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Rotational Dynamics — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsRotational Dynamics4 Marks
Question
State the expression for the acceleration of a circularly symmetric rigid body rolling without slipping down an inclined plane. Hence, deduce the acceleration of
(i) a ring
(ii) a solid cylinder
(iii) a hollow cylinder
(iv) a solid sphere, rolling without slipping down an inclined plane.
✓

Answer

A circularly symmetric rigid body, of radius $\mathrm{R}$ and radius of gyration $\mathrm{k}$, on rolling down an inclined plane of inclination $\theta$ has an acceleration
$\alpha=\frac{g \sin \theta}{1+\beta}=\frac{g \sin \theta}{1+\left(k^2 / R^2\right)}$
$
\therefore \alpha=\frac{g \sin \theta}{1+1}=\frac{1}{2} g \sin \theta \quad(=0.5 g \sin \theta)
$
(ii) Solid cylinder or disc : $I=\frac{1}{2} M R^2$, so that $\beta=\frac{1}{2}$.
$
\therefore \alpha=\frac{g \sin \theta}{1+\frac{1}{2}}=\frac{2}{3} g \sin \theta \quad(=0.667 g \sin \theta)
$
(iii) Spherical shell (hollow sphere) $: I=\frac{2}{3} M R^2$, so that
$
\begin{aligned}
& \beta=\frac{2}{3} . \quad \\
& \therefore a=\frac{g \sin \theta}{1+\frac{2}{3}}=\frac{3}{5} g \sin \theta \quad(=0.6 \mathrm{~g} \sin \theta) \quad \ldots
\end{aligned}
$
(iv) Solid sphere : $I=\frac{2}{5} M R^2$, so that $\beta=\frac{2}{5}$.
$
\therefore a=\frac{g \sin \theta}{1+\frac{2}{7}}=\frac{5}{7} g \sin \theta \quad(=0.714 g \sin \theta)
$

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