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Think about the Circleโฆ
1 m
If you walked around this circle once, what is your total distance?
C = 2ฯr = 2ฯ(1 m)
C = 2ฯ meters
C = 6.28 meters
What is a Radian??
r
We can define a circular distance in terms of a generic radius, rโฆ
r
How many radians are there in one full revolution?
C = 2ฯr
2ฯ radians
1 rad
Try thisโฆ.
If a child on a merry-go-round rotates 3.5 times, what is their angular displacement in radians?
If an ant on a record player spins for an angular displacement of 14 radians, how many revolutions has it experienced?
2ฯ 2ฯ ฯ2ฯ3.5(2ฯ) = 7 ฯ
2ฯ = 22 rad
14
2๐= 2.23 ๐๐๐ฃ๐๐๐ข๐ก๐๐๐๐
Angular Velocity
For Linear Motion: ๐ฃ =๐๐๐ ๐ก๐๐๐๐
๐ก๐๐๐=๐
๐ก
๐ =๐๐๐๐ข๐๐๐ ๐๐๐ ๐ก๐๐๐๐
๐ก๐๐๐=
2๐
๐2ฯ rad
Time for one revolution = T
If you know the time it takes for one revolution you can calculate the angular velocity
Angular Velocity [rad s-1]
Try thisโฆ.
A ferris wheel takes 40 seconds to make on full revolution, what is its angular velocity in rad/s?
A car tire rotates with an average angular velocity of 29 rad/s. In what time interval will the tire rotate 3.5 times?
2ฯ rad
๐ =2๐
๐=2๐
40= 0.157 ๐๐๐ ๐ โ1
T
29 =2๐
๐๐ = 0.217 s
0.217 ร 3.5 = ๐. ๐๐๐ ๐ฌ
29 =3.5(2๐)
๐ก
๐ก = ๐. ๐๐๐ ๐ฌ
Time for one revolution
Linear Velocity
At any given point, an object with circular motion will also have an instantaneous linear velocity.
This velocity will be in the direction _________ to the curvetangent
Calculating Linear Velocity
๐ฃ =๐ฃ =๐
๐ก
Distance of one complete rotation(circumference)
Time for one complete rotation(Period)
2ฯr
T
Try thisโฆ.
If the carousel spins at 1 complete rotation every 10 seconds, what is the angular and linear velocity of each row?2 m
3 m
A
B
A๐ = 2๐
๐
๐ = 2๐10
= 0.63 ๐๐๐ ๐ โ1
๐ฃ = ๐๐
๐ฃ = (0.63)(2)= 1.3 ๐ ๐ โ1
B๐ = 2๐
๐
๐ = 2๐10
= 0.63 ๐๐๐ ๐ โ1
๐ฃ = ๐๐
๐ฃ = (0.63)(3)= 1.9 ๐ ๐ โ1
Time for 1 Rotation:
T = 10 s
Try thisโฆ.
If you were sitting 4 m from the center of a carousel spinning at 12 rad s-1 and threw a ball in the air, how fast would the ball continue in a straight line?
A woman passes through a revolving door with a tangential speed of 1.8 m s-1. If she is 0.8 m from the center of the door, what is the doorโs angular velocity?
r = 4 m
ฯ = 12 rad s-1v = ฯr = (12)(4) = 48 m s-1
v = 1.8 m s-1
r = 0.8 m v = ฯr
1.8 = ฯ(0.8)
ฯ = 2.25 rad s-1
Remember backโฆ
There are 3 ways that an object can be experiencing acceleration?
Speeding Up Slowing Down Changing Direction
Centripetal Acceleration
Centripetal acceleration represents the rate of change of velocity and its direction
๐ =๐ฃ2
๐
Centripetal Acceleration
Centripetal acceleration can be seen when finding the change between velocity vectors
v
u
v
-u
Centripetal acceleration will always
point to the a
a
center
Calculating Tangential Velocity
๐ =๐ฃ2
๐๐ฃ =
2๐๐
๐๐ฃ = ๐๐
๐ =
2๐๐๐
2
๐=
4๐2๐2
๐2
๐=4๐2๐2
๐2๐=๐๐ ๐๐
๐ป๐
Try thisโฆ.
