At time t=0, a disk of radius 1 m starts to roll without slipping… - Vidyadip

MCQ

At time $t=0$, a disk of radius $1 m$ starts to roll without slipping on a horizontal plane with an angular acceleration of $\alpha=\frac{2}{3} rad s ^{-2}$. A small stone is stuck to the disk. At $t=0$, it is at the contact point of the disk and the plane. Later, at time $t=\sqrt{\pi} s$, the stone detaches itself and flies off tangentially from the disk. The maximum height (in $m$ ) reached by the stone measured from the plane is $\frac{1}{2}+\frac{x}{10}$. The value of $x$ is. . . . . . .[Take $g=10 m s ^{-2}$.]

A
$0.20$
B
$0.30$
✓
$0.52$
D
$0.60$

✓

Answer

Correct option: C.

$0.52$

c
(image)

$\text { At } t =0, \omega=0$

$\text { at } t =\sqrt{\pi}, \omega=\alpha t =\frac{2}{3} \sqrt{\pi}, v =\omega r =\frac{2}{3} \sqrt{\pi}$

$\theta=\frac{1}{2} \alpha t ^2$

$\theta=\frac{1}{2} \times \frac{2}{3} \times \pi=\frac{\pi}{3}$

$\theta=60^{\circ}$

$v _y= v \sin 60=\frac{\sqrt{3}}{2} V$

$h =\frac{ u _Y^2}{2 g }=\frac{\frac{3}{4} v ^2}{2 g }$

$h =\frac{\frac{3}{4} \times \frac{4}{9} \pi}{2 g }$

$h =\frac{3 \pi}{9 \times 2 g }=\frac{\pi}{6 g }$

Maximum height from plane, $H=\frac{R}{2}+h$

$H =\frac{1}{2}+\frac{\pi}{6 \times 10}$

$x =\frac{\pi}{6} ; x =0.52$

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