JEE MAIN - Physics (2019 - 9th April Morning Slot - No. 10)
The following bodies are made to roll up
(without slipping) the same inclined plane from
a horizontal plane. : (i) a ring of radius R, (ii)
a solid cylinder of radius
R/2 and (iii) a solid
sphere of radius
R/4 . If in each case, the speed
of the centre of mass at the bottom of the incline
is same, the ratio of the maximum heights they
climb is :
20 : 15 : 14
4 : 3 : 2
2 : 3 : 4
10 : 15 : 7
Explanation
Total kinetic energy of a rolling body is given as
$$ E_{\text {total }}=\frac{1}{2} m v^2\left[1+\frac{K^2}{R^2}\right] $$
where, $K$ is the radius of gyration.
Using conservation law of energy,
$$ \begin{array}{rlrl} \frac{1}{2} m v^2\left[1+\frac{K^2}{R^2}\right] =m g h \\\\ \text { or } h =\frac{v^2}{2 g}\left[1+\frac{K^2}{R^2}\right] \end{array} $$
For ring, $ \frac{K^2}{R^2}=1$
$$ \Rightarrow h_1=\frac{v^2}{2 g}[1+1]=\frac{2 v^2}{2 g}=\frac{v^2}{g} $$
For solid cylinder, $\frac{K^2}{R^2}=\frac{(R / 2 \sqrt{2})^2}{(R / 2)^2}$
$$ \begin{aligned} & =\frac{R^2}{8} \times \frac{4}{R^2}=\frac{1}{2} \\\\ \Rightarrow h_2 & =\frac{v^2}{2 g}\left[1+\frac{1}{2}\right]=\frac{3 v^2}{4 g} \end{aligned} $$
For solid sphere, $\frac{K^2}{R^2}=\frac{2}{5}$
$$ \Rightarrow h_3=\frac{v^2}{2 g}\left[1+\frac{2}{5}\right]=\frac{7 v^2}{10 g} $$
So,the ratio of $h_1, h_2$ and $h_3$ is
$$ \begin{aligned} h_1: h_2: h_3 & =\frac{v^2}{g}: \frac{3 v^2}{4 g}: \frac{7}{10} \frac{v^2}{g} \\\\ & =1: \frac{3}{4}: \frac{7}{10}=20: 15: 14 \end{aligned} $$
$$ E_{\text {total }}=\frac{1}{2} m v^2\left[1+\frac{K^2}{R^2}\right] $$
where, $K$ is the radius of gyration.
Using conservation law of energy,
$$ \begin{array}{rlrl} \frac{1}{2} m v^2\left[1+\frac{K^2}{R^2}\right] =m g h \\\\ \text { or } h =\frac{v^2}{2 g}\left[1+\frac{K^2}{R^2}\right] \end{array} $$
For ring, $ \frac{K^2}{R^2}=1$
$$ \Rightarrow h_1=\frac{v^2}{2 g}[1+1]=\frac{2 v^2}{2 g}=\frac{v^2}{g} $$
For solid cylinder, $\frac{K^2}{R^2}=\frac{(R / 2 \sqrt{2})^2}{(R / 2)^2}$
$$ \begin{aligned} & =\frac{R^2}{8} \times \frac{4}{R^2}=\frac{1}{2} \\\\ \Rightarrow h_2 & =\frac{v^2}{2 g}\left[1+\frac{1}{2}\right]=\frac{3 v^2}{4 g} \end{aligned} $$
For solid sphere, $\frac{K^2}{R^2}=\frac{2}{5}$
$$ \Rightarrow h_3=\frac{v^2}{2 g}\left[1+\frac{2}{5}\right]=\frac{7 v^2}{10 g} $$
So,the ratio of $h_1, h_2$ and $h_3$ is
$$ \begin{aligned} h_1: h_2: h_3 & =\frac{v^2}{g}: \frac{3 v^2}{4 g}: \frac{7}{10} \frac{v^2}{g} \\\\ & =1: \frac{3}{4}: \frac{7}{10}=20: 15: 14 \end{aligned} $$
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