Transcript notes5

Chapter 5: Dynamics of Uniform
Circular Motion
 An object moving on a circular path of radius r at a
constant speed v
 As motion is not on a straight line, the direction of the
velocity vector is not constant
 The motion is circular
 Compare to:
1D – straight line
2D – parabola

v
 Velocity vector is always tangent to
the circle
 Velocity direction constantly
changing, but magnitude remains
constant

v

r

v
 Vectors r and v are always perpendicular
 Since the velocity direction always changes, this
means that the velocity is not constant (though
speed is constant), therefore the object is
accelerating
 The acceleration ar points
radially inward. Like velocity its


direction changes, therefore
ar v
the acceleration is not constant
(though its magnitude is)
 Vectors ar and v are also perpendicular
 The speed does not change, since ar acceleration
has no component along the velocity direction
 Why is the acceleration direction radially inward?

ar


Since
v
1
v1
 

  v 2  v1
 v

2 ar a 
v2
t 2  t1
 This radial acceleration is called the centripetal
2
acceleration
v
ac 
r
 This acceleration implies a ``force’’
mv
Fc  mac 
r
2
The centripetal force (not a
fundamental force)
 The centripetal force is the net force required to
keep an object moving on a circular path
 Consider a motorized model airplane on a wire
which flies in a horizontal circle, if we neglect
gravity, there are only two forces, the force
provided by the airplane motor which tends to
cause the plane to travel in a straight line and the
tension force in the wire, which forces the plane to
travel in a circle – the tension is the centripetal
force


T
Consider forces in radial
motor direction (positive to center)
F
mv
 Fr  mar  T  r
2
 Time to complete a full orbit
D  2r  circumfere nce
circumfere nce 2r
T

 Period
speed
v
 The Period T is the time (in seconds) for the
object to make one complete orbit or cycle
 Find some useful relations for v and ar in terms
of T
2r
v
T
v
 2r  1 4 r
ar 

  2  ar
r  T  r
T
2
2
2
Example
A car travels around a curve which has a radius of
316 m. The curve is flat, not banked, and the
coefficient of static friction between the tires and
the road is 0.780. At what speed can the car
travel around the curve without skidding? y

v

r

FN
mg
mg
FN
fs
fs
r
F
y
 ma y
FN  mg  0
FN  mg
F
r
 mar
mv
fs 
r
2
 Now, the car will not skid as long as Fc is less than
the maximum static frictional force
2
2
mv
mv
max
 f s   s FN 
  s mg
r
r
v   s gr  (0.780)(9.80 sm2 )(316 m)  49.1 ms
 3600 s  1 mi 
mi
 49.1 

110


hr
 1 hr  1609 m 
m
s
Example
To reduce skidding, use a banked curve. Consider
same conditions as previous example, but for a
curve banked at the angle 
y
FN
mg
F
y

fs
r
 may
FN cos  f s sin   mg  0
 F
N

r
fs
mg Choose this
coordinate system
since ar is radial
Since acceleration
is radial only
 Since we want to know at what velocity the car
will skid, this corresponds to the centripetal force
being equal to the maximum static frictional force
fs  f
max
s
  s FN
Substitute into previous
equation
FN cos    s FN sin   mg
mg
FN (cos    s sin  )  mg  FN 
cos    s sin 
 Fr  mar
2
mv
FN sin   f s cos  
r 2
Substitute for FN
mv
and solve for v
FN (sin    s cos  ) 
r


mg
mv 2

(sin    s cos  ) 
r
 cos    s sin  
(sin    s cos  )
v  rg
(cos    s sin  )
 Adopt the same conditions as used in example 8
of the textbook: r = 316 m and  = 31°, and
s=0.780 from earlier
v  89.7 ms  200 mihr
 Compare to example 8 where s=0
(sin  )
v  rg
 rg tan   43.1 ms  96.5 mihr
(cos  )