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 axle and the incline are frictionless. The reel is wound counterclockwise so that the spring stretches a distance d from its un-stretched position and the reel is then released from rest. The angular speed of the reel when the spring is again un-stretched is

A
$\sqrt{\frac{2 mgd \sin θ + {kd}^{2}}{I + {mR}^{2}}}$
B
$\sqrt{\frac{mgd \sin θ + {kd}^{2}}{I + {mR}^{2}}}$
C
$\sqrt{\frac{mgd \sin θ - {kd}^{2}}{I + {mR}^{2}}}$
D
$\sqrt{\frac{2 mgd \sin θ - {kd}^{2}}{I + {mR}^{2}}}$

Solution:

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 the speed of the block are related by v = $ω$ R. The block-reel-Earth system is isolated, so

$∆ K + ∆ U = 0$

$(\frac{1}{2} {mv}^{2} - 0) + (\frac{1}{2} {Iω}^{2} - 0)$ + $(0 - mgd \sin θ) + (0 - \frac{1}{2} {kd}^{2}) = 0$

$\frac{1}{2} ω^{2} (I + {mR}^{2}) = mgd \sin θ + \frac{1}{2} {kd}^{2}$

$\Rightarrow ω = \sqrt{\frac{2 mgd \sin θ + {kd}^{2}}{I + {mR}^{2}}}$

Solution:

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