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Lecture 17 Goals: • Chapter 12 Define center of mass Analyze rolling motion Introduce and analyze torque Understand the equilibrium dynamics of an extended object in response to forces Employ “conservation of angular momentum” concept Assignment: HW7 due tomorrow Wednesday, Exam Review Physics 207: Lecture 17, Pg 1 A special point for rotation System of Particles: Center of Mass (CM) A supported object will rotate about its center of mass. Center of mass: Where the system is balanced ! Building a mobile is an exercise in finding centers of mass. m1 + m2 m1 + m2 mobile Physics 207: Lecture 17, Pg 3 System of Particles: Center of Mass How do we describe the “position” of a system made up of many parts ? Define the Center of Mass (average position): For a collection of N individual point like particles whose masses and positions we know: mi ri N i 1 RCM M RCM m2 m1 r1 r2 y x (In this case, N = 2) Physics 207: Lecture 17, Pg 4 Sample calculation: Consider the following mass distribution: m at ( 0, 0) mi ri N RCM i 1 M XCM î YCM ĵ ZCM k̂ 2m at (12,12) m at (24, 0) XCM = (m x 0 + 2m x 12 + m x 24 )/4m meters RCM = (12,6) YCM = (m x 0 + 2m x 12 + m x 0 )/4m meters (12,12) 2m XCM = 12 meters YCM = 6 meters m (0,0) m (24,0) Physics 207: Lecture 17, Pg 5 Connection with motion... So a rigid object that has rotation and translation rotates about its center of mass! And Newton’s Laws apply to the center of mass mi ri N K TOTAL K Rotation K Translatio n 2 1 K TOTAL K Rotation 2 MVCM RCM i 1 M K m v m ( r ) R p p p p For a point p rotating: p p p p 2 1 2 1 2 VCM p p p p Physics 207: Lecture 17, Pg 7 2 Work & Kinetic Energy: Recall the Work Kinetic-Energy Theorem: K = WNET This applies to both rotational as well as linear motion. K I m(V 1 2 2 f 2 i 1 2 2 CM f V 2 CM i ) WNET What if there is rolling? Physics 207: Lecture 17, Pg 8 Demo Example : A race rolling down an incline Two cylinders with identical radii and total masses roll down an inclined plane. The 1st has more of the mass concentrated at the center while the 2nd has more mass concentrated at the rim. Which gets down first? M Two cylinders with radius R and mass m h q A) Mass 1 B) Mass 2 C) They both arrive at same time M who is 1st ? Physics 207: Lecture 17, Pg 9 Same Example : Rolling, without slipping, Motion A solid disk is about to roll down an inclined plane. What is its speed at the bottom of the plane ? M h q M v? Physics 207: Lecture 17, Pg 10 Rolling without slipping motion Again consider a cylinder rolling at a constant speed. 2VCM VCM CM Physics 207: Lecture 17, Pg 11 Motion Again consider a cylinder rolling at a constant speed. Rotation only VTang = R CM Both with |VTang| = |VCM | 2VCM VCM CM If acceleration acenter of mass = - aR Sliding only VCM CM Physics 207: Lecture 17, Pg 12 Example : Rolling Motion A solid cylinder is about to roll down an inclined plane. What is its speed at the bottom of the plane ? Use Work-Energy theorem Disk has radius R M h q Mgh = ½ Mv2 + ½ ICM 2 and M v? v =R Mgh = ½ Mv2 + ½ (½ M R2 )(v/R)2 = ¾ Mv2 v = 2(gh/3)½ Physics 207: Lecture 17, Pg 13 How do we reconcile force, angular velocity and angular acceleration? Physics 207: Lecture 17, Pg 16 Angular motion can be described by vectors With rotation the distribution of mass matters. Actual result depends on the distance from the axis of rotation. Hence, only the axis of rotation remains fixed in reference to rotation. We find that angular motions may be quantified by defining a vector along the axis of rotation. We can employ the right hand rule to find the vector direction