If the carousel spins at 1 complete rotation every 10 seconds, what is the centripetal acceleration for each row?2 m
3 m
A
B
A
ฯ = 0.63 rad s-1 | v = 1.3 m s-1
B
ฯ = 0.63 rad s-1 | v = 1.9 m s-1
๐ =๐ฃ2
๐
Waitโฆ Whereโs the Force?
We know from Newtonโs 2nd Law that every time that we have acceleration, there must be a force causing that change in velocity
๐น = ๐๐
๐น =๐ =
๐ฃ2
๐
Try Thisโฆ
A 3 kg rock swings in a circle of radius 5 m. If its constant speed is 8 m s-1, what is the centripetal acceleration?
Defining Circular Motion | Summary
๐ฃ =2๐๐
๐= ๐๐๐ =
2๐
๐
Angular Velocity Tangential Velocity (linear speed)
Velocity Linearv โ m s-1
Angularฯ โ rad s-1
Centripetal Accelerationchanges direction toward center
ac โ m s-2
Centripetal Forcedirected toward center
F = maSee derived equations
Warm Up
A pilot is flying a small plane at 30.0 m s-1 with a radius of 100.0 m. If a force of 635 N is needed to maintain the pilotโs circular motion, what is the pilotโs mass?
v 30 m s-1
r 100 m
F 635 N
m ?
๐น =๐๐ฃ2
๐
635 =๐(30)2
100
๐ = 70.56 kg
Try Thisโฆ
If you swing a ball on a string above your head, and the
string breaks, what happens?
A inward facing force is required for circular motion
Travels in a straight line
tangent to the circle
Try Thisโฆ
Iโm usually running late for school and sometimes I forget my plate of pop tarts on the top of my car. What happens when I take a sharp turn to the right? Why?
Pop Tarts will keep moving forward
(in a straight line) unless an outside
force acts upon them
Think about itโฆ
If you swing a ball on a string in a vertical circle, where is the string most likely to break? Why?
F
A
B
C
D
EBecause gravity is
pulling against the
string at this point
Centripetal Force
Remember, for an object to follow a curved path, there must be an inward pointing centripetal force (Fc)
This is not really a force that shows up on a free body diagram like Fg , R, Ff , and FT .
Rather, it is more like the net force that is required to create that circular motion
If an object is in circular motion: Fnet = Fc
Vertical Circle
When you make a vertical circle the net force at all points must equal the centripetal force (Fc)
FcFc
Fc
Fc
Fc
Fc
This is the case for horizontal circles too! The main difference is that now the weight is a factorโฆ
Now with numbers!
At the Top:
20 N
At the Bottom:
Fc required is 20 NFg of object is 5 N
20 N
Fg = 5 NFT
20 N
Fg = 5N
FT
20 N
FT = 15 N
FT = 25 N
*FT is determined by comparing the
known forces (Fg) to the net force (Fc)
and finding the difference
Roller Coaster
8 m
10 m s-1
m 200 kg
vt 10 m s-1
r 8 m
Fc 2500 N
Fnet 2500 N
Fg 1962 N
R 4462 N
๐น๐ =๐๐ฃ2
๐=(200)(10)2
8= ๐๐๐๐ ๐
๐น๐ = ๐๐ = 200 9.81 = ๐๐๐๐ ๐
Fg
RFc
Fnet
or
๐น๐๐๐ก = ๐ โ ๐น๐
๐ = 2500 + 1962
๐ = 4462 N
Roller Coaster
8 m
5 m s-1 m 200 kg
vt 5 m s-1
r 8 m
Fc 625 N
Fnet 625 N
Fg 1962 N
R 1337 N
๐น๐ =๐๐ฃ2
๐=(200)(5)2
8= ๐๐๐ ๐
๐น๐ = ๐๐ = 200 9.81 = ๐๐๐๐ ๐
Fg
RFc
Fnet
or
๐น๐๐๐ก = ๐น๐ โ ๐
๐ = 1962 โ 625
๐ = 1337 N
Loop the Loop!
