MCQ
A small disk is attached to the end of a light inextensible string, which passes through a hole in a frictionless horizontal tabletop. Initially the disk moves on a circle of radius $R$ with kinetic energy $K_0$. The string is then slowly pulled so that the disk finally rotates on a circle of radius $\frac{R}{\eta }$. What is the work $W$ done in pulling the string
  • A
    $W = \eta^2K_0$
  • $W = (\eta^2 - 1)K_0 $
  • C
    $W = (\eta - 1)K_0 $
  • D
    $ W = 0$

Answer

Correct option: B.
$W = (\eta^2 - 1)K_0 $
b
Torque by tension force is zero

$\Rightarrow$ angular momentum conservation

$\mathrm{I}_{1} \mathrm{\omega}_{1}=\mathrm{I}_{2} \mathrm{\omega}_{2}$

Let $\mathrm{\omega}_{1}=\mathrm{\omega}_{0}$

$\mathrm{mR}^{2} \mathrm{\omega}_{0}=\frac{\mathrm{mR}^{2}}{\mathrm{n}^{2}} \mathrm{\omega}^{2}$

$\Rightarrow \mathrm{\omega}_{2}=\mathrm{\omega}_{0} \eta^{2}$

$=\left(\mathrm{k}_{0}=\frac{1}{2} \mathrm{I}_{1} \mathrm{\omega}^{2}\right)$

$\mathrm{k}_{0}=\frac{1}{2} \times\left(\mathrm{mR}_{0}^{2}\right) \mathrm{\omega}_{0}^{2}=\frac{\mathrm{m} \mathrm{R}_{0}^{2} \mathrm{\omega}_{0}^{2}}{2}$

Here change in $\mathrm{K} . \mathrm{E} .=(\mathrm{WD})$

$\mathrm{k}_{f}=\frac{1}{2} \mathrm{I}_{2} \mathrm{\omega}_{2}^{2}=\frac{1}{2} \times\left(\frac{\mathrm{mR}^{2}}{\eta^{2}}\right) \times \mathrm{\omega}_{0}^{2} \eta^{4}$

$=\left(\frac{1}{2} \mathrm{mR}^{2} \mathrm{\omega}_{0}^{2}\right) \eta^{2}$

$\mathrm{WD}=\mathrm{k}_{\mathrm{f}}-\mathrm{k}_{\mathrm{i}}=\frac{1}{2} \mathrm{mR}^{2} \mathrm{\omega}_{0}^{2}\left(\eta^{2}-1\right)$

$=k_{0}\left(\eta^{2}-1\right)$

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