontal circle.
(a)The 150g mass moves in a horizontal
circle of radius 1.2m. Calculate the
maximum angular velocity of the mass.
(b)The mass is now rotated at 85rpm.
Find the maximum possible radius of
the circle.
Solution
(a) F  mr 2
(b)
56  150 x10 1.2  
3
56

3
150 x10 1.2
  17.6 rads -1
2
85  2
 8.9 rads -1
85rpm 
60
F  mr 2
56  150 x10 3  r  8.9
56
 4.7m
r
3
150 x10  8.9  8.9
2