Question 270: Centre of Mass and Rotational Motion

Question

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 :

Accepted Answer

Total kinetic energy of a rolling body is given as
$$
E_{	ext {total }}=\frac{1}{2} m v^2\left[1+\frac{K^2}{R^2}ight]
$$
where, $K$ is the radius of gyration.
Using conservation law of energy,
$$
\begin{array}{rlrl}
\frac{1}{2} m v^2\left[1+\frac{K^2}{R^2}ight]  =m g h \\
 	ext { or }  h  =\frac{v^2}{2 g}\left[1+\frac{K^2}{R^2}ight]
\end{array}
$$
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, $\frac{K^2}{R^2}=\frac{(R / 2 \sqrt{2})^2}{(R / 2)^2}$
$$
\begin{aligned}
& =\frac{R^2}{8} 	imes \frac{4}{R^2}=\frac{1}{2} \\
\Rightarrow h_2 & =\frac{v^2}{2 g}\left[1+\frac{1}{2}ight]=\frac{3 v^2}{4 g}
\end{aligned}
$$
For solid sphere, $\frac{K^2}{R^2}=\frac{2}{5}$
$$
\Rightarrow h_3=\frac{v^2}{2 g}\left[1+\frac{2}{5}ight]=\frac{7 v^2}{10 g}
$$
So,the ratio of $h_1, h_2$ and $h_3$ is
$$
\begin{aligned}
h_1: h_2: h_3 & =\frac{v^2}{g}: \frac{3 v^2}{4 g}: \frac{7}{10} \frac{v^2}{g} \\
& =1: \frac{3}{4}: \frac{7}{10}=20: 15: 14
\end{aligned}
$$

Answer

20 : 15 : 14

Answer

4 : 3 : 2

Answer

2 : 3 : 4

Answer

10 : 15 : 7

Centre of Mass and Rotational Motion

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