PHYS 101 Lecture 14 - Angular Acceleration 14 - 1

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PHYS 101 Lecture 14 - Angular acceleration 14 - 1

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.

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.

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.