Physics 207: Lecture 17, Pg 17 The Angular Velocity Vector • The magnitude of the angular velocity vector is ω. • The angular velocity vector points along the axis of rotation in the direction given by the right-hand rule as illustrated above. • As increased the vector lengthens Physics 207: Lecture 17, Pg 18 From force to spin (i.e., ) ? A force applied at a distance from the rotation axis gives a torque q FTangential =|FTang| sin q r a FTangential F Fradial r Fradial If a force points at the axis of rotation the wheel won’t turn Thus, only the tangential component of the force matters With torque the position & angle of the force matters NET = |r| |FTang| ≡ |r| |F| sin q Physics 207: Lecture 17, Pg 19 Rotational Dynamics: What makes it spin? a A force applied at a distance FTangential from the rotation axis F NET = |r| |FTang| ≡ |r| |F| sin q Fradial r Torque is the rotational equivalent of force Torque has units of kg m2/s2 = (kg m/s2) m = N m NET = r FTang = r m aTang =rmra = (m r2) a For every little part of the wheel Physics 207: Lecture 17, Pg 20 For a point mass NET = m r2 a The further a mass is away from this axis the greater the inertia (resistance) to rotation (as we saw on Wednesday) NET = I a a FTangential F Frandial r This is the rotational version of FNET = ma Moment of inertia, I ≡ Si mi ri2 , is the rotational equivalent of mass. If I is big, more torque is required to achieve a given angular acceleration. Physics 207: Lecture 17, Pg 21 Rotational Dynamics: What makes it spin? A force applied at a distance from the rotation axis gives a torque a FTangential F NET = |r| |FTang| ≡ |r| |F| sin q Fradial r A constant torque gives constant angular acceleration if and only if the mass distribution and the axis of rotation remain constant. Physics 207: Lecture 17, Pg 22 Torque, like , is a vector quantity |r| |F| sin q (2) |Ftangential | |r| (3) |F| |rperpendicular to line of action | Direction is parallel to the axis of rotation with respect to the “right hand rule” r sin q line of action F cos(90°q) = FTang. r a 90°q q F F F Fradial Magnitude is given by (1) r r r And for a rigid object = I a Physics 207: Lecture 17, Pg 23 Example : Rolling Motion Newton’s Laws: N SFx Max Mg sin q f SFy Ma y 0 N Mg cos q S CM Ia CM MR a CM fR 1 2 f Mg x dir q 2 a CM R ax Notice rotation CW (i.e. negative) when ax is positive! Combining 3rd and 4th expressions gives f = Max / 2 Top expression gives Max + f = 3/2 M ax = Mg sin q So ax =2/3 Mg sin q Physics 207: Lecture 17, Pg 24 Statics Equilibrium is established when Translatio nal motion Rotational motion SFNet 0 S Net 0 In 3D this implies SIX expressions (x, y & z) Physics 207: Lecture 17, Pg 29 Example Two children (60 kg and 30 kg) sit on a horizontal teeter-totter. The larger child is 1.0 m from the pivot point while the smaller child is trying to figure out where to sit so that the teeter-totter remains motionless. The teeter-totter is a uniform bar of 30 kg its moment of inertia about the support point is 30 kg m2. Assuming you can treat both children as point like particles, what is the initial angular acceleration of the teeter-totter when the large child lifts up their legs off the ground (the smaller child can’t reach)? For the static case: Rotational motion S Net 0 Physics 207: Lecture 17, Pg 30 Example: Soln. Use S Net 0 N 30 kg 30 kg 0.5 m 300 N 300 N 60 kg 1m 600 N Draw a Free Body diagram (assume g = 10 m/s2) 0 = 300 d + 300 x 0.5 + N x 0 – 600 x 1.0 0= 2d + 1 – 4 d = 1.5 m from pivot point Physics 207: Lecture 17, Pg 31 Recap Assignment: HW7 due tomorrow Wednesday: review session Physics 207: Lecture 17, Pg 32