Wednesday, 11/05/14 TEKS: P.4C: Analyze and describe accelerated motion in two dimensions using...
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Transcript of Wednesday, 11/05/14 TEKS: P.4C: Analyze and describe accelerated motion in two dimensions using...
Wednesday, 11/05/14
TEKS:
P.4C: Analyze and describe accelerated motion in two
dimensions using equations, including projectile and circular
examples.
By the end of today, IWBAT…
Analyze characteristics of 2D circular motion.
Essential Question:
What are the components of
circular motion?
Topic:Circular Motion
C-Notes!
+
Unit 4 - Circular MotionSlides with gray speckled background are extra and do not need to be incorporated into your C-Notes for today. Use them for review.
+Rotation & Revolution
Rotation
Revolution
Axis
Axis
+Period & Frequency
Period (T): seconds/cycle
Radius (r)
Frequency (f): cycles/second (Hz)
+Period & Frequency
Radius (r)
T = 1/f
f = 1/T
+Period & Frequency Exercise
Radius (r)
If the frequency is 40 Hz, what’s the period?
1/40
+Period & Frequency Exercise
If the period is 0.05 s, what’s the frequency?
Radius (r)
Frequency = No. of cycles per second Therefore F=1/0.05 or F=100/5=20
Hertz
+Rotational & Linear Speed
R
r
A A
BB 2πR
2πr A
B
+Rotational & Linear Speed
R
r
A A
BB
??? 2πR = 2πr ???
2πR
2πr A
B
+Rotational & Linear Speed
Linear speed:
distance moved per unit of time
v = Δd / Δt
The linear speed is greater on the outer edge of a rotational object than it is closer to the axis
Rr
+Rotational & Linear Speed
Tangential speed:
The speed of an object moving along a circular path can be called tangential speed because the direction of motion is always tangent to the circle
v
v
v
v
+Rotational & Linear Speed
For circular motion,
tangential speed = linear speed
+Rotational & Linear Speed
Circumference = 2πr
Radius (r)
Linear/Tangential Speed = 2πr / T = 2πrf
Period = T
Linear / Tangential Speed (v):
+Rotational & Linear Speed Exercise
Tangential Speed ?
3 m
Period = 2 s
Linear / Tangential Speed (v):
+Rotational & Linear Speed Exercise
Tangential Speed ?
4 m
Frequency = 2 Hz
Linear / Tangential Speed (v):
+Rotational & Linear Speed Exercise
Tangential Speed = 12π m/s
2 m
Frequency = ?
Period = ?
Linear / Tangential Speed (v):
+Rotational & Linear Speed
Rotational / Angular speed ():
The number of rotations per unit of time
All parts of a rotational object have the same rate of rotation, or same number of rotations per unit of time
Unit of rotational speed: Degrees/second or radians/second Revolutions per minute (RPM)
+
Radius (r)
Rotational & Linear Speed
Rotational / Angular speed ():
Rotational Speed = 2π/T = 2πf (rads/s)
1 revolution = 2π
Period = T
+Rotational & Linear Speed Exercise
Rotational / Angular speed ():
Period = 2 s Rotational Speed = ?
5 m
+Rotational & Linear Speed Exercise
Rotational / Angular speed ():
Frequency = 2 Hz Rotational Speed = ?
5 m
+Rotational & Linear Speed
Rotational / Angular speed ():
Rotational Speed = 2πf (rads/s)
Tangential Speed v = 2πrf (m/s)
v = r
(Tangential speed) = (Radial distance) x (Rotational speed)
+Rotational & Linear Speed
Rotational / Angular speed ():
At the center (or axis) of the rotational platform, there is no tangential speed, but there is rotational speed
+Rotational & Linear Speed Exercise
Rotational / Angular speed ():
Rotational Speed = 4π
3 m
Linear Speed = ?
+Rotational & Linear Speed Exercise
Rotational / Angular speed ():
Linear Speed = 6π m/s
2 m
Rotational Speed = ?
+Rotational & Linear Speed Exercise
Rotational / Angular speed ():
Period = 3 s
4 m
Rotational Speed = ? Linear Speed = ?
