Question

At t = 0, a flywheel has an angular velocity of 4.7 rad/s, a constant angular acceleration of −0.25 rad/s², and a reference line at θ₀ = 0. (a) Through what maximum angle θmax will the reference line turn in the positive direction? What are the (b) first and (c) second times the reference line will be at θ = 1 2 θmax? At what (d) negative time and (e) positive time will the reference line be at θ = -9.9 rad?

Answer 1

Answer:

Given that

ωo = 4.7 rad/s

α =  −0.25 rad/s²

θ₀ = 0

a)

For maximum turn angle ,the final angular (ω)speed of wheel should be zero.

ω² = ω²o - 2 α θ

0² = 4.7² - 2 x 0.25 x θ

θ=44.18 rad ≅44 rad  ( max)

b) and c)

We know that

$\theta=\omega _ot-\dfrac{1}{2}\alpha t^2$

θ = 1 /2 θmax

θ = 22 rad

$\theta=\omega _ot-\dfrac{1}{2}\alpha t^2$

$22=4.7t-\dfrac{1}{2}\times 0.25\times t^2$

By solving above equation we get

t= 32.12 s

t=5.47 s

d) and e)

θ = -9.9 rad

$\theta=\omega _ot-\dfrac{1}{2}\alpha t^2$

$-9.9=4.7t-\dfrac{1}{2}\times 0.25\times t^2$

By solving above equation we get

t= - s

t= 39.6 s