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

Correct Answer: 0.52

\(\theta=\frac{1}{2}\alpha t^2\)
\(=\frac{1}{2}\times\frac{2}{3}\pi=\frac{\pi}{3}=60\degree\)
\(V_{cm}=\alpha t\)
The resultant velocity of point P is represented as \(V = αt\), making an angle of 60° with the horizontal, where \(u_y = αt\ sin 60°.\)

\(y_{max}=\frac{1}{2}+\frac{u_y^2}{2g}\)
\(=\frac{1}{2}+\frac{\alpha^{2}t^23}{20\times4}\)

\(=\frac{1}{2}+\frac{\pi}{60}\)
\(=0.52\)

Answer: 0.52