2 mA
B
+Rotational & Linear Speed
R
r
A
B
+Rotational & Linear Speed
R
r
2πR
2πR
2πRA
B
A
B
A
B
+Centripetal Force & Acceleration
+Centripetal Force & Acceleration
Centripetal Force
Inertia
+Centripetal Force & Acceleration
+Centripetal Force & Acceleration
+Centripetal Force & Acceleration
Centripetal Force
Inertia
Circular Motion
Suppose you drive a go cart in a perfect circle at a constant speed. Even though your speed isn’t changing, you are accelerating. This is because acceleration is the rate of change of velocity (not speed), and your velocity is changing because your direction is changing! Remember, a velocity vector is always tangent to the path of motion.
v v
v
Tangential vs. Centripetal Acceleration
10 m/s
15 m/s
So how do we calculate the centripetal acceleration ? ? ? Stay tuned!
18 m/s
start
finish
Suppose now you drive your go cart faster and faster in a circle. Now your velocity vector changes in both magnitude and direction. If you go from start to finish in 4 s, your average tangential acceleration is:
at = (18 m/s - 10 m/s) / 4 s = 2 m/s2
So you’re speeding up at a rate of 2 m/s per second. This is the rate at which your velocity changes tangentially. But what about the rate at which your velocity changes radially, due to its changing direction? This is your centripetal (or radial) acceleration.
+Centripetal Force & AccelerationCentripetal Acceleration
Acceleration is a vector quantity
a = Δv / Δt
Velocity can be changed by increasing/ decreasing the magnitude of v, or changing the direction
+Centripetal Force & AccelerationCentripetal Acceleration
A
B
C
D
A
B
C
D
Change Speed Change Direction
+Centripetal Force & AccelerationCentripetal Acceleration
An object moves around in a circle with constant speed has acceleration, because its direction is constantly changing
This acceleration is called centripetal acceleration (Ac)
Centripetal Acceleration
v0 = v
r
r
vf = v Let’s find a formula for centripetal acceleration by considering uniform circular motion. By the definition of acceleration, a = (vf - v0) / t. We are subtracting vectors here, not speeds, otherwise a would be zero. (v0 and vf have the same magnitudes.) The smaller t is, the smaller will be, and the more the blue sector will approximate a triangle. The blue “triangle” has sides r, r, and v t (from d = v t ). The vector triangle has sides v, v, and | vf - v0 |. The two triangles are similar (side-angle-side similarity).
vf
v0
vf - v0
r r
v t
continued on next slide
Centripetal Acceleration (cont.)
v0 = v
r
r
vf = v By similar triangles,
vv
| v f - v0 |
r
r
v t
v r =
| vf - v0 | v t
So, multiplying both sides above by v, we have
| vf - v0 |
t=
rv
2ac =
Unit check:(m/s)2
m = m2 / s2
m
m s2=
Centripetal acceleration vector always points toward center of circle.
vac
at
vac
at
moving counterclockwise; speeding up
moving counterclockwise; slowing down
“Centripetal” means “center-seeking.” The magnitude of ac
depends on both v and r. However, regardless of speed or tangential acceleration, ac always points toward the center. That is,
ac is always radial (along the radius).
+Centripetal Force & AccelerationCentripetal Acceleration
Centripetal acceleration is directed toward the center of the circle
Ac
Ac
Ac
Ac
+Centripetal Force & AccelerationCentripetal Acceleration
An acceleration that is directed at a right angle to the path of a moving object and produces circular motion
Centripetal acceleration (Ac)
Ac = v 2 / r
+Centripetal Force & AccelerationCentripetal Acceleration
Ac = v 2 / r = (r) 2 / r = r 2
Ac = v 2 / r = r 2
+Centripetal Acceleration Exercise
Centripetal Acceleration (Ac):
Linear speed = 6 m/s
3 m
Centripetal Acceleration = ?
+Centripetal Acceleration Exercise
Centripetal Acceleration (Ac):
Rotational speed = 2 rad/s
3 m
Centripetal Acceleration = ?
+Centripetal Acceleration Exercise
Centripetal Acceleration (Ac):
Period = 2 s Centripetal Acceleration = ?
5 m
+Centripetal Force & AccelerationCentripetal Force
Centripetal force is a force directed toward the center of the circle
Fc
Fc
Fc
Fc
Centripetal Force, Fc
From F = ma, we get Fc = mac = mv2 / r.
