JEE MAIN - Physics (2023 - 13th April Morning Shift - No. 10)
A disc is rolling without slipping on a surface. The radius of the disc is $$R$$. At $$t=0$$, the top most point on the disc is $$\mathrm{A}$$ as shown in figure. When the disc completes half of its rotation, the displacement of point A from its initial position is
_13th_April_Morning_Shift_en_10_1.png)
$$R\sqrt {({\pi ^2} + 1)} $$
$$2R$$
$$R\sqrt {({\pi ^2} + 4)} $$
$$2R\sqrt {(1 + 4{\pi ^2})} $$
Explanation
_13th_April_Morning_Shift_en_10_2.png)
$$ \begin{aligned} & \text { Displacement of } C M=R \omega \frac{T}{2}=\pi r \\\\ & \Rightarrow \text { Displacement of } A =\sqrt{(2 R)^2+(\pi R)^2} \\\\ & =R \sqrt{\pi^2+4} \end{aligned} $$
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