An inclined plane makes an angle 3 0 ∘ with the horizontal. A solid sphere rolling down an inclined plane from rest without slipping has linear acceleration ( g = acceleration due gravity) ( sin 3 0 ∘

Question

An inclined plane makes an angle 3 0 ∘ with the horizontal. A solid sphere rolling down an inclined plane from rest without slipping has linear acceleration ( g = acceleration due gravity) ( sin 3 0 ∘ = 0.5 )

TestHub · Accepted Answer

The acceleration of a body rolling down an inclined plane without slipping is given by

$a = \frac{g sin θ}{1 + \frac{K ^{2}}{R ^{2}}}$

For a solid sphere, the radius of gyration $K$ is given by $K^{2} = \frac{2}{5} R^{2}$ , so $\frac{K ^{2}}{R ^{2}} = \frac{2}{5}$ .

Given $θ = 3 0^{\circ}$ , substituting the values we get

$a = \frac{g sin 3 0 ^{\circ}}{1 + \frac{2}{5}}$

$a = \frac{g \times \frac{1}{2}}{\frac{7}{5}}$

$a = \frac{5 g}{14}$

Answer: $\frac{5 g}{14}$

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