JEE Advanced 2017 Paper 2 Physics: Paragraph 1

The string unwinds from the axle, so the instantaneous axis of rotation is the point where the string leaves the axle, at distance $r$ from the centre. Hence $v=\omega r$ and $a=\alpha r$. With tension $T$: $Mg-T=Ma$ and $Tr=I\alpha=\tfrac12MR^2\,a/r$, so $T=\dfrac{MR^2}{2r^2}a=2Ma$ because $R^2/r^2=4$. Therefore $Mg=3Ma$ and $a=g/3=3.33\ \mathrm{m\,s^{-2}}$ with tension $T=2Mg/3$. The value $6.67$ comes from using $r=R$, and $2.00$ comes from taking $I=MR^2$ for the disc. The tension $T=2Mg/3$ is less than the weight, as it must be for downward acceleration, and the answer is independent of the mass of the yo-yo. A thinner axle slows the descent further because the rotational inertia then weighs more heavily against the gravitational pull.

From the previous result the acceleration is constant at $a=g/3=10/3\ \mathrm{m\,s^{-2}}$. Starting from rest, $v^2=2as=2\times\tfrac{10}{3}\times0.60=4$, so $v=2.0\ \mathrm{m\,s^{-1}}$. The string unwinds from the axle, so $\omega=v/r=2.0/0.10=20\ \mathrm{rad\,s^{-1}}$. Energy conservation confirms this: $Mgs=\tfrac12Mv^2+\tfrac12I\omega^2=\tfrac12Mv^2(1+2)$, so $v^2=\tfrac{2gs}{3}=4$. Dividing by $R$ instead of $r$ gives $10\ \mathrm{rad\,s^{-1}}$, and $24.5$ arises from using $a=5\ \mathrm{m\,s^{-2}}$. As a cross-check, the ratio of rotational to translational kinetic energy is $\dfrac{I\omega^2}{Mv^2}=\dfrac{\tfrac12MR^2\,(v/r)^2}{Mv^2}=\dfrac{R^2}{2r^2}=2$. Only one third of the lost potential energy becomes translational kinetic energy; the remaining two thirds is stored as rotational kinetic energy, which explains the slow descent compared with free fall.