AP EAPCET 07-Jul-2022 Shift 2 Paper

Given, $\omega=314\;\frac{\text{rad}}{\text{s}}$ at $t=2\text{ s}$ Let $\alpha=$ constant angular acceleration Then using $\omega_1=\omega_0+\alpha t$ We have, $\omega_1=\alpha t$ $\cdots [\because \omega_0=0]$ $\therefore 314=\alpha(2)$ $\Rightarrow \alpha=\frac{314}{2}\;\frac{\text{rad}}{\text{s}^2}$ Now, angle covered in 20 seconds is given by, $\theta_1=\omega_0 t+\frac{1}{2}\alpha t^2$ $\Rightarrow\theta_1=\frac{1}{2}\alpha t^2$ So, $\theta_1=\frac{1}{2}\times\frac{314}{2}\times(20)^2=3.14\text{ rad}$ At $t=20\text{ s},\text{ angular speed of wheel,}$ $\Rightarrow\omega_2=\omega_0+\alpha t$ $\Rightarrow\omega_2=0+\frac{314}{2}\times 20$ As, acceleration does not operates after $t=20\text{ s}$. So, wheel now rotates freely (with constant angular speed) upto $40\text{ s}$. Angular displacement in radians covered in another 20s of freewheeling is $\theta_2\;=\omega_1\times t$ $=314\times 20=628\text{ rad}$ Total angle covered by wheel, $\theta=(314+628)\text{ rad}$ $=942\text{ rad}$ $\therefore$ Number of revolutions of wheel $n=\frac{\theta}{2\pi}=\frac{942}{2\pi}$ $=\frac{314\times 3}{2\times 314}=150$