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Section 8-2: Kinematic Equations

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

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Section 8-2: Kinematic Equations

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  1. 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!

  2. 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!!

  3. 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?

  4. Example: Rotating Wheel • A wheel rotates with constant angular accelerationα = 3.5 rad/s2. It’s angular speed at time t = 0 is ω0= 2.0 rad/s. (A) Calculate the angular displacement Δθit 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 Δθfrom radians to revolutions: A full circle = 360º = 2πradians = 1 revolution 11.0 rad = 630º = 1.75 rev (C) Find the angular speed ωafter t = 2 s. Use: ω= ω0 + αt = 2 + (3.5)(2) = 9 rad/s

  5. Consider a CD player playing a CD. For the player to read a CD, the angular speed ωmust 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 Example: CD Player

  6. 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

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

  8. Example 8-7 r = 0.34m v0 = 8.4 m/s  Bicycle: v0 = 8.4 m/s. Comes to rest after 115 m. Diameter = 0.68 m (r = 0.34m) 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

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