Phys 250 Ch9 p1 Angular velocity an object which rotates about a fixed axis has an average angular velocity  av : usually rad/s but sometime rpm, rps instantaneous angular velocity is given by:  r s Chapter 9: Rigid Bodies and Rotational Motion

Phys 250 Ch9 p2 Angular Acceleration: the rate of change of angular speed ac=2rac=2r aT=raT=r Example: In a hammer throw, a 7.25 kg shot is swung in a circle 5 times and then released. The shot moves with an average radius of 2.1 m and an average angular acceleration of 2.3 rad/s 2. What is the average tangential force and what is the maximum centripetal force on the hammer?

Phys 250 Ch9 p3 Rotation with constant angular acceleration (just like linear 1-d) watch units consistency!!!

Phys 250 Ch9 p4 Example: The wheel on a moving car slows uniformly from 70m rad/s to 42 rad/s in 4.2 s. What is the angular acceleration of the wheel? What angle does the wheel rotate in those 42 s? How far does the car go if the radius of the wheel is 0.32 m.

Phys 250 Ch9 p5 Torque: the rotational analogue of force Torque = force x moment arm  = FL=F r sin  moment arm = perpendicular distance through which the force acts L F L F L F Example: The bolts holding a head gasket are to be “torqued down” to 90 N-m. If a 45 cm wrench is used, what force should be applied perpendicular to the wrench handle?

Phys 250 Ch9 p6 Example: The crank arm of a bicycle pedal is 16.5 cm long. If a 52.0 kg woman puts all her weicht on one pedal, how much torque is developed when the crank is horizontal? How much torque is developed when the pedal is 15º from the top?

Phys 250 Ch9 p7 Equilibrium: stability, steadiness, balance etc. Mechanical Equilibrium: absence of change in motion => Net Force = 0 ! (usually, no motion) With Rotational Equilibrium Rotational Equilibrium: absence of change in rotation (usually: no rotation) => net torque is zero  i = 0 about any axis! Positive torque for counterclockwise rotation:  = F L Watch signs for torque L F L F Negative torque for clockwise rotation:  =  F L

Phys 250 Ch9 p8 Center of Gravity (CG) aka Center of Mass: the point of an object from which it could be suspended without tending to rotate. The point where all the mass of an object can be considered to be located. CG does not need to be located within the physical object! Horseshoe, for example usually easily identified from symmetry. Example: A 5 kg mass hangs from the 5 cm mark on a 1 meter long rod. An unknown mass hangs from the 85 cm mark. The rod has a mass of 2.0 kg and is balanced at the 35 cm mark. What is the unknown mass?

Phys 250 Ch9 p9 Example: A sign weighing 400 N is suspended from the end of a 350 N horizontal uniform beam. What is the tension in the cable? w 35 o

Phys 250 Ch9 p10 Elasticity “stretchiness/springiness” -how materials respond to stress compression tension shear “Stretch-ability” = amount of stress (applied force) produces a strain (elongation/compression/shear) Hooke’s Law: the amount of stretching is proportional to the applied force. F = k x 1 Ton 2 Tons x x The details of such springiness depends upon the size and shape of the material as well as how the forces are applied

Phys 250 Ch9 p11 Elastic Limit: the maximum stress (force) which can be applied to an object without resulting in permanent deformation. Plastic Deformation: the permanent deformation which results when a materials elastic limit has been exceeded. Ultimate strength: greatest tension (or compression or shear) the material can withstand. *snap* A malleable or ductile material has a large range of plastic deformation. Fatigue: small defects reduce materials strength well below original strength.

Phys 250 Ch9 p12 Young’s Modulus: how things stretch (elastically) stress: force per area = F/A strain: fractional change in length = change in length per original length =  L/Lo Elastic modulus = stress/strain Young’s modulus (for stretching in one direction) compression tension L A L0L0 A L0L0 A L A

Phys 250 Ch9 p13 Example: A steel elevator cable supports a load of 900 kg. The cable has a diameter of 2.0 cm and an initial length of 24 m. Find the stress and the strain on the cable and the amount that it stretches under this load.

Phys 250 Ch9 p14 Torque and Moment of Inertia For a single mass: ac=2rac=2r F T = ma T F T r =ma T r  = m  r r = mr 2  moment of inertia I = mr 2  = I  looks like F = ma for a system of objects ( a rigid object) I =  m i r i 2

Phys 250 Ch9 p15 LL R R2R2 R2R2 R a b a b RR

Phys 250 Ch9 p16 Example: A cylindrical winch of radius R and moment of inertia I is free to rotate without friction. A cord of negligible mass is wrapped about the shaft and attached to a bucket of mass m. What is the acceleration of the bucket when it is released?

Phys 250 Ch9 p17 Angular Momentum L = I  (like p = m v ) + Angular momentum is conserved in the absence of external torques L start = L end for a point mass moving in a circle L = mvr = mr 2  conservation of angular momentum implies Kepler’s 3 rd law! Example: Ann ice skater starts spinning at a rate of 1.5 rev/s with arms extended. He then pulls his arms close to his body, decreasing his moment of inertia to ¾ of its initial value. What is the skater’s final angular velocity?

Phys 250 Ch9 p18 Rotational Kinetic Energy for a single point particle for a solid rotating object

Phys 250 Ch9 p19 Combined Translation and Rotation KE = KE translation + KE rotation when rolling without slipping s =  r v =  r a =  r,  : angular acceleration The Great Race lost PE = gained KE same radius, object with the smallest I has most v => wins race