Transcript Motion in a Plane - Dallas School District

Motion in a Plane
Chapter 8
Centripetal Acceleration
• Centripetal Acceleration – acceleration
that points towards the center of a circle.
– Also called Radial Acceleration (aR)
v
Ball rolling in a
straight line (inertia)
v
aR
aR
Same ball, hooked to
a string
v
aR = v2
r
If you are on a carousel at constant speed, are
you experiencing acceleration?
If you twirl a yo-yo and let go of the string,
what way will it fly?
Period and Frequency
Period (T)
– Time required for one complete (360o) revolution
– Measured in seconds
Frequency
– Number of revolutions per second
– Measured in rev/s or Hertz (Hz)
T=1
f
Formulas
v = 2pr
T
v = wr
aR = v2
r
a = w2r
A 150-kg ball is twirled at the end of a 0.600
m string. It makes 2.00 revolutions per
second. Find the period, velocity, and
acceleration.
(0.500 s, 7.54 m/s, 94.8 m/s2)
The moon has a radius with the earth of about
384,000 km and a period of 27.3 days.
A. Calculate the acceleration of the moon
toward the earth. (2.72 X 10-3 m/s2)
B. Calculate the previous answer in “g’s” (2.78 X
10-4 g)
Centripetal Force – the “center seeking” force
that pulls an object in a circular path.
– Yo-yo
– Planets
– Merry-go-round
– Car rounding a curve
Centrifugal Force
A word about Centrifugal
Force
• Doesn’t really exist.
• “apparent outward force”
• Water in swinging cup
example
Direction water
wants to go
Centripetal
Force of
string
Centripetal Motion
SF = maR = mv2
r
A 0.150 kg yo-yo is attached to a 0.600 m string
and twirled at 2 revolutions per minute.
a. Calculate the velocity in m/s ()
b. Calculate the centripetal force in the string
(14.2 N)
Thor’s Hammer (mjolnir)
has a mass of 10 kg
and the handle and
loop have a length of
50 cm. If he can swing
the hammer at a
speed of 3 m/s, what
force is exerted on
Thor’s hands?
(Ans: 180 N)
Can Thor swing his hammer so that it is
perfectly parallel to the ground?
FR
What angle will the hammer take with the
horizontal?
Let’s resolve the FR vector into it’s components:
FRx = FRsinq
q
FRy = FRcosq
SFy = 0 (the hammer is not rising or falling)
SFy = 0 = FRcosq – mg
mg
FRcosq = mg
cosq = mg/FR
q= 57o
How about if he swings faster?
A father places a 20.0 kg child on a 5.00 kg
wagon and twirls her in a circle with a 2.00 m
rope of tension 100 N. How many rpms does
the wagon make (w)? (14 rpm)
A 0.150 kg ball is swung on a 1.10-m string in a
vertical circle. What minimum speed must it
have at the top of the circle to keep moving in
a circle?
mg
FT
At the top of the circle, both the weight and
the tension in the string contribute to the
centripetal force
SF = FT + mg
SF = FT + mg
FR = FT + mg
mv2 = FT + mg
r
(tricky part: assume FT = 0, just as the cord
goes slack, but before the ball falls)
mv2 = mg
r
v2 = gr
v = 3.28 m/s
Note: this equation is also the minumum
velocity for orbit of a satellite
v = \/rg
What is the tension in the cord at the bottom
of the arc if the ball moves at twice the
minimum speed? (v = 6.56 m/s)
At the bottom of the circle, the weight opposes
the centripetal force.
FT
mg
SF = FT – mg
mv2 = FT - mg
r
FT = mv2 + mg
r
FT = 7.34 N
Car Rounding a Turn
• Friction provides the centripetal force
• Use the coefficient of static friction (ms). The
wheels are turning, not sliding, across the
surface
• Wheel lock = kinetic friction takes over. mk is
always less than ms, so the car is much more
likely to skid.
A 1000-kg car rounds a curve (r=50 m) at a speed of
14 m/s. Will the car skid if the road is dry and
ms=0.60?
Let’s first solve for the Normal Force
FN
Ffr = FR
FN = mg = (1000 kg)(9.8 m/s2)
FN = 9800 N
mg
SFx = Ffr
FR = Ffr
mv2 = msFN
r
(1000 kg)(14m/s)2 = (0.60)(9800 N)
(50 m)
3920 N < 5800 N
The car will make it. 3920 N are required,
and the frcition provides 5800 N.
Will the car make it if it is icy and the ms = 0.25
SFx = Ffr
FR = Ffr
mv2 = msFN
r
(1000 kg)(14m/s)2 = (0.25)(9800 N)
(50 m)
3920 N > 2450 N
The car will not make it. 3920 N are required, and the
friction only provides 2450 N.
What is the maximum speed a 1500 kg car can
take a flat curve with a radius of 50 m (ms =
0.80)
BANKED CURVES
• Banked to reduce the reliance on friction
• Part of the Normal Force now contributes
to the centripetal force
FR = Ffr + FNsinq
(ideally, we bank the
road so that no
friction is required: Ffr
= 0)
Banked Curves: Example 1
A 1000-kg car rounds a 50 m radius turn at 14
m/s. What angle should the road be
banked so that no friction is required?
FN
q
mg
FN = mgcosq
q
Now we will simply work with the Normal
Force to find the component that points to
the center of the circle
First consider the y forces.
SFy = FNcosq - mg
Since the car does not move up or
down:
SFy = 0
FNcosq
FN
q
0 = FNcosq – mg
FNsinq
FNcosq = mg
FN = mg/cosq
q
mg
q
mv2 = FNsinq
r
mv2 = mgsinq
r
cosq
v2 = gtanq
r
v2 = gtanq
r
v2 = tanq
gr
tan q =
q = 22o
(14 m/s)2
(50 m)(9.8m/s2)
=
0.40
Fred Flintstone places a 1.00 kg rock in a 1.00 m long
sling. The vine breaks at a tension of 200 N.
a. Calculate the angle below the horizontal plane that
the rock will take. (2.81o)
b. Calculate the maximum linear velocity the rock can
twirl. (14.1 m/s)
c. Calculate the angular velocity in rpm’s. (135 rpm)
Circular Orbits
• Orbits are freefall (not true weightlessness)
• Orbital velocity must match the weight
mg = mv2
r
g = v2
r
v = √ gr
A satellite wishes to orbits at a height of 200
miles above the earth’s surface.
a. Calculate the height above the center of the
earth if Rearth = 6.37 X 106 m. (6.69 X 106 m)
b. Calculate the orbital velocity. (8098 m/s)
c. Calculate the period in minutes. (86.5 min)
Review of Angular Kinematics
A motor spins a 2.0 kg block on an 80.0 cm arm at 200
rpm. The coefficient of kinetic friction is 0.60.
a. Draw a free body diagram of the block.
b. Calculate the tangential acceleration of the block
(due to friction). (-5.88 m/s2)
c. Calculate the angular acceleration. (-7.35 rad/s2)
d. Calculate the time until the block comes to a rest.
(2.8 s)
e. Calculate the number of revolutions. (4.7 rev)