circular motion
DESCRIPTION
Circular Motion. 1. Rotation and Revolution. Rotation. the motion or spin on an internal axis. Revolution . the motion or spin on an external axis. Rotational Speed. Rotational Speed. Number of rotations per unit of time Rpm or Rps All objects that rotated on same axis have the - PowerPoint PPT PresentationTRANSCRIPT
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Circular MotionCircular Motion
the motion or spin the motion or spin on an internal axison an internal axis
the motion or spin the motion or spin on an external axison an external axis
• Number of rotations per unit of timeRpm or Rps
• All objects that rotated on same axis have the same rotational speed. • Also called Frequency
(cycles/s or Hertz) • Period (seconds) is the inverse of Frequency
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Gymnast on a High BarA gymnast on a high bar swings through two rotations or cycles in a time of 1.90s. Find the average rotational speed (in rps) or frequency (in Hz) of the gymnast.
Given: t = 1.90 s & 2 rotation (cycle)Given: t = 1.90 s & 2 rotation (cycle)Find the average rotational speed (in rps) RpsRps = = rotationsrotations secondsecond
= = 2 rotation 2 rotation 1.90 seconds1.90 seconds
= 1.05 rps = 1.05 cycles/second= 1.05 rps = 1.05 cycles/second
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A Helicopter BladeA Helicopter BladeFind the rotational speed or frequency at #1 if it takes 0.154 s for one rotation (cycle)?
Given: t = 0.154 s & 1 rotation (cycle)Given: t = 0.154 s & 1 rotation (cycle)Find the average rotational speed (in rps) RpsRps = = rotationsrotations secondsecond
= = 1 rotation 1 rotation 0.154 seconds0.154 seconds
= 6.49 rps = 6.49 cycles/second= 6.49 rps = 6.49 cycles/second
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A Helicopter BladeA Helicopter BladeFind the rotational speed or frequency at #2 if takes 0.154 s for one rotation (cycle)?
Given: t = 0.154 s & 1 rotation (cycle)Given: t = 0.154 s & 1 rotation (cycle)Find the average rotational speed (in rps) RpsRps = = rotationsrotations secondsecond
= = 1 rotation 1 rotation 0.154 seconds0.154 seconds
= 6.49 rps = 6.49 cycles/second= 6.49 rps = 6.49 cycles/second
Do Frequency/ Do Frequency/ Rotational Speed Rotational Speed
ProblemsProblems
• The speed in m/s of something moving along a circular path.
• It always tangent to the circle.
• The distance moved The distance moved per unit of time.per unit of time.
• Linear speed is Linear speed is greater on the outer greater on the outer edge of a rotating edge of a rotating object than it is object than it is closer to the axis.closer to the axis.
Linear SpeedLinear Speed
Distance traveled in one period is the
circumference 22ππrr
Time for one “cycle” is the “period” (TT)
Tangential Speed Tangential Speed = = CircumferenceCircumference / Period / Period
Tangential Speed Tangential Speed = = 22ππrr TT
But remember that period is the inverse of
frequencySo instead of dividing by period you multiply by
frequency
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A Helicopter BladeA Helicopter BladeA helicopter blade has an angular speed of 6.50 rps. For points 1 on the blade, find the tangential speed
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A Helicopter BladeA Helicopter BladeA helicopter blade has an angular speed of 6.50 rps. For points 2 on the blade, find the tangential speed
Do Tangential SpeedDo Tangential SpeedLinear VelocityLinear Velocity
Problems Problems
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Centripetal AccelerationCentripetal Acceleration
rva T
c
2
(centripetal acceleration)
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A Helicopter BladeA Helicopter BladeA helicopter blade has an angular speed of 6.50 rps. For points 1 on the blade, find the tangential acceleration
Given: r = 3.00 m & Given: r = 3.00 m & Tangential Speed Tangential Speed = 122 m/s= 122 m/s
Tangential Tangential Acceleration Acceleration == (122 m/s) (122 m/s)22 / 3.00m / 3.00m
== 4,960 m/s 4,960 m/s22
== 4.96 x 10 4.96 x 1033 m/s m/s22
rva T
c
2
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A Helicopter BladeA Helicopter BladeA helicopter blade has an angular speed of 6.50 rps. For points 2 on the blade, find the tangential acceleration
Given: r = 6.70 m & Given: r = 6.70 m & Tangential Speed Tangential Speed = = 273 m/s 273 m/s
Tangential Tangential Acceleration Acceleration == (273 m/s) (273 m/s)22 / 6.70m / 6.70m
== 11,200 m/s 11,200 m/s22
== 1.12 x 10 1.12 x 1044 m/s m/s22
rva T
c
2
Do CentripetalDo CentripetalAccelerationAcceleration
Problems Problems
Centripetal ForceCentripetal ForceFFcc = ma = macc
FFcc = = mvmvTT22
rr
Vertical drum rotates, you’re pressed against wallFriction force against wall matches gravitySeem to stick to wall, feel very heavy
The forces real and perceived
Real Forces:
Friction; upCentripetal; inwardsGravity (weight); down
Perceived Forces:
Centrifugal; outwardsGravity (weight); down
Perceived weight; down and out
Weight the force due to gravity on an objectWeight = Mass Acceleration of GravityW = m g
Weightlessness - a conditions wherein gravitational pull appears to be lackingExamples:
AstronautsFalling in an ElevatorSkydivingUnderwater
From 2001: A Space Odysseyrotates like bicycle tire
• Just like spinning drum in amusement park, create gravity in space via rotation
• Where is the “floor”?• Where would you
still feel weightless?• Note the windows on
the face of the wheel
Do CentripetalDo CentripetalForce Problems Force Problems
What makes something rotate?What makes something rotate?
