Q6.5 Angular acceleration (S)
A flywheel completes 40 revolutions as it slows from an angular speed of 1.5 rad s-1 to a complete stop. Assuming uniform acceleration- What is the angular acceleration?
- What is the time required for it to come to rest?
- How much time is required to complete the first 20 of the 40 revolutions?
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Solution
The angular acceleration is constant, therefore we can use the constant acceleration equations (Key Point 5.11).The angular acceleration can be calculated using the third constant acceleration equation, with ω=0, ω0=1.5 rad s-1 and the displacement Δθ=40 revolutions
![\[ \omega^2 = \omega_0^2 + 2 \alpha \Delta \theta. \]](mastermathpng-0.png)
This gives α=-4.47×10-3 rad s-2 (3.s.f)
Using the first constant acceleration equation
ω=ω0+αt,and solving for t we get t=336 s (3 s.f).
To complete the first 20 of the 40 revolutions we need to use the second constant acceleration equation. The total displacement is Δθ=20×2π.
![\[ \Delta \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \]](mastermathpng-1.png)
Solving this quadratic for t (and choosing the smallest root, the one less than t=336 s) we obtain t=98.1 s (3.s.f).