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  4. A solid sphere rolls down without slipping on an inclined plane at angle 60° over a distance of 10 m. The acceleration (in ms -2 ) is

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Q. A solid sphere rolls down without slipping on an inclined plane at angle $60^{\circ}$ over a distance of $10\, m$. The acceleration (in $ms ^{-2}$ ) is

System of Particles and Rotational Motion

Solution:

Here, $\theta=60^{\circ}, l=10\, m ,\, a =?$
For solid sphere, $K^{2}=\frac{2}{5} R^{2}$
$\therefore a=\frac{9.8 \sin 60^{\circ}}{1+\frac{2}{5}}$
$\left(\because a =\frac{g \sin \theta}{1+\frac{K^{2}}{R^{2}}}\right)$
Or $a=\frac{5}{7} \times 9.8 \times \frac{\sqrt{3}}{2}=6.06\, ms ^{-2}$