TORQUE
AXLE
How do I apply a force to make the rod rotateabout the axle? Not just anywhere!
Torque = force times lever armTorque = force times lever arm
Torque = F Torque = F LL
Torque exampleTorque example
FF
LL
What is the torque on a boltWhat is the torque on a boltapplied with a wrench that applied with a wrench that has a lever arm of 30 cmhas a lever arm of 30 cmwith a force of 30 N?with a force of 30 N?
Torque = F x LTorque = F x L = 30 N x 0.30 m= 30 N x 0.30 m = 9 N m= 9 N m
For the same force, you get more torqueFor the same force, you get more torquewith a bigger wrench with a bigger wrench the job is easier! the job is easier!
Net Force = 0 , Net Torque ≠ 0Net Force = 0 , Net Torque ≠ 0
10 N10 N 10 N10 N
> The net force = 0, since the forces are applied in> The net force = 0, since the forces are applied in opposite directions so it will not accelerate.opposite directions so it will not accelerate.. .
Net Force = 0 , Net Torque ≠ 0Net Force = 0 , Net Torque ≠ 0
10 N10 N
10 N10 N
> However, together these forces will make the rod> However, together these forces will make the rod rotate in the clockwise direction. rotate in the clockwise direction.
Net torque = 0, net force ≠ 0Net torque = 0, net force ≠ 0
The rod will accelerate upward under theseThe rod will accelerate upward under thesetwo forces, but will not rotate.two forces, but will not rotate.
Balancing torquesBalancing torques
10 N10 N 20 N20 N
1 m1 m 0.5 m0.5 m
Left torque = 10 N x 1 m = 10 n mLeft torque = 10 N x 1 m = 10 n mRight torque = 20 N x 0.5 m = 10 N mRight torque = 20 N x 0.5 m = 10 N m
Balancing torquesBalancing torques
10 N10 N 20 N20 N
1 m1 m 0.5 m0.5 mLeft torque = 10 N x 1 m = 10 n mLeft torque = 10 N x 1 m = 10 n m
Right torque = 20 N x 0.5 m = 10 N mRight torque = 20 N x 0.5 m = 10 N mHow much force is exerted up by the Fulcrum? How much force is exerted up by the Fulcrum?
Torque = force times lever armTorque = force times lever arm
Torque = F Torque = F LL
EquilibriumEquilibrium• To ensure that an object To ensure that an object
does not accelerate or does not accelerate or rotate two conditions must rotate two conditions must be met:be met:
net force = 0net force = 0 net torque = 0 net torque = 0
Example 1Given M = 120 kg.Neglect the mass of the beam.
