Question

The flywheel of an engine is rotating at 35.0 rad/s. When the engine is turned off, the flywheel slows at a constant rate and stops in 19.0 s. Calculate (a) the angular acceleration of the flywheel, (b) the angle through which the flywheel rotates in stopping, and (c) the number of revolutions made by the flywheel in stopping.

Answer 1

Answer:

a.$-1.8rad/s^2$

b.$332.5 rad$

c.52.9 rev

Explanation:

We are given that

Initial angular speed of wheel=$\omega_0=35 rad/s$

Time=t=19 s

Final angular speed=$\omega=0 rad/s$

a.We have to find the angular acceleration of the wheel.

We know that

Angular acceleration$,\alpha=-\frac{\omega_0}{t}$

Using the formula

$\alpha=-\frac{35}{19}=-1.8rad/s^2$

b.$\theta=\frac{1}{2}\omega_0t$

Using the formula

$\theta=\frac{1}{2}(35)\times 19=332.5 rad$

$\theta=332.5 rad$

c.$2\pi$ rad=1 rev

1 rad=$\frac{1}{2\pi}$ rev

332.5 rad=$\frac{332.5}{2\pi}=52.9 rev$

Number of revolutions made by the flywheel in stopping=52.9 rev