“WE COME SPINNING OUT OF NOTHINGNESS, SCATTERING STARS LIKE...
Transcript of “WE COME SPINNING OUT OF NOTHINGNESS, SCATTERING STARS LIKE...
CIRCULAR MOTION“WE COME SPINNING OUT OF NOTHINGNESS, SCATTERING STARS LIKE DUST” — JALAL AD-DIN RUMI
HOW DO YOU MAKE AN OBJECT TURN?
• 1st Law: An object in motion will tend to stay in motion at a constant speed in a straight line unless acted on by an external, unbalanced force.
• Ok, so we need an external, unbalanced force
• But where should we apply it?
v FA
UNIFORM CIRCULAR MOTION
• An object that moves in a circle at a constant speed v is said to experience uniform circular motion
• The magnitude of the velocity remains constant, but the direction is continuously changing
v
v
UNIFORM CIRCULAR MOTION
• Rotation (or spin) — when a body turns about an internal axis
• Revolution — when a body turns about an external axis
• The Earth rotates around an axis passing through its geographic poles once every 24 hours
• The Earth revolves around the Sun once every 365.25 days
UNIFORM CIRCULAR MOTION
radial - behavior toward and away from the center of the circle
tangential - behavior along the edge of the circle
e.g. v is the tangential velocity
(vt)
v
v
UNIFORM CIRCULAR MOTION
Which way must the force be acting to make the object move in a circle?
Forces which point toward the center of rotation are called centripetal forces ( Fc ), meaning “center seeking” forces
v
v
F
F
CENTRIPETAL FORCE
What forces act as centripetal forces?
ball on a string swung in a circle: tension
car rounding a bend: friction
orbit of planets: gravity
rollercoaster car going around a loop: normal
force
v
v
Fc
Fc
CENTRIPETAL FORCE
As I swing a ball in a circle overhead, three quantities determine how big a force I’ll need to keep it in a circle:
How big is the ball: m
How big is the circular path: r
How fast is the ball moving through the circle: v
v
v
Fc
Fc
CENTRIPETAL FORCE
Fc = mac
ac = v2/r
(Proof: https://
www.youtube.com/
watch?v=TNX-
Z6XR3gA)
Fc = mv2/r
v
v
Fc
Fc
FREQUENCY & PERIOD
Frequency ( f ) is the number of revolutions per second
measured in Hertz (Hz)
1 Hz = 1 rev/s
Period (T ) is the time required to make one full revolution
measured in seconds
T = 1/f
v
r
FREQUENCY & PERIOD
You swing a ball from a sting of length r and, using a stopwatch, measure a period of T
How fast must you be swinging the ball?
v = 2πr/T
v
r
EXAMPLE 1
• You got the Ball and Chain! Set it to y or x and swing to do serious damage.
• You can swing the 18.0 kg ball from its 1.50-m-long chain at 18.8 m/s. What is the force of tension in the chain?
• Ans. FT = 4,240 N
EXAMPLE 2
• The Moon’s nearly circular orbit about the Earth has a radius of about 384,000 km and a period T of 27.3 days. How fast does the Moon orbit the Earth (in m/s)?
• Ans. 𝑣 = 1,020 m/s (or 2,290 mph)
SANITY CHECK
• You’re driving along in your car when you make a hard and quick left
• In what direction is the (net) force acting on your body? • To the left!
• But your body feels like it’s being pushed to the right because your inertia resists changes to its motion • The so-called centrifugal (“center fleeing”) force is a
fictitious force experienced because your reference frame (the car, in this case) is accelerating
SANITY CHECK
• A ball is swung in a horizontal circle and released as shown to the right. Which way will the ball travel once it’s released?
• On a tangential path
SANITY CHECK
• A rider on a Ferris wheel moves in a vertical circle of radius r at constant speed v. Is the normal force the seat exerts on the rider at the top of the ride greater than, less than, or equal to the normal force the seat exerts at the bottom?
• Ans. FN,top < FN,bottom
QUESTION 1
• A centripetal force acts
A.along the edge of motion
B. toward the center of motion
C.away from the center of motion
D.downward
QUESTION 1
• A centripetal force acts
A.along the edge of motion
B. toward the center of motion
C.away from the center of motion
D.downward
QUESTION 2
• When an object moves in a circle, there is no force pushing the object outward from the circle.
A.True
B. False
QUESTION 2
• When an object moves in a circle, there is no force pushing the object outward from the circle.
A.True
B. False
EXAMPLE 3
a) A 1000.-kg car rounds a curve on a flat road of radius 50. m at a speed of 50. kph. What is the force of friction allowing the car to make the turn?
