Section 8-2: Kinematic Equations • Recall: 1 dimensional kinematic equations for uniform (constant) acceleration (Ch. 2). • We’ve just seen analogies between linear & angular quantities: Displacement & Angular Displacement: x θ Velocity & Angular Velocity: v ω Acceleration & Angular Acceleration: a α • For α = constant, we can use the same kinematic equations from Ch. 2 with these replacements! The equations of motion for constant angular acceleration are the same as those for linear motion, substituting the angular quantities for the linear ones. For α = constant, & using the replacements x θ, v ω, a α we get the equations: NOTE These are ONLY VALID if all angular quantities are in radian units!! Example 8-6: Centrifuge Acceleration A centrifuge rotor is accelerated from rest to frequency f = 20,000 rpm in 30 s. a. Calculate its average angular acceleration. b. Through how many revolutions has the centrifuge rotor turned during its acceleration period, assuming constant angular acceleration? Example: Rotating Wheel • A wheel rotates with constant angular acceleration α = 3.5 rad/s2. It’s angular speed at time t = 0 is ω0 = 2.0 rad/s. (A) Calculate the angular displacement Δθ it makes after t = 2 s. Use: Δθ = ω0t + (½)αt2 = (2)(2) + (½)(3)(2)2 = 11.0 rad (630º) (B) Calculate the number of revolutions it makes in this time. Convert Δθ from radians to revolutions: A full circle = 360º = 2π radians = 1 revolution 11.0 rad = 630º = 1.75 rev (C) Find the angular speed ω after t = 2 s. Use: ω = ω0 + αt = 2 + (3.5)(2) = 9 rad/s Example: CD Player • Consider a CD player playing a CD. For the player to read a CD, the angular speed ω must vary to keep the tangential speed constant (v = ωr). A CD has inner radius ri = 23 mm = 2.3 10-2 m & outer radius ro = 58 mm = 5.8 10-2 m. The tangential speed at the outer radius is v = 1.3 m/s. (A) Find angular speed in rev/min at the inner & outer radii: ωi = (v/ri) = (1.3)/(2.3 10-2) = 57 rad/s = 5.4 102 rev/min ωo = (v/ro) = (1.3)/(5.8 10-2) = 22 rad/s = 2.1 102 rev/min • (B) Standard playing time for a CD is 74 min, 33 s (= 4,473 s). How many revolutions does the disk make in that time? θ = (½)(ωi + ωf)t = (½)(57 + 22)(4,473 s) = 1.8 105 radians = 2.8 104 revolutions Section 8-3: Rolling Motion • Without friction, there would be no rolling motion. • Assume: Rolling motion with no slipping Can use static friction • Rolling (of a wheel) involves: – Rotation about the Center of Mass (CM) PLUS – Translation of the CM • Wheel, moving on ground with axle velocity v. Rolls with no slipping! ω Relation between axle speed v & angular speed ω of the wheel: v = rω Example 8-7 Bicycle: v0 = 8.4 m/s. Comes to rest after 115 m. Diameter = 0.68 m (r = 0.34m) r = 0.34m v0 = 8.4 m/s a) ω0 = (v0/r) = 24.7rad/s b) total θ = (/r) = (115m)/(0.34m) = 338.2 rad = 53.8 rev c) α = (ω2 - ω02)/(2θ). Stopped ω = 0 α = 0.902 rad/s2 d) t = (ω - ω0)/α. Stopped ω = 0 t = 27.4 s v=0 d = 115m vg = 8.4 m/s