PHYS 101 Lecture 14 - Angular acceleration 14 - 1 Lecture 14 - Angular acceleration What’s important: • angular acceleration • angular velocity and acceleration as vectors Demonstrations: • bike wheel, rotating stool Circular Motion with Variable Speed Continuing with circular motion, let us consider the case where ω is not a constant. This means that there are two components to the acceleration, a centripetal component towards the centre and a tangential component along the edge: a tangential = at net acceleration a centrip = ac • Just as we define linear acceleration in terms of a change in velocity, we can define an angular acceleration α by a change in angular velocity [angular acceleration] ≡ α = [change in ω] / [change in time] α ≡ Δω / Δt = dω /dt as Δt → 0 Link between α and a Starting with v = ωr, the angular acceleration is α = d (v/r) / dt But if r is constant (for circular motion) α = (dv/ dt) / r. Again, with constant r, the rate dv/dt must be the tangential acceleration at, so α = at / r or at = αr. © 2001 by David Boal, Simon Fraser University. All rights reserved; further copying or resale is strictly prohibited. PHYS 101 Lecture 14 - Angular acceleration 14 - 2 Thus, we have the parallel equations (for constant radius) arc length = θ r speed = ω r tangential acceleration = α r Thus, we expect to find relations amongst θ, ω, α and t just as with x, v, a and t. Suppose we have uniform angular acceleration angular acceleration ! take t the area angular frequency angle " # ! area = ( " - "o)t/2 "o area ! area = "ot t t (assume # = 0 at t = 0) The area of the angular acceleration graph gives a linear time dependence for the angular velocity: ωf = ωo + αt ---> αt = ωf - ωo The area of the angular velocity graph gives a quadratic time dependence for the angle: θ = ωot + (ωf - ωo)t / 2 or 2 θ = ωot + αt /2. Finally, an alternate expression for the angle which does not explicitly show the time dependence can be found by rearranging the above expressions: 2 2 θ = (ωf - ωo ) / 2α. © 2001 by David Boal, Simon Fraser University. All rights reserved; further copying or resale is strictly prohibited. PHYS 101 Lecture 14 - Angular acceleration 14 - 3 Example A carousel, starting from rest, attains a frequency of one revolution every 5 sec after 10 sec. Assuming uniform acceleration, what is the angular acceleration? If the carousel is 10 m in diameter, what is at after 10 sec? Solution After 10 seconds, T = 5 s f = 1/T = 1/5 s-1 ω = 2πf = 2π/5 rad/sec We can use ω = ωo + αt -1 ωo = 0 plus ω = 2π/5 s (at t = 10 s) gives 2π/5 = 0 + 10α. The angular acceleration is then α = 2π / (5 • 10) = π/25 radians/sec2. The tangential acceleration at (time-independent magnitude if α = const), can be found from the angular acceleration: 2 2 at = αr = (π/25) • 5 = π/5 m/s (or 0.63 m/s ) NOTE: even though at is constant, atot is not constant, since ac changes as the carousel accelerates. © 2001 by David Boal, Simon Fraser University. All rights reserved; further copying or resale is strictly prohibited. .