• Ans. Ffr = 3,900 N
b) What must be the coefficient of friction between the road and tires?
• Ans. µs = 0.39
EXAMPLE 4
• In 2001: A Space Odyssey, the astronauts use a giant centrifuge to simulate gravity while in space.
• If the centrifuge has a radius of 65 m, how fast should the outer edge be rotating in order to properly simulate gravity?
• Ans. 𝑣 = 25 m/s
Rotational Mechanics“ We dance round in a ring and suppose, but the secret sits in the middle and knows.” — Robert Frost
Why do you hold your arms out when trying to hold your balance (on a balance beam, tightrope, slack line, curb etc.)?
Rotational Motion• Motion of a rigid body is
usually broken up into translational motioncenter of mass and motionrotation
• A has definite and unchanging shape
Angular Quantities• Angular quantities are
analogous to corresponding quantities in linear motion
• Instead of asking how far and how fast an object travels, we can ask how much and how quickly it rotates
θO
Pr l
𝑥
Angular Quantities
θO
Pr l
Quantity Linear Angular Relationship
position l in meters θ in radians θ = l/r
velocity v in m/s ω in rad/s ω = v/r= ∆θ/∆t
acceleration a in m/s2 α in rad/s2 α = a/r= ∆ω/∆t
Note: 2π rad = 360º
Example 1• A particular bird’s eye can just
distinguish objects that subtend an angle no smaller than about 3×10
a) How many degrees is this?
b) How small an object can the bird just distinguish when flying at a height of 100 m?
• Ans. a) 0.017º
• b) l = 3 cm
Sanity Check• A rotating carousel has one
child sitting on a horse near the outer edge and another child on a lion halfway out from the center.
a) Which child has greater linear speed?
b) Which child has greater angular speed?
• Ans. a) the child on the horse
• b) both are the same
Circular Motion & Angular Quantities
• Centripetal acceleration in terms of angular velocity
• ac
• Frequency in terms of angular velocity
• ω = 2π
Example 2a) What is the angular speed of a
child seated 1.2 m from the center of a steadily rotating merry-go-round that makes one complete revolution in 4.0 s?
b) What is her tangential speed?
c) What is her centripetal acceleration?
• a) ω = 1.6 rad/s
• b)
• c) a
Example 3• The platter of the hard disk of a
computer rotates at 5400 rpm.
a) What is the angular velocity of the disk?
b) If the reading head is of the drive is located 3.0 cm from the axis of rotation, what is the speed of the of the disk below it?
c) What is the centripetal acceleration of this point?
Example 3• The platter of the hard disk of a
computer rotates at 5400 rpm.
a) What is the angular velocity of the disk?
b) If the reading head is of the drive is located 3.0 cm from the axis of rotation, what is the speed of the of the disk below it?
c) What is the linear acceleration of this point?
d) If a single bit requires 5 µm of length along the motion direction, how many bits per second can the writing head write when it is 3.0 cm from the axis?
• a) ω = 570 rad/s
• b)
• c) a
Kinematic Equations
Angular Linear
ωf = ωi + α∆t vf = vi + a∆t
∆θ = ωi∆t + ½α∆t2 ∆𝑥 = vi∆t + ½a∆t2
ωf2 = ωi2 + 2α∆θ vf2 = vi2 + 2a∆𝑥
Note: remember! the kinematic equations only work for constant
Example 4• A centrifuge rotor is accelerated from rest to 20,000 rpm
in 5.0 min.
a) What is the angular acceleration of the rotor in (rad/s
b) How many revolutions has it turned through in this time?
