Question

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 $g=10 \,m s ^{-2}$]

Answer 1

Answer: 0.52

Answer 2

Given,
At t=0, the angular velocity \(\omega\) of the disk is 0
\(t=0, \omega=0\)
at \(t=\sqrt{\pi}, \omega=\alpha t=\frac{2}{3} \sqrt{\pi}\),
The linear velocity v of a point on the disk is given by \(v = \omega r\).
Substituting the given values we find, \(v=\omega r=\frac{2}{3} \sqrt{\pi}\)
Using the formula for angular displacement, \(\theta=\frac{1}{2} \alpha t^2\)
\(\theta=\frac{1}{2} \times \frac{2}{3} \times \pi=\frac{\pi}{3}\)
\(\theta=60^{\circ}\)

\(\begin{aligned} &\text{The vertical velocity component}\ v_y=v \sin 60=\frac{\sqrt{3}}{2} V \\ & h=\frac{u_y^2}{2 g}=\frac{\frac{3}{4} v^2}{2 g} \\ & h=\frac{\frac{3}{4} \times \frac{4}{9} \pi}{2 g} \\ & h=\frac{3 \pi}{9 \times 2 g}=\frac{\pi}{6 g} \end{aligned}\)

Total maximum height from the plane, \(H=\frac{R}{2}+h\)
Substituting \(h = \frac{\pi}{6g}\)​, where \(g = 10 \, \text{m/s}^2\)
\(\begin{aligned} & H=\frac{1}{2}+\frac{\pi}{6 \times 10} \\ & x=\frac{\pi}{6} ; x=0.52 \end{aligned}\)

So, the answer is 0.52