First slide

Rolling motion

Question

A small object of uniform density roll up a curve surface with an initial velocity v. It reaches up to a maximum height of $\frac{3 v^{2}}{4 g}$ with respect to the initial position. The object is

Easy

A

Ring
B

Solid sphere
C

Hollow sphere
D

Disc

Solution

Conserving energy $\large \frac{1}{2}m{v^2}\left( {1 + \frac{{{K^2}}}{{{R^2}}}} \right) = mgh$
Given $\large h = \frac{{3{v^2}}}{{4g}}$
So, $\large v^2=\frac {2gh}{1+\frac {k^2}{r^2}}=\frac {2g3v^2}{4g\left ( 1+\frac {k^2}{r^2} \right )}=\frac {6gv^2}{4g\left ( 1+\frac {k^2}{r^2} \right )}$
$\large 1=\frac {3}{2\left ( 1+\frac {k^2}{r^2} \right )}\;or\;1+\frac {k^2}{r^2}=\frac 32$
$\large or\;\frac{{{k^2}}}{{{r^2}}} = \frac{3}{2} - 1 = \frac{1}{2}$
$\large {k^2} = \frac{1}{2}{r^2}$ For disc, $\large {k^2} = \frac{\frac 12mr^2}{m}=\frac {r^2}{2}$
Hence the object is disc.

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