The velocity needs to be fast enough that the R is greater than 0 N
Minimum velocity required = ๐๐
๐น๐ = ๐น๐
๐๐ฃ2
๐= ๐๐
๐ฃ = ๐๐
Fc Fg
R=0
What is the tension?
m = 2 kg
Fc
Fc
m 2 kg
vt 5 m/s
r 0.5 m
Fc 100 N
Fnet 100 N
Fg 19.62 N
FT 80.38 N
m 2 kg
vt 5 m/s
r 0.5 m
Fc 100 N
Fnet 100 N
Fg 19.62 N
FT 119.62 N
Top Bottom
Fg
FT
Fc
Fg
FT
Fc
Top Bottom
Remember โThe Rotorโ
We can use this example to discuss that there must be an inward force (centripetal force) acting towards the center. But why donโt they fall down?!?
R
Ff
Fg
Friction
You canโt forget about Friction!
Remember that friction is related to the normal force and the coefficient of friction (ฮผ). The only thing that is different here is that the normal force is the centripetal force.
R
Fg
Ff
๐น๐ = ๐๐2๐
๐น๐ = ๐
๐น๐ = ๐๐
๐น๐๐๐ก = ๐
๐น๐๐๐ก = ๐น๐ = ๐
R ฮผ
Give it a Shot!
The "Rotor" ride is the one which presses you against the walls of the spinning rotor as the floor drops away. The coefficient of static friction between the wall and the 75-kg rider is ฮผ = 0.06. If the ride is rotating at an angular velocity of 5.2 rad s-1, what must be the radius of the rotor?
R
Fg
Ff
๐น๐ = ๐ = ๐๐2๐
๐น๐ = ๐น๐๐๐ = ๐๐
(0.06)๐ = (75)(9.81)
๐ = 12,263 N
12,263 = (75)(5.2)2๐๐ = 6.05 m
*Friction and weight
are equal and opposite
*Normal Reaction Force is
equal to the Centripetal Force
You didnโt need the mass โบ
The "Rotor" ride is the one which presses you against the walls of the spinning rotor as the floor drops away. The coefficient of static friction between the wall and the 75-kg rider is ฮผ = 0.06. If the ride is rotating at an angular velocity of 5.2 rad s-1, what must be the radius of the rotor?
R
Fg
Ff
๐น๐ = ๐๐
๐น๐ = ๐น๐
๐น๐ = ๐๐
๐๐ = ๐๐
๐๐๐2๐ = ๐๐
๐น๐ = ๐๐2๐
๐น๐ = ๐น๐๐๐ก = ๐
๐๐2๐ = ๐ solve for r
Skidding Around a Curve
r
vWhat is providing the centripetal force causing the car to move around the curve?
Ff Friction
Fg = mg
R
Ff =ฮผR๐น๐๐๐ก = ๐น๐ = ๐น๐
Skidding Around a Curve
A car of mass 1240 kg moves around a bend of radius 63 m on a horizontal road at a speed of 18 m s-1. If the car was to be driven any faster there would not be enough friction and it would begin to skid.
What is the coefficient of friction between the road and the tires?
๐ = 1240 kg
๐ = 63 m
๐ฃ = 18 m sโ1
๐น๐๐๐ก = ๐น๐ = ๐น๐ ๐๐ฃ2
๐= ๐๐
(1240)(18)2
63= ๐(12,164)
๐ = 0.52Fg = mg
R
Ff =ฮผR
๐ = ๐น๐ = ๐๐
= 1240 9.81 = 12164 N
Banked Curve
r
v
R Ry
Rx
Rx
Ff
*The centripetal force
creating the circular
motion doesnโt rely
entirely on friction when a
component of the normal
force is horizontal
Pendulum Circle
What is centripetal force required to cause a 0.12 kg mass to swing in a horizontal circle with the string at an angle of 30ยฐ?
๐๐ฆ = ๐น๐ = ๐๐ = 0.12 9.81 = 1.18 N
Tx
Ty
T30ยฐ
tan ๐ =๐๐ฅ๐๐ฆ
๐๐ฅ = ๐๐ฆtan ๐ = 1.18tan 30ยฐ
๐๐ฅ = 0.68 N
๐น๐๐๐ก = ๐น๐ = ๐๐ฅ
๐น๐ = 0.68 N