From last time:
For motion at speed v in a circular path of radius r:
v2
Centripetal acceleration, ac =
r
7
5.6
Lettuce drier: spin a container containing the lettuce, water
forced out through holes in the sides of the container.
Radius = 12 cm, rotated at 2 revolutions/second. What is
the centripetal acceleration of the wall of the container?
v2
Centripetal acceleration, ac =
r
v = 2 × 2pr m/s = 1.51 m/s
1.512
ac =
= 18.9 m/s2 = 1.9g
0.12
8
Centripetal Force
• the force that causes the centripetal acceleration
• acts toward the centre of the circular path – in the direction of
the acceleration
• generated by tension in a string, gravity (planetary motion),
friction (driving around a curve)...
As F = ma, centripetal force is:
mv2
Fc = mac =
r
9
A 0.9 kg model airplane
moves at constant speed
in a circle parallel to the
ground.
Find the tension in the
guideline if r = 17 m and
v = 19 m/s and 38 m/s.
r = 17 m
T
v
Speed = 19 m/s,
mv2 0.9 × 192
=
= 19.1 N
T = Fc =
r
17
Speed = 38 m/s,
mv2 0.9 × 382
=
= 76.4 N
T = Fc =
r
17
10
A 0.6 kg and a 1.2 kg airplane fly
at the same speed using the same
type of guideline.
The smallest circle the 0.6 kg
plane can fly in without the line
breaking is 3.5 m
T
How small a circle can the 1.2 kg plane
fly in?
mv2
Tension in the line is T =
r
2
0.6v
0.6 kg plane: T =
(3.5 m)
1.2v2
1.2 kg plane: T =
r
0.6v2 1.2v2
→ r = 3.5 × 1.2/0.6 = 7 m
If the tensions are equal:
=
3.5
r
11
How fast can you go around a curve?
µs = 0.9, r = 50 m
FN = mg
FN
r
v
Fs
mg
mv2
Centripetal force =
r
Provided by static friction force,
Fs = µsFN = µsmg
√
mv2
√
So,
= µsmg → v = µsrg = 0.9 × 50 × 9.8 = 21 m/s (76 km/h)
r
On ice µs = 0.1 → v = 7 m/s (25 km/h)
12
Flying around in circles
Lift: L/2 + L/2 = mg
mg
Plane banking to turn in a
horizontal circular path of
radius r:
mv2
L sin q =
r
mg
L cos q = mg
v2
tan q =
rg
13
Driving around in circles – banked road
2
mv
FN sin q =
r
FN cos q = mg
(in absence of friction)
mv2
1
v2
→ tan q =
×
=
r
mg rg
→ best angle of banking
14
Driving around in circles – banked curve
If you drive slower, you slide
down the slope.
If you drive faster, you skid up
the slope.
If q = 31◦ and r = 316 m,
and there is no friction, what is
the best speed to drive around
the banked curve?
!
v2
tan q = , so v = rg tan q
rg
√
v = 316 × 9.8 tan 31◦ = 43.1 m/s = 155 km/h
15
5.18
A car travels at 28 m/s around a curve of radius 150 m. A
mass is suspended from a string.
Roof of car
v
v2
ac =
r
q
T
v2
ac =
r
What is the angle q ?
mg
Force toward centre of circular path due to tension in the string:
mv2
= T sin q
r
Forces in the vertical direction: mg = T cos q
16
mv2
= T sin q
r
mg = T cos q
T sin q mv2
1
So,
=
×
T cos q
r
mg
v2
282
tan q =
=
= 0.5333
rg
150g
q = 28.1◦
17
Orbiting the Earth
“The secret to flying is to throw yourself at the earth and miss”
Hitch Hiker’s Guide to the Galaxy
The centripetal force on the satellite
is provided by the gravitational force
from the earth.
mv2 GME m
Fc =
=
r
r2
!
GME
the smaller the radius,
So v =
the greater the speed
r
v
ME
r
Fc
m
Earth
Synchronous orbit: period = 24 hours
– satellite stays above same part of the earth (at the equator)
– communication satellites
18
Synchronous Orbit – what is its radius?
circumference of orbit
The period of an orbit is: T =
speed of satellite
From previous page, v =
So T = 2pr ×
r3 = GME
!
!
!
GME
r
r
= 2p × r3/2
GME
T
2p
"2
2pr
=
v
!
1
GME
(Kepler’s 3rd law of planetary motion:
T 2 µ r3)
19