Question

A wheel rotates with a constant angular acceleration ${\omega }_0=-0.75$ rad/s at time t=0.Consider angular position at t=0 as ${\theta }_0=0$.Find out
(a)angular speed at t=5s
(b)it's angular position at time t=5s
(c)how much time will it take to reach an angular position of $\theta =10$ radian?

Answer 1

The kinematics law for rotational motion are:

$\omega ={\omega }_0+\alpha t$ ..(1)

$\theta ={\theta }_0+{\omega }_{0}t+\frac{1}{2}\alpha t^2$ ..(2)

${\omega }^2-{\omega }_0^2-2\alpha \theta$ ...(3) Here,

$\omega$ is angular velocity at time $t$,

${\omega }_0$ is angular velocity at time $t=0$ ,

$\alpha$ is angular acceleration ,

$\theta$ is angular position at time $t$ and ${\theta }_0$ is angular position at time $t=0$

(a) $\omega =-0.75+(0.5\times 5)$ from(1)

$=-0.75+2.5=1.75rad/s$

(b) $\theta =0+(-0.75\times 50+\frac{1}{2}\times (0.5\times 25)$ from (2)

$=-3.75+6.25=2.5rad$

(c) $10=0+(-0.75\times t)+\frac{1}{2}\times (0.5\times t^2)$ from(2)

$10=-0.75+0.25t^2$

$0.25t^2-0.75t-10=0$

$t^2-3t-40-0$ (divinding by 0.25)

$t=\frac{3\pm \sqrt{9+4\times 40}}{2}=\frac{3\pm \sqrt{169}}{2}=\frac{3+13}{2}=8s$ (taking positive root)