Question

To start a lawn mower, you must pull on a rope wound around theperimeter of a flywheel. After you pull the rope for 0.95 s, theflywheel is rotating at 4.5 revolutions per second, at which pointthe rope disengages. This attempt at starting the mower does notwork, however, and the flywheel slows, coming to rest 0.24 s afterthe disengagement. Assume constant acceleration during both spin-upand spin-down.
(a) Determine the average angularacceleration during the 4.5-s spin-up and again during the 0.24-sspin-down.

(b) What is the maximum angular speed reachedby the flywheel?

(c) Determine the ratio of the number ofrevolutions made during spin-up to the number made duringspin-down.

Answer 1

Answer:

29.76245 rad/s², -117.80972 rad/s²

28.2743 rad/s

3.95833

Explanation:

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation

t = Time taken

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{4.5\times 2\pi-0}{0.95}\\\Rightarrow \alpha=29.76245\ rad/s^2$

Angular acceleration during speed up is 29.76245 rad/s²

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-4.5\times 2\pi}{0.24}\\\Rightarrow \alpha=-117.80972\ rad/s^2$

Angular acceleration during spin down is -117.80972 rad/s²

Angular speed is given by

$\omega=2\pi 4.5=28.2743\ rad/s$

Maximum angular speed reached by the flywheel is 28.2743 rad/s

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=0\times t+\frac{1}{2}\times 29.76245\times 0.95^2\\\Rightarrow \theta=13.4303\ rad$

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=2\pi 4.5\times 0.24+\frac{1}{2}\times -117.80972\times 0.24^2\\\Rightarrow \theta=3.39292\ rad$

The ratio would be $\dfrac{13.4303}{3.39292}=3.95833$