Find the Torque Find the Torque exerted by the massexerted by the mass
Torque Torque = = F F LL == 120 kg (9.8 m/s 120 kg (9.8 m/s22) ) (7 (7 m)m)
= = 8232 N m8232 N m
W beam
T le ft T right
W box
8 m2 m
A B C D
Example
Given:Given: WWboxbox=300 N=300 NFind: Find: FFTRTR = = FFCC CC = = ? N? N
ACD
TorqueTorqueCC == TorqueTorqueCCCC
FFCCL L == FFCCCCLL 300 N (6 m) 300 N (6 m) == FFCCCC (8 m) (8 m)
225 N 225 N == FFCCCC
W beam
T le ft T right
W box
8 m2 m
A B C D
Example
Given:Given: WWboxbox=300 N=300 NFind: Find: FFTLTL = = FFCCCC = = ? N? N
ACD
TorqueTorqueCC == TorqueTorqueCCCC
FFCCL L == FFCCCCLL FFC C (8 m) (8 m) == 300N (2 m) 300N (2 m)
FFC C = 75 N= 75 N
Example
W beam
T le ft T right
W box
8 m2 m
A B C D
ACD
Does this make sense?Does this make sense?FFTL TL = 75 N= 75 NFFTR TR = 225 N= 225 NDoes the Does the FFUP UP = = FFDOWN DOWN ?? FFUP UP = 75 N + 225 N = 300 N == 75 N + 225 N = 300 N =FFDOWN(Box)DOWN(Box)
Given:Given: WWboxbox=300 N=300 N
Another ExampleGiven: W=50 N, Given: W=50 N,
L=0.35 m, L=0.35 m, x=0.03 mx=0.03 m
Find the tension in the muscleFind the tension in the muscle xL
W
TorqueTorqueCC == TorqueTorqueCCCC
FFCCL L == FFCCCCLL 50N (0.350 m) 50N (0.350 m) == FFCCCC (0.030m) (0.030m) 50N (0.350 m) / 50N (0.350 m) / (0.030m) (0.030m) == FFCCCC
583 N = F583 N = FCCCC
StabilityStabilityCM &TorqueCM &Torque
Condition for stability Condition for stability
If the CG is above If the CG is above the edge, the objectthe edge, the objectwill not fallwill not fallCGCG
when does it fall over?
CG CG
STABLE NOT STABLE
If the vertical lineextending down fromthe CG is inside theedge the object willreturn to its uprightposition the torquedue to gravity bringsit back.
Stable and UnstableStable and Unstable
stablestable unstableunstabletorque due to gravitytorque due to gravitypulls object backpulls object back
torque due to gravitytorque due to gravitypulls object downpulls object down
Stable structuresStable structures
Structures areStructures arewider at theirwider at their
base to lower theirbase to lower theircenter of gravitycenter of gravity
If the center of gravity If the center of gravity is supported, the is supported, the blocks do not fall overblocks do not fall over
Playing with your blocksPlaying with your blocks
CGCG
Rotational InertiaRotational InertiaThe rotational “laziness” of an object
Recall : Recall : inertiainertiaA measure of the “laziness” of an A measure of the “laziness” of an
object because of object because of
Quantified by the mass (kg) of objectQuantified by the mass (kg) of object
a Fnet
m
Rotational Inertia (Rotational Inertia (II))A measure of an object’s “laziness” to A measure of an object’s “laziness” to
changes in rotational motionchanges in rotational motion
Depends on mass Depends on mass ANDAND
distance of mass from axis of rotationdistance of mass from axis of rotation
QuickTime™ and aTIFF (Uncompressed) decompressor
are needed to see this picture.
Balancing Pole increases Rotational InertiaBalancing Pole increases Rotational Inertia
Angular MomentumAngular Momentum
Angular MomentumAngular MomentumMomentum resulting Momentum resulting from an object moving in from an object moving in linear motion is called linear motion is called linear momentumlinear momentum. . Momentum resulting Momentum resulting from the rotation (or from the rotation (or spin) of an object is spin) of an object is called called angular angular momentummomentum..
Conservation of Angular Conservation of Angular MomentumMomentum
Angular momentum is Angular momentum is important because it important because it obeys a conservation obeys a conservation law, as does linear law, as does linear momentum. momentum. The total angular The total angular momentum of a closed momentum of a closed system stays the same.system stays the same.
Calculating angular momentumCalculating angular momentum
Angular momentum is calculated in a similar way to linear Angular momentum is calculated in a similar way to linear momentum, except the mass and velocity are replaced by the momentum, except the mass and velocity are replaced by the moment of inertia and angular velocity.moment of inertia and angular velocity.
AngularAngularvelocityvelocity(rad/sec)(rad/sec)
AngularAngularmomentummomentum(kg m/sec(kg m/sec22))
L = I L = I ww
Moment of Moment of inertiainertia(kg m(kg m22))
Gyroscopes Angular MomentumGyroscopes Angular Momentum
A A gyroscopegyroscope is a device that contains a spinning object is a device that contains a spinning object with a lot of angular momentum. with a lot of angular momentum.
Gyroscopes can do amazing tricks because they Gyroscopes can do amazing tricks because they conserve angular momentum. conserve angular momentum.
For example, a For example, a spinning spinning gyroscope can easily balance gyroscope can easily balance on a pencil point. on a pencil point.
A gyroscope on the space shuttle is mounted at the center of A gyroscope on the space shuttle is mounted at the center of mass, allowing a computer to measure rotation of the mass, allowing a computer to measure rotation of the spacecraft in three dimensions.spacecraft in three dimensions.
An on-board computer is able to accurately measure the An on-board computer is able to accurately measure the rotation of the shuttle and maintain its orientation in rotation of the shuttle and maintain its orientation in space.space.
Gyroscopes Angular MomentumGyroscopes Angular Momentum
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