Wednesday, Dec. 1, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 1 PHYS 1441 – Section 002 Lecture #22 Wednesday, Dec. 1, 2010 Dr. Jaehoon Yu • Rotational Kinetic Energy • Angular Momentum • Angular Momentum Conservation • Similarities Between Linear and Rotational Quantities • Conditions for EquilibriumAnnouncements • The Final Exam – Date and time: 11am, Monday Dec. 13 – Place: SH103 – Comprehensive exam • Covers from CH1.1 – what we finish Wednesday, Dec. 8 • Plus appendices A.1 – A.8 • Combination of multiple choice and free response problems • Bring your Planetarium extra credit sheet to the class next Wednesday, Dec. 8, with your name clearly marked on the sheet! • Colloquium this today Wednesday, Dec. 1, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 2Wednesday, Dec. 1, 2010 3 PHYS 1441-002, Fall 2010 Dr. Jaehoon YuWednesday, Dec. 1, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 4 Rotational Kinetic Energy What do you think the kinetic energy of a rigid object that is undergoing a circular motion is? Since a rigid body is a collection of masslets, the total kinetic energy of the rigid object is Since moment of Inertia, I, is defined as Kinetic energy of a masslet, mi, moving at a tangential speed, vi, is ri mi θ"O x y vi The above expression is simplified asWednesday, Dec. 1, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 5 Example for Moment of Inertia In a system of four small spheres as shown in the figure, assuming the radii are negligible and the rods connecting the particles are massless, compute the moment of inertia and the rotational kinetic energy when the system rotates about the y-axis at angular speed w. Since the rotation is about y axis, the moment of inertia about y axis, Iy, is Thus, the rotational kinetic energy is Find the moment of inertia and rotational kinetic energy when the system rotates on the x-y plane about the z-axis that goes through the origin O. x y This is because the rotation is done about y axis, and the radii of the spheres are negligible. Why are some 0s? M M l l m m b b O = mii∑ri2 = Ml2= 2Ml2=12Iω2=122Ml2( )ω2= Ml2ω2= mii∑ri2 = Ml2= 2 Ml2+ mb2( )=12Iω2=122Ml2+ 2mb2( )ω2= Ml2+ mb2( )ω2 + Ml2 +m ⋅ 02 +m ⋅ 02 + Ml2 +mb2 +mb2Wednesday, Dec. 1, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 6 Kinetic Energy of a Rolling Sphere Since vCM=Rω Let’s consider a sphere with radius R rolling down the hill without slipping. R h q vCM ω"Since the kinetic energy at the bottom of the hill must be equal to the potential energy at the top of the hill What is the speed of the CM in terms of known quantities and how do you find this out?Wednesday, Dec. 1, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 7 Example for Rolling Kinetic Energy For solid sphere as shown in the figure, calculate the linear speed of the CM at the bottom of the hill and the magnitude of linear acceleration of the CM. Solve this problem using Newton’s second law, the dynamic method. Gravitational Force, Since the forces Mg and n go through the CM, their moment arm is 0 and do not contribute to torque, while the static friction f causes torque M h θ We know that What are the forces involved in this motion? Mg f Newton’s second law applied to the CM gives Frictional Force, Normal Force n We obtain Substituting f in dynamic equations = fRWednesday, Dec. 1, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 8 Work, Power, and Energy in Rotation Let’s consider the motion of a rigid body with a single external force F exerting on the point P, moving the object by Δs. The work done by the force F as the object rotates through the infinitesimal distance Δs=rΔθ is What is Fsinϕ? The tangential component of the force F. ΔWSince the magnitude of torque is rFsinϕ, F φ"O r Δθ"Δs What is the work done by radial component Fcosϕ? Zero, because it is perpendicular to the displacement. ΔWThe rate of work, or power, becomes How was the power defined in linear motion? The rotational work done by an external force equals the change in rotational Kinetic energy. The work put in by the external force then = F⋅ Δs=τΔθ=ΔWΔt =τ ΔθΔt= IΔωΔt⎛⎝⎜⎞⎠⎟ΔW= rF sinφ( )Δθ= F sinφ( )rΔθ τΔθ∑= IωΔωWednesday, Dec. 1, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 9 Angular Momentum of a Particle If you grab onto a pole while running, your body will rotate about the pole, gaining angular momentum. We’ve used the linear momentum to solve physical problems with linear motions, the angular momentum will do the same for rotational motions. L = mvr = mr2ϖ= IϖLet’s consider a point-like object ( particle) with mass m located at the vector location r and moving with linear velocity v L≡ r× pThe 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? Note that L depends on origin O. Why? Because r changes The direction of L is +z. What else do you learn? Since p is mv, the magnitude of L becomes If the linear velocity is perpendicular to position vector, the particle moves exactly the same way as a point on a rim. z x yWednesday, Dec. 1, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 10 Angular Momentum of a System of Particles The total angular momentum of a system of particles about some point is the vector sum of the angular momenta of the individual particles L=Since the individual angular momentum can change, the total angular momentum of the system can change. τext∑=Δ LΔtThus the time rate change of the angular momentum of a system of particles is equal to only the net external torque acting on the system Let’s consider a two particle system where the two exert forces on each other. Since these forces are the 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. + L2 + Ln = Li∑ L1τext∑=ΔLzΔt=Δ Iϖ( )Δt=I ΔϖΔt= IαFor a rigid body, the external torque is written F∑=Δ
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