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UT Arlington PHYS 1443 - Lecture Notes

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Monday, Nov. 15, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu11. Work, Power and Energy in Rotation 2. Angular Momentum 3. Angular Momentum and Torque4. Conservation Angular Momentum5. Similarity of Linear and Angular Quantities6. Conditions for EquilibriumPHYS 1443 – Section 003Lecture #20Monday, Nov. 15, 2004Dr. Jaehoon YuQuiz #3 next Monday!!Monday, Nov. 15, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu2Work, Power, and Energy in RotationLet’s consider a motion of a rigid body with a single external force F exerting on the point P, moving the object by ds.The work done by the force F as the object rotates through the infinitesimal distance ds=rdθ is What is Fsinφ?The tangential component of force F.dWSince the magnitude of torque is rFsinφ,FφOrdθdsWhat is the work done by radial component Fcosφ?Zero, because it is perpendicular to the displacement.dWThe rate of work, or power becomesPHow was the power defined in linear motion?The rotational work done by an external force equals the change in rotational energy. ∑τThe work put in by the external force thendWsdF ⋅=()cos( )Frdπφθ=−θτd=dtdW=dtdθτ=τω=αI=⎟⎠⎞⎜⎝⎛=dtdIω⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛=dtdddIθθωθτd∑=ωωdI=W∫∑=fdθθιθτ∫=fdIωωιωω222121ifIIωω−=()θφdrFsin=⎟⎠⎞⎜⎝⎛=θωωddI()θφrdF sin=Monday, Nov. 15, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu3Angular Momentum of a ParticleIf you grab onto a pole while running, your body will rotate about the pole, gaining angular momentum. We’ve used linear momentum to solve physical problems with linear motions, angular momentum will do the same for rotational motions.φsinmvrL=Let’s consider a point-like object ( particle) with mass m located at the vector location r and moving with linear velocity vprL ×≡The angular momentum L of this particle relative to the origin O is What do you learn from this?If the direction of linear velocity points to the origin of rotation, the particle does not have any angular momentum.What is the unit and dimension of angular momentum? 2/kg m s⋅Note that L depends on origin O. Why? Because r changesThe direction of L is +zWhat else do you learn? Since p is mv, the magnitude of L becomesIf the linear velocity is perpendicular to position vector, the particle moves exactly the same way as a point on a rim.21[]MLT−Monday, Nov. 15, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu4Angular Momentum and TorqueTotal external forces exerting on a particle is the same as the change of its linear momentum.Can you remember how net force exerting on a particle and the change of its linear momentum are related?∑τThus the torque-angular momentum relationshipThe same analogy works in rotational motion between torque and angular momentum. Net torque acting on a particle is The net torque acting on a particle is the same as the time rate change of its angular momentumdtpdF =∑dtLddtLd=∑τxyzOpφL=rxprmWhy does this work?Because v is parallel to the linear momentum()dtprd ×=dtpdrpdtrd×+×=dtpdr ×+= 0∑×= Frdtpdr ×=Monday, Nov. 15, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu5Angular Momentum of a System of ParticlesThe total angular momentum of a system of particles about some point is the vector sum of the angular momenta of the individual particlesinLLLLL∑=+++= ......21Since the individual angular momentum can change, the total angular momentum of the system can change.dtLdext=∑τThus the time rate change of the angular momentum of a system of particles is equal to the net external torque acting on the systemLet’s consider a two particle system where the two exert forces on each other.Since these forces are action and reaction forces with directions lie on the line connecting the two particles, the vector sum of the torque from these two becomes 0.Both internal and external forces can provide torque to individual particles. However, the internal forces do not generate net torque due to Newton’s third law.Monday, Nov. 15, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu6Example for Angular MomentumA particle of mass m is moving on the xy plane in a circular path of radius r and linear velocity v about the origin O. Find the magnitude and direction of angular momentum with respect to O.rxyvOLUsing the definition of angular momentumSince both the vectors, r and v, are on x-y plane and using right-hand rule, the direction of the angular momentum vector is +z (coming out of the screen)The magnitude of the angular momentum isLSo the angular momentum vector can be expressed askmrvL =Find the angular momentum in terms of angular velocity ω.LUsing the relationship between linear and angular speed pr ×=vmr ×=vrm ×=vrm ×=φsinmrv=D90sinmrv=mrv=kmrv= kmrω2=ω2mr=ωI=Monday, Nov. 15, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu7Angular Momentum of a Rotating Rigid BodyLet’s consider a rigid body rotating about a fixed axisiiiivrmL =Each particle of the object rotates in the xy plane about the z-axis at the same angular speed, ωατIdtdLzext==∑Thus the torque-angular momentum relationship becomesWhat do you see?Since I is constant for a rigid bodyMagnitude of the angular momentum of a particle of mass miabout origin O is mivirixyzOpφL=rxprmSumming over all particle’s angular momentum about z axis∑=iizLL()ω∑=iiizrmL2dtdLzα is angular accelerationThus the net external torque acting on a rigid body rotating about a fixed axis is equal to the moment of inertia about that axis multiplied by the object’s angular acceleration with respect to that axis.ω2iirm=()∑=iiirmω2ωI=dtdIω=αI=Monday, Nov. 15, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu8Example for Rigid Body Angular MomentumA rigid rod of mass M and length l is pivoted without friction at its center. Two particles of mass m1and m2are attached to either end of the rod. The combination rotates on a vertical plane with an angular speed of ω. Find an expression for the magnitude of the angular momentum.IThe moment of inertia of this system isαFirst compute net external torqueθτcos21lgm=1m1gxyOlm1m2θm2gIf m1= m2, no angular momentum because net torque is 0. If θ=+/−π/2, at equilibrium so no angular momentum.⎟⎠⎞⎜⎝⎛++==212314mmMlILωωFind an expression for the magnitude of the angular acceleration of the system when the rod makes an angle θ with the horizon.2τττ+=1extThus α becomes21mmrodIII++=222124141121lmlmMl ++=⎟⎠⎞⎜⎝⎛++=212314mmMl


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UT Arlington PHYS 1443 - Lecture Notes

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