• a) α = 7.0 rad/s
• b) Δθ = 3.15×10
Rolling Motion
• A bicycle slows down uniformly from a distance of 115 m. Each wheel and tire has an overall diameter of 68.0 cm. Determine
a) the angular velocity of the wheels at the initial instant
b) the total number of revolutions each wheel rotates in coming to rest
c) the angular acceleration of the wheel
d) the time it took to come to a stop
Rolling Motion• A bicycle slows down uniformly from
over a distance of 115 m. Each wheel and tire has an overall diameter of 68.0 cm. Determine
a) the angular velocity of the wheels at the initial instant
• ω0
• ω0
• ω0
Rolling Motion• A bicycle slows down uniformly from
distance of 115 m. Each wheel and tire has an overall diameter of 68.0 cm. Determine
b) the total number of revolutions each wheel rotates in coming to rest
• Revs =
• Revs =
• Revs = (115 m)/(2π
• Revs = 53.8 rev
Rolling Motion• A bicycle slows down uniformly from
over a distance of 115 m. Each wheel and tire has an overall diameter of 68.0 cm. Determine
c) the angular acceleration of the wheel
• α = (ω
• α = (0 - (24.7 rad/s)
• α = -0.902 rad/s
Rolling Motion• A bicycle slows down uniformly from
over a distance of 115 m. Each wheel and tire has an overall diameter of 68.0 cm. Determine
c) the time it took to come to a stop
• t = (ω
• t = (0 - 24.7 rad/s)/(-0.902 rad/s
• t = 27.4 s
Torque
• Rotational kinematics —
• Rotational dynamics —
• To make an object rotate, we need a force
• But the direction of the force, and where we apply it, matters
Torque• Apply force
• The bigger the force, the more quickly the door opens
• Apply the same force closer to the hinge, at
• Door will not open as quickly
• The angular acceleration of the door is proportional to the magnitude of the force applied that force is from the axis of rotation
• That distance is called the arm
r1r2
FF
Torque (or why hobbit doors are a dumb design)• The “twisting force” which
cause rotation is called the torque (
• 𝜏
• Measure in Nm
• α
Torque• F
same magnitude and the same distance
• but they will the same twisting motion
• only the component of the force will contribute to rotation
r
F F
F
Torque• F
same magnitude and the same distance
• but they will the same twisting motion
• only the component of the force will contribute to rotation
r
F
• 𝜏
θF
F
Example 5• Hercules and the Hulk are in
competition for some reason. They’ve matched each other in every test of strength, so Bruce Banner devises the following challenge.
• The two push on opposite ends of a lever fixed by a pivot at the center. The Hulk applies 4.3 million Newtons of force 5.0 m from the pivot. Hercules apply 5 million N of force 3.0 m from the pivotWhat’s the net torque?
• Ans. The Hulk wins.
FF
5.0 m 3.0 m
Rotational Inertia• Linear acceleration
• a
• Angular acceleration
• α = ∑
• Moment of inertiarotational inertia) is a measure of a body’s resistance to changes in its rotation
• Rotational “laziness”
Rotational Inertia• Picture a particle of mass
revolving in a circle of radius
• Initially at rest, we want this particle to start rotating, so we give it a push
• F
• F
• 𝜏 = mr
• mrinertia of the particle (measured in kg
rm
F
Rotational Inertia• Consider a rotating rigid body
• basically a collection of particles all at varying distances from the axis of rotation
• ∑
• I = ∑
• ∑
• Newton’s 2
Things to Note• Moment of inertia (
motion that mass plays for translational motion
• The rotational inertia of an object depends not only on its mass, but also on how that mass is distributed with respect to the axis of rotation
• A large-diameter cylinder will have greater rotational inertia than a smaller-diameter cylinder of equal mass
• The former will be harder to start rotating and harder to stop
Example 6• Two weights of mass 5.0 kg and
7.0 kg are mounted 4.0 m apart on a massless rod. Calculate the moment of inertia of the system when rotated about an axis halfway between the weights
• Ans. I = 48 kg
4.0 m
5.0 kg 7.0 kgAxis
Moment of Inertia
• Don’t memorize
• Do
• roughly how they rank from greatest to least
• what that implies about their behavior
Example 7• A 15.0 N force is applied to a cord wrapped
around a pulley of mass radius accelerate uniformly from rest to reach an angular speed of 30.0 rad/s in 3.00 s. Calculate the pulley’s
a) net torque
b) angular acceleration
c) momentum of inertia
a) ∑𝜏
b) α = 10.0 rad/s
c) I = 0.495 kg F
R
Rotational Kinetic Energy• Translational kinetic energy
• KE
• Rotational kinetic energy
• KE
• Make sure, do the units check out?
• KE
Example 8• A solid sphere of mass
radius unknown height without slipping and picks up a speed 7.5 m/s by the time it reaches the bottom.
a) What are the sphere translational and rotational kinetic energies at the bottom?
b) What must have been the sphere’s gravitational potential energy at the top?
a) KE
b) PE
Angular Momentum• Linear momentum
• p
• Angular momentum
• L
• Measured in kg
• Newton’s 2
• ∑
• Newton’s 2
• ∑
Conservation of Angular Momentum
• The total angular momentum of a rotating body remains constant if the net torque acting on it is zero
• I1
Example 9• A mass
string revolves in a circle on a frictionless tabletop. The other end of the string passes through a hole in the table. Initially, the mass revolves with an angular speed ω= 3.0 rad/s in a circle of radius 0.80 m. The string is then pulled slowly through the hole so that the radius is reduced to What is the angular speed, ωthe mass now?
• Ans. ω
• Why do you hold your arms out when trying to hold your balance (on a balance beam, tightrope, slack line, curb etc.)?
• Holding your arms out increases your rotational inertial, making it harder for you to tip over.