A solid disc of radius 'a' and mass ' $m$ ' rolls down without slipping on anclined plane making an angle

Question:

A solid disc of radius 'a' and mass ' $m$ ' rolls down without slipping on anclined plane making an angle $\theta$ with the horizontal. The

acceleration of the disc will be $\frac{2}{b} g \sin \theta$ where $b$ is__________ (Round off to the Nearest Integer)

$(g=$ acceleration due to gravity)

$(\theta=$ angle as shown in figure $)$

Solution:

Ans. (3)

$\mathrm{a}=\frac{\mathrm{g} \sin \theta}{1+\frac{\mathrm{I}}{\mathrm{mR}^{2}}}=\frac{\mathrm{g} \sin \theta}{1+\frac{1}{2}}=\frac{2}{3} \mathrm{~g} \sin \theta$

$\mathrm{b}=3$

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