2. 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!

3. NOTE These are ONLY VALID if all angular quantities are in radian units!! 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:

4. 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?

5. Example: Rotating Wheel A wheel rotates with constant angular acceleration α = 3.5 rad/s 2. It’s angular speed at time t = 0 is ω 0 = 2.0 rad/s. (A) Calculate the angular displacement Δθ it makes after t = 2 s. Use: Δθ = ω 0 t + (½)αt 2 = (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 = ω 0 + αt = 2 + (3.5)(2) = 9 rad/s

6. 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 r i = 23 mm = 2.3  10 -2 m & outer radius r o = 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: ω i = (v/r i ) = (1.3)/(2.3  10 -2 ) = 57 rad/s = 5.4  10 2 rev/min ω o = (v/r o ) = (1.3)/(5.8  10 -2 ) = 22 rad/s = 2.1  10 2 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)(ω i + ω f )t = (½)(57 + 22)(4,473 s) = 1.8  10 5 radians = 2.8  10 4 revolutions

7. 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

8. Wheel, moving on ground with axle velocity v. Relation between axle speed v & angular speed ω of the wheel: v = r ω Rolls with no slipping! ω

9. Example 8-7 Bicycle: v 0 = 8.4 m/s. Comes to rest after 115 m. Diameter = 0.68 m (r = 0.34m) a) ω 0 = (v 0 /r) = 24.7rad/s b) total θ = ( /r) = (115m)/(0.34m) = 338.2 rad = 53.8 rev c) α = (ω 2 - ω 0 2 )/(2θ). Stopped  ω = 0  α = 0.902 rad/s 2 d) t = (ω - ω 0 )/α. Stopped  ω = 0  t = 27.4 s r = 0.34m v 0 = 8.4 m/s  v = 0 d = 115m   v g = 8.4 m/s