PHY 101 1nd Edition Lecture 19Outline of Last Lecture I. Elastic Collision in 1DII. 6.4 Glancing CollisionsIII. Problem Solving: 2D CollisionsIV. Chapter 7: Rotational Motion V. 7.1 Angular Speed and Angular AccelerationVI. Rigid BodyVII. Sign of Angular Displacement, Speed & AccelerationOutline of Current Lecture VIII. 7.2 Rotational Motion Under Constant Angular AccelerationIX. 7.3 Angular & Linear Quantitiesa. Tangential Speed & AccelerationX. 7.4 Centripetal Accelerationa. Centripetal & Tangential AccelerationCurrent Lecture7.2 Rotational Motion Under Constant Angular Acceleration- Kinematic equations for rotational motion read similar to those of linear motiono ω = ωi + ato ΔΘ = ωit + ½ at2o ω2 = ωi2 + 2aΔΘ- exampleo a wheel rotates with a constant angular acceleration of 3.5 rad/s2o at t = 0, the wheel’s angular speed is 2.0 rad/s.o a) what is the wheel’s angular speed at t = 2s?o through what angle does the wheel rotate from t = 0 and t = 2s?o solution:  a) ω = ωi + at ω = 2.0 + 3.5(2) = 9.0 rad/s b) ΔΘ = ωit + ½ at2 ΔΘ = 2.0(2)+ ½ (3.5)(2)2 = 11.0 rad7.3 Angular & Linear QuantitiesThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.- Rigid body: every point has the same angular displacement, speed and acceleration- When an object rotates:o The arc length (linear distance) Δs = rΔΘ depends on the position r of a point I the rigid body- The velocity is tangential to its trajectory- Tangential speedo Vi =Δs/Δt = rΔΘ/Δt = rω- Tangential accelerationo Αr = Δvi/Δt = rΔω/Δt Tangential Speed & Acceleration- Tangential speed: v = rω- Acceleration: αi = rα- The farther away a point from the rotational axis, the greater its tangential speed & acceleration will be- example: tangential speed and accelerationo a CD rotates from rest with a constant angular acceleration to an angular speed of 31.4 rad/s in 0.9 s.o A) what is the angular acceleration?o B) through what angle does the disk rotate within the 0.9 s?o Solution: A) ω = ωi + at 31.4 = at; a = 31.4/0.9 = 34.9 rad/s2 B) ΔΘ = ωit + ½ at2 ½ (34.9)(0.9)2 = 14.1 rado C) the radius of the disk is 4.45cm. a tiny bug rides the rim of the disk. What is the tangential acceleration of the bug at t = 0.9 s? V = rω = (31.4)(4.45 x 10-2) = 1.4 m/s Tangential acceleration: A = αr = 34.9 (4.45 x 10-2) = 1.55 m/s2o What is the distance the bug rides during the 0.9s? Solution: from part (b) we have at =0.9s, ΔΘ = 14.1 rad the distance the bug rides:- Δx = rΔΘ = 4.45 x 10-2 (14.1) = 0.63 m7.4 Centripetal Acceleration- An object travelling in a circle, even though it moves with constant speed, it will have acceleration- The centripetal acceleration is due to the change in the direction of the velocityo A = vf – vi/Δt- Centripetal: “center – seeking” it can be shown that the direction of the velocity change points toward the center- Therefore, the acceleration is directed toward the center of the circular motion- For a constant speed circular motion, the magnitude of the centripetal acceleration iso Ac = v2/ r (v: tangential speed)- Do not confuse the centripetal acceleration with the tangential acceleration that we discussed earlier- Tangential acceleration is due to the change in the tangential speed- Centripetal acceleration is due to the change in the direction of velocityCentripetal & Tangential Acceleration- Tangential speed v = rωo Ac = v2/r = (rω)2/r = (rω)2- If the object is doing varying speed circular motion, then we have both tangential acceleration at & centripetal acceleration ac- The magnitude of the total acceleration satisfies the Pythagorean