Slide 1AnnouncementsMoment of InertiaCalculation of Moments of InertiaEx.10 – 11 Rigid Body Moment of InertiaSlide 6Torque & Angular AccelerationRolling Motion of a Rigid BodyMore Rolling Motion of a Rigid BodyRotational Kinetic EnergyKinetic Energy of a Rolling SphereEx. 10 – 16: Rolling Kinetic EnergyWork, Power, and Energy in RotationAngular Momentum of a ParticleExample for Rigid Body Angular MomentumConservation of Angular MomentumEx. 11 – 3 Neutron StarSimilarity Between Linear and Rotational MotionsConditions for EquilibriumMore on Conditions for EquilibriumHow do we solve equilibrium problems?Ex. 12 – 3: Seesaw BalancingSeesaw Example Cont’dEx. 12.4 for Mechanical EquilibriumExample 12 – 6Example 12 – 6 cont’dWednesday, July 6, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu1PHYS 1443 – Section 001Lecture #16Wednesday, July 6, 2011Dr. Jaehoon Yu•Calculation of Moment of Inertia•Torque and Angular Acceleration•Rolling Motion & Rotational Kinetic Energy•Work, Power and Energy in Rotation•Angular Momentum & Its Conservation•EquilibriumThe final homework is homework #9, due 10pm, Saturday, July 9!!Wednesday, July 6, 2011 2Announcements•Quiz #3 results–Class average: 20.1/35•Equivalent to 59.7/100•Extremely consistent: 62.5 and 61.4–Top score: 33/35•Quiz #4 tomorrow, Thursday, July 7–Beginning of the class–Covers CH10.1 through what we learn today (CH12.2?)•Please to not forget the planetarium special credit sheet submission tomorrow•Final Comprehensive Exam–8 – 10am, Monday, July 11 in SH103–Covers CH1.1 through what we learn Thursday, July 7–Mixture of multiple choice and free response problems•Bring your two special projects during the intermissionPHYS 1443-001, Summer 2011 Dr. Jaehoon YuWednesday, July 6, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu3Moment of Inertia Rotational Inertia:What are the dimension and unit of Moment of Inertia? I ≡kg ⋅m2ML2⎡⎣⎤⎦Measure of resistance of an object to changes in its rotational motion. Equivalent to mass in linear motion.Determining Moment of Inertia is extremely important for computing equilibrium of a rigid body, such as a building.I �For a group of particlesFor a rigid body miri2i∑2r dm�Dependent on the axis of rotation!!!Wednesday, July 6, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu4Calculation of Moments of InertiaMoments of inertia for large objects can be computed, if we assume the object consists of small volume elements with mass, Δmi.It is sometimes easier to compute moments of inertia in terms of volume of the objects rather than their massUsing the volume density, ρ, replace dm in the above equation with dV.The moment of inertia for the large rigid object isHow can we do this?I = limΔmi→ 0ri2Δmii∑ dmr2ρ =dmdVThe moments of inertia becomesI = ρr2dV∫Example: Find the moment of inertia of a uniform hoop of mass M and radius R about an axis perpendicular to the plane of the hoop and passing through its center.xyROdmThe moment of inertia isI = r2dm∫What do you notice from this result?The moment of inertia for this object is the same as that of a point of mass M at the distance R.=R2dm∫=MR2dm =ρdVWednesday, July 6, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu5Ex.10 – 11 Rigid Body Moment of InertiaCalculate the moment of inertia of a uniform rigid rod of length L and mass M about an axis perpendicular to the rod and passing through its center of mass.The line density of the rod is What is the moment of inertia when the rotational axis is at one end of the rod.xyLxdxλ =MLso the masslet is dmThe moment of inertia is II = r2dm∫Will this be the same as the above. Why or why not?Since the moment of inertia is resistance to motion, it makes perfect sense for it to be harder to move when it is rotating about the axis at one end.=λdx=MLdx= r2dm∫=x2ML− L / 2L /2∫dx=ML13x3⎡⎣⎢⎤⎦⎥− L /2L /2=M3LL2⎛⎝⎜⎞⎠⎟3− −L2⎛⎝⎜⎞⎠⎟3⎡⎣⎢⎢⎤⎦⎥⎥=M3 LL34⎛⎝⎜⎞⎠⎟=ML212=x2ML0L∫dx=ML13x3⎡⎣⎢⎤⎦⎥0L=M3LL( )3− 0⎡⎣⎤⎦=M3LL3( )=ML23Wednesday, July 6, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu6Check out Figure 10 – 20 for moment of inertia for various shaped objectsWednesday, July 6, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu7Torque & Angular AccelerationLet’s consider a point object with mass m rotating on a circle.What does this mean?The tangential force Ft and the radial force FrThe tangential force Ft isWhat do you see from the above relationship?mrFtFrWhat forces do you see in this motion? Ft=matThe torque due to tangential force Ft is τ =Ftr τ =IαTorque acting on a particle is proportional to the angular acceleration.What law do you see from this relationship?Analogs to Newton’s 2nd law of motion in rotation.How about a rigid object?rdFtδmOThe external tangential force δFt is δFt= τ =limδτ→ 0δτ∑= dτ∫=The torque due to tangential force Ft isThe total torque isdt =What is the contribution due to radial force and why?Contribution from radial force is 0, because its line of action passes through the pivoting point, making the moment arm 0.mra= =matr =mr2αIa=tmad =mrd atF rd = r2δm( )α α limδm→ 0r2δm∑=α r2dm∫=IaWednesday, July 6, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu8Rolling Motion of a Rigid BodyWhat is a rolling motion?To simplify the discussion, let’s make a few assumptionsLet’s consider a cylinder rolling on a flat surface, without slipping. A more generalized case of a motion where the rotational axis moves together with an objectUnder what condition does this “Pure Rolling” happen?The total linear distance the CM of the cylinder moved isThus the linear speed of the CM isA rotational motion about a moving axis1. Limit our discussion on very symmetric objects, such as cylinders, spheres, etc2. The object rolls on a flat surfaceRθ ss=Rθs=dtdsvCMThe condition for a “Pure Rolling motion”dtdRRRqWednesday, July 6, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu9More Rolling Motion of a Rigid BodyAs we learned in rotational motion, all points in a rigid body moves at the same angular speed but at different linear speeds.At any given time, the point that comes to P has 0 linear speed while the point at P’ has twice the speed of CMThe magnitude of the linear acceleration of the CM isA rolling motion can be interpreted as the sum of Translation and
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