Question

The reel shown in the figure has radius $R$ and the moment of interval $I$. One end of the block of mass $m$ is connected to a spring of force constant $k$, and the other end is fastened to a cord wrapped around the reel. The reel axel and the incline are frictionless. The reel is wound counterclockwise so that a spring stretches a distance $d$ from its unstretched position and the reel is then released from rest. The angular speed of the reel when the spring is again unstretched is

• A. $\sqrt{\frac{2mgd\sin \theta +k{d}^2}{1+m{R}^2}}$
• B. $\sqrt{\frac{mgd\sin \theta +k{d}^2}{1+m{R}^2}}$
• C. $\sqrt{\frac{mgd\sin \theta -k{d}^2}{1+m{R}^2}}$
• D. $\sqrt{\frac{2mgd\sin \theta -k{d}^2}{1+m{R}^2}}$

Answer 1

The correct option is A $\sqrt{\frac{2mgd\sin \theta +k{d}^2}{1+m{R}^2}}$

$x=d$

$mg\sin \theta +Kx-T=a$

$a=R\alpha$

$T.R=I\alpha$

$\omega ={\omega }_0+\alpha t,{\omega }_0=0$

$mg\sin \theta +Kx=\alpha [MR+\frac{I}{R}]$ $\alpha ={\left[\frac{mg\sin \theta +Kx}{M{R}^2+I}\right]}^R$

$\frac{{\omega }^2}{2}=\frac{I}{R}\left(\frac{mg\sin \theta d+\frac{K{d}^2}{2}}{M{R}^2+I}\right){\omega }^2=\frac{I}{R}\left(\frac{2mg\sin \theta d+K{d}^2}{M{R}^2+I}\right)\omega =\sqrt{\frac{2mg\sin \theta d+K{d}^2}{M{R}^2+I}}$

Now for a displacement $dx$

$Rd\theta =dx$

${\omega }_f^2={\omega }_i^2+2\alpha \theta$

$({\omega }_f-{\omega }_i)({\omega }_f+{\omega }_i)2\alpha \frac{dx}{R}$

${\int }_0^{\omega }2\omega d\omega ={\int }_0^{d}2\alpha dc$