Question

Portion AB of the wedge shown in figure is rough and BC is smooth, A solid cylinder rolls without slipping from A to B. The ratio of transnational kinetic energy to rotational kinetic energy. when the cylinder reaches point C is

• A. 3/4
• B. 5
• C. 7/5
• D. 8/3

Answer 1

The correct option is B 5

$\frac{mgH}{2}=\frac{1}{2}m{v}^2+\frac{1}{2}\frac{m{r}^2}{2}\frac{V^2}{R^2}$

$=\frac{1}{2}m{v}^2+\frac{1}{4}m{v}^2$

$K.E_{T}K.E_R$

As cylinder comes $\frac{mgH}{2}$ downward-

Loss in PE $\frac{mgH}{2}$=gain in KE.

$\frac{mgH}{2}=\frac{3}{4}m{v}^2$

$\frac{1}{2}m{v}^2+\frac{3}{4}m{v}^2+\frac{1}{4}m{v}^2$

$\frac{5}{2}m{v}^2+\frac{1}{4}m{v}^2$

$K.E{)}_{Trans.}$$K.E_{Rotational}$

$\frac{K.E{)}_{Transation}}{K.E{)}_{Rotational}}=\frac{\frac{5}{4}m{v}^2}{\frac{1}{4}m{v}^2}=5$