Question

A block slides down a plane, inclined at ${30}^o$ to the horizontal, with an acceleration $\alpha$. A disc rolling without slipping down the same inclined plane would have an acceleration

• A. $\frac{1}{3}\alpha$
• B. $\frac{1}{2}\alpha$
• C. $\frac{2}{3}\alpha$
• D. $\frac{5}{7}\alpha$

Answer 1

The correct option is C $\frac{2}{3}\alpha$

$\begin{array}{c}\text{acceleration of the block}\alpha =g\sin {30}^{\circ }=\frac{g}{2}\\ \text{Let the mass of the dise be}\left(m\right)\text{and Radius}\left(R\right)\\ \text{Therefore moment of Inertia will be}\\ I=\frac{m{R}^2}{2}\end{array}$

$\begin{array}{c}\text{By linear Translatory motion equation-}\\ \begin{array}{c}mg\sin {30}^{\circ }-f=m{\alpha }^{\prime }\\ \frac{mg}{2}-f=m{\alpha }^{\prime }-1\end{array}\\ \text{By Rotatory motion equation of torgue}\\ fR=I\cdot \frac{{\alpha }^{\prime }}{R}=\frac{mR{\alpha }^{\prime }}{2}\end{array}$

$\begin{array}{c}\Rightarrow f=\frac{m{\alpha }^{\prime }}{2}-\left(2\right)\\ \text{From}ea\left(0\text{and}\right(2)\text{we get}\\ {\alpha }^{\prime }=\frac{g}{3}=\frac{2}{3}\alpha \Rightarrow {\alpha }^{\prime }=\frac{2}{3}\alpha \end{array}$