Question

The reel shown in figure has radius $R$ and moment of inertia $I$. 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's axle and incline are frictionless. The reel is wound counterclockwise so that the 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. $\sqrt{\frac{2mgd\sin \theta +k{d}^2}{I+m{R}^2}}$
• B. $\sqrt{\frac{mgd\sin \theta +k{d}^2}{I+m{R}^2}}$
• C. $\sqrt{\frac{2mgd\sin \theta -k{d}^2}{I+m{R}^2}}$
• D. $\sqrt{\frac{mgd\sin \theta -k{d}^2}{I+m{R}^2}}$

Answer 1

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

The block and end of the spring are pulled a distance $d$ up the incline and then released. The angular speed of the reel and speed of the block are related by $v=\omega R.$ The block-reel-earth system is isolated, so
$\Delta K+\Delta U=0.$
$\left(\frac{m{v}^2}{2}-0\right)+\left(\frac{I{\omega }^2}{2}-0\right)+(0-mgd\sin \theta )+\left(0-\frac{k{d}^2}{2}\right)=0$
$\frac{{\omega }^2}{2}\left(I+m{R}^2\right)=mgd\sin \theta +\frac{k{d}^2}{2}$
$\Rightarrow \omega =\sqrt{\frac{2mgd\sin \theta +k{d}^2}{I+m{R}^2}}$