Question

A disc of radius 0.1 m rolls without sliding on a horizontal surface with a velocity of 6 m/s. It then ascends a smooth continuous track as shown in figure. The height upto which it will ascend is (g=10 m/ $s^2$)

• A. 2.4 m
• B. 0.9 m
• C. 2.7 m
• D. 1.8 m

Answer 1

The correct option is D

1.8 m

Let m be the mass of the disc. then translational energy of the disc is $K_T$ = $\frac{1}{2}$ m $V^2$

When it ascends on a smooth track its rotational kinetic energy will remains same while traslational kinetic energy will go on decreasing . At highest point.

$K_T$ = mgh $\Rightarrow$ $\frac{1}{2}$ m $V^2$ = mgh

$\Rightarrow$ h = $\frac{V^2}{2g}$ = $\frac{{\left(6\right)}^2}{2\ast 10}$ = 1.8 m