Question 440: Centre of Mass and Rotational Motion

Question

A solid cylinder length is suspended symmetrically through two massless strings, as shown in the figure. The distance from the initial rest position, the cylinder should be unbinding the strings to achieve a speed of $$4 \mathrm{~ms}^{-1}$$, is ____________ cm. (take g = $$10 \mathrm{~ms}^{-2}$$)

Accepted Answer

In case of rotational motion of a rigid body like this, the kinetic energy is not just due to the linear motion, but also due to its rotation.
For a solid cylinder, the moment of inertia I is given by $\frac{1}{2} m r^2$. The kinetic energy due to rotation is given by $\frac{1}{2} I \omega^2$. But we also know that $\omega = \frac{v}{r}$, hence the rotational kinetic energy can be written as $\frac{1}{2} \frac{1}{2} m v^2 = \frac{1}{4} m v^2$, where $\frac{1}{2}$ is from the moment of inertia of the solid cylinder.
Therefore, the total kinetic energy (linear + rotational) when the string snaps is $\frac{1}{2} m v^2 + \frac{1}{4} m v^2 = \frac{3}{4} m v^2$.
Equating this to the potential energy $mgh$ and solving for h gives the result :
$$
h = \frac{v^2}{2g} 	imes \frac{3}{2} = 1.2 \, \mathrm{m} = 120 \, \mathrm{cm}
$$
Alternate Method :
Applying COE, we get
$$
m g h=\frac{1}{2} m v^2\left(1+\frac{\mathrm{K}^2}{r^2}ight)
$$
$\mathrm{K}=$ radius of gyration
For a solid cylinder, $\frac{\mathrm{K}^2}{r^2}=\frac{1}{2}$
$$
\begin{aligned}
	herefore h & =\frac{v^2}{2 g}\left(1+\frac{1}{2}ight) \\
& =\frac{16}{2 	imes 10} 	imes \frac{3}{2} \\
& =1.2 \mathrm{~m} \\
& =120 \mathrm{~cm}
\end{aligned}
$$

Centre of Mass and Rotational Motion

A solid cylinder length is suspended symmetrically through two massless strings, as shown in the figure. The distance from the initial rest position, the cylinder should be unbinding the strings to achieve a speed of $4 \mathrm{~ms}^{-1}, is ____________ cm. (take g = 10 \mathrm{~ms}^{-2}$)

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