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# Modern Mechanics PHYS 17200

Purdue

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This 26 page Class Notes was uploaded by Mrs. Justyn Cremin on Saturday September 19, 2015. The Class Notes belongs to PHYS 17200 at Purdue University taught by Gabor Csathy in Fall. Since its upload, it has received 91 views. For similar materials see /class/207910/phys-17200-purdue-university in Physics 2 at Purdue University.

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Date Created: 09/19/15

PHYS 172 Modern Mechanics Fall 2009 Lecture 19 Angular Momentum Read 101 104 CLICKER QUESTION 1 Reading Question Sections 101 104 LArAxp 1ArAxF Consider the falling block of mass m as shown in the figure The force of gravity is mg there are no other forces acting on it With respect to the point A A There is angular momentum and a torque l W There is no angular momentum or torque because nothing is rotating There can be angular momentum but no torque because that would require a rigid connection to A w 0 Spinning Objects Child runs towards and jumps onto a n i m merrygoround mm we a 5pm motion with the subsequent rotation g gt 7 7 How can we easily relate the child s Nsm Imlnisr Most convenient method is to 3 Spin mum consider angular momentum of the merrygoround i Max rottw The Rotation of a Point Particle Consider a particle fixed to the edge of a rotating disk Particle is rotating it s moving through some angle Consider another particle moving with speed v to the right relative to the axis shown This particle also moves through some angle D V A6 A quot f This particle is also rotating about the specific point A Angular Momentum of Point Particle EA 2 angular momentum of point particle relative to point A This is a vector it has magnitude and direction Magnitude LA E ZA ip rApsin6 This definition seems sensible since larger values of both ri and p result in faster spinning of the merrygoround Direction of Angular Momentum The red particle on the disk has a position and momentum that changes continuously But one thing doesn t change the X axis of rotation Use the rotation axis to define the direction of rotation and hence A This convention along with our definition of the magnitude of LA is built into the vector crossproduct gt LA rAXp The Cross Product Kx E La nitude sz 21 sine Direction always i to plane containing the 2 vectors How to determine whether it s into or out of plane is shown below counter Clockwise Equotrlt I39lj l clockwise II A l 7 I quot l39 I Ax K x E is avector with magnitude X x E AB sin6 and direction given by a quotlighthand 1111a Evaluating Cross Product Algebraically Ax 1sz AZBy AZBX AXBZ AxBy AyBxgt NOTE If you ve studied determinants here s a useful way to remember the above expression is X w H Ru an N am 92k k 0 the A 1sz AZBy AZBX AxB Z j39 AxBy AyBx Angular Momentum of Point Particle l N its II n X El NOTE we need to move vectors tail to tail before we graphically evaluate the cross product 17A is the position of the particle relative to point A NEVER FORGET Angular momentum is NOT DEFINED unless you specify Picking different A s different reference points defines different values of angular momentum CLICKER QUESTION 2 What is the direction of lt0 0 3 x lt 0 4 0 Algebraic answer 1sz AszAZBX AXBZAXBy AyBX 00 34 30 00 04 00 1200 CLICKER QUESTION 3 What is the direction of A X y B X 0 4 0X0 0 3 C y ZLX D V E zero magnitude Algebraic answer 1sz Asz AZBX AXBZAXBy AyBX 43 00 00 03 00 40 1200 CLICKER QUESTION 4 What is the direction of A X y B X I 0 0 6X0 0 3 C y z X D V E zero magnitude Algebraic answer 1sz AszAZBX AXBZ AXBy AyBX 0 3 60 60 0 3 00 00 000 CLICKER QUESTION 5 A ball falls straight down in the xy plane Its momentum is shown by the red arrow What is the direction of the ball39s angular momentum about location A i y o L l 9 x zaxis points out of page 03gt lt lt UO E zero magnitude N CLICKER QUESTION 6 A ball falls straight down in the xy plane Its momentum is shown by the red arrow What is the ball39s 2 component of angular momentum about location A y y A 10 kgm2s O B 10kgm2s 110 kg m C 40 kgmzls z X D 40 kgmzls zaxis points E 0 A 4 m x out ofpage CLICKER QUESTION 7 A planet orbits a star in a circular orbit in the xy plane Its momentum is shown by the red arrow What is the direction of the angular momentum of the planet A same direction as B opposite to C into the page D out of the page E zero magnitude Bquot xA Vi I i Angular Momentum for multibarticle systems Recall from Ch 7 KtotKtran5 Krel Ktrans Krot wa Jr nlrr 1 mm By analogy we also have LA Luml Lmt J 939 For example the Earth rotates about its axis spin and moves about the Sun orbital 11 Angular Velocity I 65 of a particle is a vector magnitude lim Al gt0 At In a time At the rotating particle direction same as L sweeps out an angle A6 ccvv gt Aegt0 cvv gt A6lt0 Another method of RiseEgg finding direction of K 39 a A L or a Right hand Right hand rule A rigid body will have a welldefined angular velocity with a direction defined by this righthand rule All particles in the body have the same 00 Si w my La t For one particle in a rotating system l x l rimv rimwri mrfa 1quot m I x f Summing over all particles Ijmt mlrf1 mzrf2 mquot It is useful to define the Moment of m 0 Inertia 2 2 2 I mlri1mzri2 m3ri3 This allows us to write the angular momentum of a rigid body about a defined axis as Zrot and the kinetic energy can be written as 2 I 2 Km 4102 l 2 2 I 21 Example 10X13 A system with angular momentum of and about its center of mass Let s consider what the time derivative of the angular momentum of a single particle is Recall the Momentum Principle all A E Ever maxi da a div X P 1 X Product rule for derivatives dt dt A dt d A A A A x p V x p 0 Vector product of parallel vectors dt 7A X L A X rm Linear Momentum Principle I f FA x 7m Definition ofa new quantity torque Z A A A A A A A ath IquotA XFnezTA OI ALAVA X FnetAZTAAl The Angular Momentum Principle for a Point Particle dEA A A L39A or ALA TAAZ d1 The Linear Momentum Principle for a Point Particle or A AZ net We will soon see a large class of problems where the Angular Momentum Principle will be of fundamental importance TORQUE mg Tume torque rAFsine TORQUE xi TU 4r torque rAFsine

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