JEE MAIN - Physics (2023 - 8th April Evening Shift - No. 3)
For particle P revolving round the centre O with radius of circular path $$\mathrm{r}$$ and angular velocity $$\omega$$, as shown in below figure, the projection of OP on the $$x$$-axis at time $$t$$ is
_8th_April_Evening_Shift_en_3_1.png)
$$x(t)=\operatorname{cos}\left(\omega t-\frac{\pi}{6} \omega\right)$$
$$x(t)=\operatorname{cos}(\omega t)$$
$$x(t)=r \cos \left(\omega t+\frac{\pi}{6}\right)$$
$$x(t)=r \sin \left(\omega t+\frac{\pi}{6}\right)$$
Explanation
_8th_April_Evening_Shift_en_3_2.png)
After time $t$, the angular displacement will be
$\theta=\omega t$
Total angular displacement from $x$-axis.
$$ \theta_{\text {total }}=\omega t+\frac{\pi}{6} \quad\quad\left(\because 30^{\circ}=\frac{\pi}{6}\right) $$
Now, OP has two component
The horizontal component will be the projection along $x$-axis
$$ =r \cos \left(\theta_{\text {Total }}\right)=r \cos \left(\omega t+\frac{\pi}{6}\right) $$
Comments (0)
