Question

An object of uniform density rolls up a curved surface with an initial velocity $v$. It reaches upto a maximum height of $\frac{3v^2}{4g}$ with respect to the initial position. The object might be

• A. hollow sphere
• B. disc
• C. ring
• D. solid sphere
  

Answer 1

The correct option is B disc

The $K.E.$ of the rolling object is converted into potential energy at height
$h=\frac{3v^2}{4g}$
So, by the law of conservation of mechanical energy,
$\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=mgh\frac{1}{2}m{v}^2+\frac{1}{2}I{\left(\frac{v}{R}\right)}^2=mg\left(\frac{3v^2}{4g}\right)\frac{1}{2}I{\left(\frac{v}{R}\right)}^2=\frac{\mathrm{3m}}{4}{v}^2-\frac{1}{2}m{v}^2\frac{1}{2}I{\left(\frac{v}{R}\right)}^2=\frac{1}{4}m{v}^{2}orI=\frac{1}{2}m{R}^2$