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