# Applications of Newton's Laws Walker_ Chapter 6

Document Sample

```					  Chapter (6)
Circular Motion

1
Circular Motion
Consider an object moving at constant speed in a circle.
The direction of motion is changing, so the velocity is
changing (even though speed is constant).
Therefore, the object is accelerating.
The direction of the acceleration is toward the center of
the circle and so we call it centripetal acceleration. The
magnitude of the acceleration is
v2
ac 
r
2
Centripetal Acceleration
• The best estimate of the acceleration at P
is found by calculating the average
acceleration for the symmetric interval
12.


v1  (v cos ) x  (v sin  ) y
ˆ              ˆ

v2  (v cos ) x  (v sin  ) y
ˆ                ˆ
Distance travelled  2 r
3
Elapsed time t  d/v  2θr/v
Components of Acceleration

ax   
v cos   v cos   0
2θr/v

ay 
 v sin    v sin     v 2 sin 
2θr/v                 r 
v2
ay  
r          if     0

Then,

v2
ac 
r
4
Example: What is the centripetal acceleration of the Earth as it
moves in its orbit around the Sun?

Solution:
r (earth)  1.496 1011 m
2
ac                  t  1year
r
2r
But        
t

Then
4 r2
ac  2  5.93  10 3 m / s
t
5

Tangential acceleration
The tangential acceleration component
causes the change in the speed of the
particle. This component is parallel to the
instantaneous velocity, and is given by
Note: If the speed is
d                 constant then the tangential
at                       acceleration is zero
dt
(uniform Circular Motion)
The radial acceleration component arises from the change in
direction of the velocity vector and is given by

2
a r   ac                                          6
r
Total acceleration
The total acceleration vector a can be written as the vector
sum of the component vectors:
  
a  at  a r
a  a 2t  a 2 r

Since the component perpendicular to other

7
Example: A car exhibits a constant acceleration of 0.300 m/s2
parallel to the roadway. The car passes over a rise in the roadway
such that the top of the rise is shaped like a circle of radius 500 m.
At the moment the car is at the top of the rise, its velocity vector
is horizontal and has a magnitude of 6.00 m/s. What is the
direction of the total acceleration vector for the car at this instant?

8
2     36
ar            0.072 m / s 2
r    500

a  a t a     2   2
r    0.309m / s   2

If      the angle between

ar
  tan   1
 13 .5 
at

9
Problem: A train slows down as it rounds a sharp horizontal
turn, slowing from 90km/h to 50km/h in the 15s that it takes to
round the bend. The radius of the curve is 150m. Compute the
acceleration at the train.

Problem: A particle moves in a circular path 0.4m in radius
with constant speed. If the particle makes five revolution in
each second of its motion, find (a) the speed of the particle
and (b) its acceleration.

10
Centripetal Force
• A string cannot push
sideways or
lengthwise.
• A string in tension only
pulls.
• The string pulls the ball
inward toward the
center of the circle

11
What if we cut the sting?
The ball should move off with
constant velocity
This means the ball will
continue along the tangent to
the circle.

12
Centripetal Force
If there is a centripetal acceleration, then the net
force must also be a centripetal force:
v2
Fc  ma c  m
r

13
Example:

• The Conical Pendulum
• As the ball revolves
faster, the angle increases
• What’s the speed for a
given angle?

14
2
mv
T sin               .......... 1)
..........       .....(
r
T cos  mg..........       ........( )
..........       2
then
v2
tan  
rg
v  rg tan 
but (r  L sin  )
 Lg sin  tan 
15
Problem: I rotate a ball at an angle of 30o. What is the centripetal
acceleration? If the string is 1 meter long, how fast is it rotating?

16
Problem
Driving in your car with a constant speed of 12 m/s, you encounter
a bump in the road that has a circular cross section, as indicated in
the Figure. If the radius of curvature of the bump is 35 m, find the
apparent weight of a 67-kg person in your car as you pass over the
top of the bump.

a=v2/r

N
mg

17

```
DOCUMENT INFO
Categories:
Tags:
Stats:
 views: 3 posted: 12/5/2011 language: English pages: 17
How are you planning on using Docstoc?