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STD 11 - 6. system of particles and rotational motion — NEET STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceNEETSTD 11 - 6. system of particles and rotational motion4 Marks
MCQ
$A$ disc of radius $r$ is rotating about its centre with an angular speed $\omega_0$. It is gently placed on a rough horizontal surface. After what time it will be in pure rolling ?
  • A
    $\frac{{{\omega _0}r}}{{2\mu g}}$
  • ✓
    $\frac{{{\omega _0}r}}{{3\mu g}}$
  • C
    $\frac{{{\omega _0}r}}{{\mu g}}$
  • D
    $\frac{3}{2}\frac{{{\omega _0}r}}{{\mu g}}$
✓

Answer

Correct option: B.
$\frac{{{\omega _0}r}}{{3\mu g}}$
b
The acceleration due to friction is given as,

$a=\frac{f}{m}$

$=\frac{\mu m g}{m}$

$=\mu g$

Angular acceleration due to friction is given as,

$\alpha=\frac{\tau}{I}$

$=\frac{f R}{\frac{1}{2} m R^{2}}$

$=\frac{2 \mu g}{R}$

The final angular velocity is given as,

$\omega=\omega_{0}-\alpha t$

$\omega=\omega_{0}-\frac{2 \mu g t}{R}$

The final velocity is given as,

$v=a t$

$v=\mu g t$

we know that,

$v=R \omega$

$\mu g t=R\left(\omega_{0}-\frac{2 \mu g t}{R}\right)$

$t=\frac{\omega_{0} R}{3 \mu g}$

Thus, the time is $\frac{\omega_{0} R}{3 \mu g}$ after that the disc will be in pure rolling.

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