Kinematic Equations

Kinematic Equations Displacement & Angular Displacement: x  θ Recall: 1 dimensional kinematic equations for uniform (constant) acceleration (Ch. 2). We’ve jseen 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!

These are ONLY VALID if all angular quantities are 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: 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? Solution: a. The final angular velocity is 2100 rad/s, so the acceleration is 70 rad/s2. b. The total angle is 3.15 x 104 rad, which is 5000 rev.

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 = 1.8  105 radians = 2.8  104 revolutions 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

 Can use static friction 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 the ground with axle velocity v. Relation between axle speed v & angular speed ω of the wheel: v = rω Rolls with no slipping! ω

Example 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