Question

A solid cylinder of mass $M$ and radius $R$ is rolling without slipping on a horizontal plane with a speed $v$. it the rolls up an inclined plane of inclination $\theta$ to a maximum height given by

• A. $\frac{v^2}{2g}$
• B. $\frac{v^2}{2g}\sin \theta$
• C. $\frac{3v^2}{4g}$
• D. $\frac{3v^2}{4g}\sin \theta$

Answer 1

Answer: (C)

$\frac{3v^2}{4g}$

$\begin{array}{c}\text{Given -}\ast \text{Solid cylinder mass}=M\text{and}\\ \text{radius}R\text{rolling without slip.}\end{array}$

$\begin{array}{c}\Rightarrow \text{kE kinetic energy}=\text{Translation}kE\\ +\text{Rotational}kE\\ \Rightarrow k{E}_T=\frac{1}{2}M{v}^2+\frac{1}{2}I{w}^{2}I\to \text{Moment of Inertia}\\ kE+=\frac{1}{2}M{v}^2+\frac{1}{2}\left(\frac{M{R}^2}{2}\right)w^2=\frac{3}{4}M{v}^2(v=\text{tuw})\end{array}$

$\begin{array}{c}\text{For maximum Height by energy conservation}\\ MgH=\frac{3}{4}M{v}^2\Rightarrow H=\frac{3v^2}{4g}\end{array}$