Question

A wheel initially has an angular velocity of 18 rad/s, but it is slowing at a constant rate of 2 rad/s 2 . By the time it stops, it will have turned through approximately how many revolutions?

Answer 1

Answer:13 revolution

Step-by-step explanation:

Given  data

Wheel initial angular velocity$\left ( \omega \right )$=18 rad/s

Contant angular deaaceleration$\left ( \alpha \right )$=2$rad/s^2$

Time required to stop wheel completely=t sec

$\omega =\omega_0 + \aplha t$

0 =18 +$\left ( -2\right )t$

t=9 sec

Therefore angle turn in 9 sec

$\theta$=$\omega_{0} t$+$\frac{1}{2}$$\left ( \alpha\right )t^{2}$

$\theta$=$18\times 9$+$\frac{1}{2}$$\left ( -2\right )\left ( 9\right )^2$

$\theta$=81rad

therefore no of turns(n) =$\frac{81}{2\times \pi}$

n=12.889$\approx$13 revolution