Transcript Motion, Forces and Energy Lecture 5: Circles and Resistance

Motion, Forces and Energy
Lecture 5: Circles and Resistance
Fr
m
r
Fr
A particle moving with uniform speed v in a
circular path of radius r experiences an
acceleration which has magnitude a = v2/r
(see Serway Section 4.4 for derivation).
Acceleration is ALWAYS directed towards the
centre of the circle, perpendicular to the velocity
vector.
Fr
2
mv
Fr 
 mar
r
m
When string breaks, ball
moves along tangent.
NB The centripetal force is a familiar force acting in the role of a force that
causes a circular motion.
Tension as a centripetal force
Fr
r
Fr
m
If the ball has a mass of 0.5kg and the cord is 1.5m
long, we can find the maximum speed of the ball
before the cord breaks (for a given tension). We
assume the ball remains perfectly horizontal during
its motion:.
The TENSION in the cord PROVIDES the
CENTRIPETAL FORCE:
mv2
T
r
so
Tr
v
m
For a tension of say 100 N (breaking strain), the
Maximum speed attainable is (100x1.5/0.5)1/2
= 17.3 ms-1.
Conical Pendulum
A conical pendulum consists of a “bob” revolving
in a horizontal plane.
The bob doesn’t accelerate in the vertical direction:
T cos  mg  may  0

T
(1)
The centripetal force is provided by the tension component:
mv2
T sin   mar 
r
L
(1) / (2) gives
v2
tan  
rg
(2)
so v  rg tan 
Finally we note that r = L sin so that:
mg
v  gL sin  tan 
A car moving round a circular curve
Fr
The centripetal force here is provided by
the frictional force between the car tyres
and the road surface:
mv
Fr  f s 
r
2
The maximum possible speed corresponds to the
maximum frictional force, fsmax = ms n.
Since here n = mg, fsmax = ms mg
vmax  m s gr
So for ms = 0.5 and r = 35 m, vmax = 13.1 ms-1.
Non-uniform circular motion
Sometimes an object may move in a circular path with varying speed. This
corresponds to the presence of two forces: a centripetal force and a tangential
force (the latter of which is responsible for the change in speed of the object as
a function of time.
An example of such motion occurs if we whirl a ball tied to the end of a string
around in a VERTICAL circle. Here, the tangential force arises from gravity
acting on the ball.
vtop
m
mg
Ttop
R
centre

T
Tbot
vbot
mgsin
mgcos
mg
mg
We must consider both
radial and tangential
forces!
mg sin   mat
at  g sin 
mv 2
T  mg cos  
R
 v2

T  m   g cos  
R

Analysis
Tangential acceleration
General equation
At the bottom of the path where  = 0o:
 vbottom2

Tbottom  m 
 g 
 R

At the bottom of the path where  = 180o:
Ttop
 vtop 2

m
 g
 R



Therefore, the maximum tension
occurs at the bottom of the circle,
where the cord is most likely to break.
Motion with resistive forces
We’ll take a quick look at fluid resistance (such as air resistance or
Resistance due to a liquid). Resistive forces can depend on speed
In a complex way, but here we will look at the simplest case:
R=bv
ie the resistance is proportional to the speed of the moving object.
R
v
mg
With fluid (air) resistance, the falling object does
not continue to accelerate, but reaches terminal
Velocity (see Terminal Velocity Analysis notes).