Question:
A sphere of mass $2 \mathrm{~kg}$ and radius $0.5 \mathrm{~m}$ is rolling with an initial speed of $1 \mathrm{~ms}^{-1}$ goes up an inclined plane which makes an angle of $30^{\circ}$ with the horizontal plane, without slipping. How low will the sphere take to return to the starting point $\mathrm{A}$ ?
Correct Option: , 3
Solution:
(3) $\mathrm{a}=\frac{\mathrm{g} \sin \theta}{1+\frac{\mathrm{I}}{\mathrm{mR}^{2}}}=\frac{5}{7} \times \frac{10}{2}=\frac{25}{7}$
$\mathrm{t}=\frac{2 \mathrm{v}_{0}}{\mathrm{a}}=\frac{2 \times 1 \times 7}{25}$
$=0.56$