An inclined plane makes an angle of 3 0 ∘ with the horizontal. A solid sphere rolling down this inclined plane from rest without slipping has a linear acceleration ( g = acceleration due to gravity, s

Question

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

TestHub · Accepted Answer

The acceleration of a body rolling down an inclined plane without slipping is given by $a = \frac{g sin θ}{1 + \frac{I}{M R ^{2}}}$

For a solid sphere, the moment of inertia about its center is $I = \frac{2}{5} M R^{2}$

Given $θ = 3 0^{\circ}$ , we have $sin 3 0^{\circ} = \frac{1}{2}$

Substituting the values into the formula:

$a = \frac{g ( \frac{1}{2} )}{1 + \frac{2}{5}}$

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

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

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

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