Lecture 17 Goals: Chapter 12 Define center of mass Analyze rolling motion Use Work Energy relationships Introduce torque Equilibrium of objects in response to forces & torques Assignment: HW7 due tomorrow Wednesday, Exam Review 1

Combining translation and rotation Objects can have translational energy Objects can have rotational energy Objects can have both K = ½ m v2 + ½ I w2

1st: A special point for rotation Center of Mass (CM) A free object will rotate about its center of mass. Center of mass: Where the system is balanced ! A mobile exploits this centers of mass. m1 m2 + m1 m2 + mobile

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: RCM m2 m1 r2 r1 y x (In this case, N = 2)

Consider the following mass distribution: m at ( 0, 0) 2m at (12,12) Sample calculation: Consider the following mass distribution: m at ( 0, 0) 2m at (12,12) m at (24, 0) XCM = (m x 0 + 2m x 12 + m x 24 )/4m meters YCM = (m x 0 + 2m x 12 + m x 0 )/4m meters XCM = 12 meters YCM = 6 meters (24,0) (0,0) (12,12) m 2m RCM = (12,6)

Connection with motion... An unconstrained rigid object with rotation and translation rotates about its center of mass! Any point p rotating: VCM  p

Work & Kinetic Energy: Work Kinetic-Energy Theorem: K = WNET Applies to both rotational as well as linear motion. What if there is rolling without slipping ?

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 q h v ?

Rolling without slipping motion Again consider a cylinder rolling at a constant speed. 2VCM CM VCM

Motion Again consider a cylinder rolling at a constant speed. Rotation only VTang = wR Both with |VTang| = |VCM | Sliding only 2VCM VCM CM CM CM VCM If acceleration acenter of mass = - aR

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 M q h v ? Disk has radius R Mgh = ½ Mv2 + ½ ICM w2 and v =wR Mgh = ½ Mv2 + ½ (½ M R2 )(v/R)2 = ¾ Mv2 v = 2(gh/3)½

How do we reconcile force, angular velocity and angular acceleration?

From force to spin (i.e., w) ? A force applied at a distance from the rotation axis gives a torque a FTangential F q Fradial r r FTangential Fradial =|FTang| sin q 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

Rotational Dynamics: What makes it spin? FTangential a r Fradial F NET = |r| |FTang| ≡ |r| |F| sin q 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 = r m r a = (m r2) a For every little part of the wheel

Torque The further a mass is away from this axis the greater the inertia (resistance) to rotation FTangential a r Frandial F NET = I a 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.

NET = I a ≡ |r| |F| sin q Rotational Dynamics F FTangential F NET = I a ≡ |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.

Torque, like w, has pos./neg. values Magnitude is given by (1) |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” And for a rigid object  = I a r sin q line of action F cos(90°-q) = FTang. r a 90°-q q F F F Fradial r r r

Statics Equilibrium is established when In 3D this implies SIX expressions (x, y & z) 1

Example Two children (30 kg & 60 kg) sit on a horizontal teeter-totter. The larger child is at the end of the bar and 1.0 m from the pivot point. The smaller child is trying to figure out where to sit so that the teeter-totter remains horizontal and motionless. The teeter-totter is a uniform bar of length 3.0 m and mass 30 kg. 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)? The moment of inertia of the bar about the pivot is 30 kg m2. For the static case:

Draw a Free Body diagram (assume g = 10 m/s2) 60 kg 30 kg 0.5 m 300 N 600 N N Example: Soln. 30 kg 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

Recap Assignment: HW7 due tomorrow Wednesday: review session 1