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  4. A rupee coin starting from rest rolls down a distance of 1 m on an inclined plane at angle of 30° with the horizontal. Assuming that g=9.81 ms -2, time taken is

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Q. A rupee coin starting from rest rolls down a distance of $1\, m$ on an inclined plane at angle of $30^{\circ}$ with the horizontal. Assuming that $g=9.81\, ms ^{-2}$, time taken is

System of Particles and Rotational Motion

Solution:

Here, $l=1\, m,\, \theta=30^{\circ},\, g =9.81\, ms ^{-2},\, t=$ ?
$\because t=\sqrt{\frac{2 l\left(I+K^{2} / R^{2}\right)}{g \sin \theta}}$
For a rupee coin, $K^{2}=\frac{1}{2} R^{2}$
$\Rightarrow t=\sqrt{\frac{2 \times 1(1+1 / 2)}{9.81 \sin 30^{\circ}}}$
$=\sqrt{\frac{6}{9.81}}=0.78\, s$