Download 12/9 Circular Motion

12/9 Circular Motion
Text: Chapter 5 Circular Motion
HW 12/9 “Rotating Drum”
HW 12/9 “Rotating Disk”
These two are for practice (will not be
collected) and they also review friction forces.
Course Evaluation today
The Pendulum
Stationary
Swinging
(moving in a circular arc)
TS,M
TS,M
?
WE,M
WE,M
If the Tension is
greater, then
acceleration must
point up.
What happens to the Tension?
Is there acceleration?
Find the acceleration
a = v/t as usual
Draw vi, vf, and v as usual
(Place v’s tail to tail and draw tip to tip.
In one second:
v
In one second:
v
In one second:
v
What is the acceleration?
In one second:
v = a if t = 1 second
a
What is the acceleration?
In one second:
What if the speed is
doubled?
v
In one second:
a
In one second:
4 times the acceleration
for twice the velocity,
same radius.
For v
For 2v
What if the radius is halved?
In one second:
What if the radius is halved?
In one second:
v
What if the radius is halved?
In one second:
2 times the acceleration
for half the radius,
same velocity.
r
r/2
a
In one second:
4 times the acceleration
for twice the velocity,
same radius.
2 times the acceleration
for half the radius,
same velocity.
For v
a
For 2v
So:
2
v
ac = r
For r
For r/2
a
Centripetal Acceleration
perpendicular to the velocity, points
towards the center of the circle
ac = v2/r v is the instantaneous velocity
tangent to the path, r is the radius.
Some Equations and Definitions
The period, T, is the time for one revolution.
The distance for one revolution is 2r, the
circumference.
The speed, v, is distance ÷ time or 2r/T
Fnet = ma works for centripetal acceleration also and
free body diagrams are handled the same way.
The Pendulum
Stationary
Swinging
(moving in a circular arc)
TS,M
TS,M
?
WE,M
a and Fnet
both point up WE,M
If the Tension is
greater, then
acceleration must
point up.
a = v2/r and we can get v from
energy.
Example
A 1kg ball is swung in a horizontal circle at constant speed at the
end of a string. Draw a FBD. Which way does a point?
Fnet,y = Ty - WE,B = 0 so Ty = WE,B
Fnet,x = Tx = ma = mv2/r
y
TS,B
Ty
x Tx
WE,B
v = 2r/T
What about the centrifugal force?
There is no such force, regardless of what
Mr. Wizard says.
Bucket of water problem
You swing a bucket of water (m=3kg) in a vertical
circle at constant speed of 5 m/s. The radius of the
circle is 2 m. What is the normal force by the bottom
of the bucket on the water at:
a. the top of the circle, and
b. the bottom of the circle.
Bucket of water problem: at the
top
v
mw = 3 kg
r=2m
v = 5 m/s
(constant v)
Acceleration points
Draw a FBD of the
water.
Apply Newton’s
2nd law.
Fnet = WE,W + NB,W = ma
Want NB,W, find WE,W and ma
WE,W = mg = 3(9.8) = 29.4 N
ma = mv2/r = 3(52)/2 = 37.5 N
Fnet = 37.5 = 29.4 + NB,W
NB,W
NB,W = 8.1 N
WE,W
If you swing slow enough the
water will come out. How
slow do you have to swing?
Slow enough so that NB,W
becomes zero.
Bucket of water problem: at the
bottom
mw = 3 kg
r=2m
v = 5 m/s
(constant v)
Acceleration points
Draw a FBD of the
water.
Apply Newton’s
2nd law.
Fnet = NB,W - WE,W = ma
Want NB,W, find WE,W and ma
NB,W
Fnet = NB,W - 29.4 = 37.5
WE,W = mg = 3(9.8) = 29.4 N
ma = mv2/r = 3(52)/2 = 37.5 N
NB,W = 66.9 N
WE,W
v