Fc =mv
2
r
If a body is turning, look at all forces acting on it, and find the net force. The component of the net force that acts toward the center of curvature (perpendicular to the body’s motion) is the centripetal force. The component that acts parallel to its motion (forward or backwards) is the tangential component of the net force.
Forces that can provide a centripetal force• Friction, as in the turning car example• Tension, as in a rock whirling around while attached to a
string, or the tension in the chains on a swing at the park.*• Normal Force, as in a “round-up ride” at an amusement
park (that spins & the floor drops out), or the component of normal force on a car on a banked track that acts toward the center.*
• Gravity: The force of gravity between the Earth and sun keeps the Earth moving in a nearly circular orbit.
• Any force directed toward your center of curvature, such as an applied force.
* Picture on upcoming slides
+Centripetal Force & Acceleration In linear motion
Fnet = m a In circular motion
Fc = m Ac
+Centripetal Force & AccelerationCentripetal Force
Centripetal force is a force directed toward the center of the circle
Fc = m Ac = mv 2/r = mr 2
+Centripetal Force Exercise
Centripetal Force (Fc):
Linear speed = 4 m/s
2 m
Centripetal Force = ?
2 kg
+Centripetal Force Exercise
Centripetal Force (Fc):
Angular speed = 3 rad/s
2 m
Centripetal Force = ?
5 kg
+Centripetal Force & AccelerationCentripetal Force
Centripetal force is directly proportional to mass (m)
Fc ~ m
(Fc = m Ac = mv 2/r = mr 2)
+Centripetal Force & AccelerationCentripetal Force
Centripetal force is directly proportional to radius (r)
Fc ~ r
(Fc = m Ac = mv 2/r = mr 2)
+Centripetal Force & AccelerationCentripetal Force
Centripetal force is directly proportional to linear speed squared (v2)
Fc ~ v2
(Fc = m Ac = mv 2/r = mr 2)
+Centripetal Force & AccelerationCentripetal Force
Centripetal force is directly proportional to angular speed squared (2)
Fc ~ 2
(Fc = m Ac = mv 2/r = mr 2)
+Centripetal Force Example
For a circular motion, what if mass is doubled? Fc will be …………
For a circular motion, what if radius is doubled? Fc will be …………
For a circular motion, what if linear speed is doubled? Fc will be …………
For a circular motion, what if angular speed is doubled? Fc will be …………
+Centripetal Force Example
For a circular motion, what if mass is halved? Fc will be …………
For a circular motion, what if radius is halved? Fc will be …………
For a circular motion, what if linear speed is halved? Fc will be …………
For a circular motion, what if angular speed is halved? Fc will be …………
+Centripetal Force Example
A 280 kg motorcycle traveling at 32 m/s enters a curve of radius = 130 m. What force of friction is required from the contact of the tires with the road to prevent a skid?
+Centripetal Force Example
A 280 kg motorcycle traveling at 32 m/s enters a curve of radius = 130 m. What force of friction is required from the contact of the tires with the road to prevent a skid?
Fc = 280kg x (32 m/s)2/130m = 2205 N
+Centripetal Force Exercise
Astronauts are trained to tolerate greater acceleration than the gravity by using a spinning device whose radius is 10.0 m. With what linear speed and rotational speed would an astronaut have to spin in order to experience an acceleration of 3 g’s at the edge of the device?
+Centripetal Force Exercise
To swing a pail of water around in a vertical circle fast enough so that the water doesn’t spill out when the pail is upside down. If Mr. Lin’s arm is 0.60 m long, what is the minimum speed with which he can swing the pail so that the water doesn’t spill out at the top of the path?
+Centripetal Force Exercise
At the outer edge of a rotating space station, 1 km from its center, the rotational acceleration is 10.0 m/s2. What is the new weight of a 1000 N object being moved to a new storage room which is 500 m from the center of the space station?
+Summary
Rotation & revolution
Period & frequency
Linear/tangential speed: v = Δd / Δt = 2πr / T = 2πrf (m/s)
Rotational/angular speed: = 2π/T = 2πf (rads/s)
Tangential speed = Radius x Rotational speed: v = r
+Summary
Centripetal force & acceleration
Centripetal acceleration: Ac = v 2 / r = r 2
Centripetal force: Fc = m Ac = mv 2/r = mr 2
Centripetal force: Fc ~ m
Centripetal force: Fc ~ r
Centripetal force: Fc ~ v2
Centripetal force: Fc ~ 2