Question 388: Centre of Mass and Rotational Motion

Question

Two masses $m$ and 2 m are connected by a light string going over a pulley (disc) of mass 30 m with radius $r=0.1 \mathrm{~m}$. The pulley is mounted in a vertical plane and it is free to rotate about its axis. The 2 m mass is released from rest and its speed when it has descended through a height of 3.6 m is
$\_\_\_\_$ $\mathrm{m} / \mathrm{s}$. (Assume string does not slip and $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

Accepted Answer

When the mass $(2 \mathrm{~m})$ descends by a height $\mathrm{h}=3.6 \mathrm{~m}$. Its potential energy change is $-(2 \mathrm{~m})$ gh.
Since they are connected by a taut string, the mass ( m ) is pulled upward by the exact same distance $\mathrm{h}=3.6 \mathrm{~m}$.
So, its potential energy change is +mgh .
The net change in potential energy for the entire system is:
$$ \Delta \mathrm{U}_{\mathrm{net}}=\mathrm{mgh}-2 \mathrm{mgh}=-\mathrm{mgh} $$
This means the system loses an amount of potential energy equal to mgh.
Initially, the system is released from rest, so the initial kinetic energy is zero ( $\mathrm{K}_{\mathrm{i}}=0$ ).
Finally, when the 2 m mass has a speed v , the m mass also has a speed v . The pulley will be rotating with an angular speed $\omega$. Because the string does not slip, the linear speed of the string equals the tangential speed of the pulley rim:
$$ \omega=\frac{\mathrm{v}}{\mathrm{r}} $$
The total final kinetic energy ( $\mathrm{K}_{\mathrm{f}}$ ) is the sum of the translational kinetic energies of the two masses and the rotational kinetic energy of the pulley:
$$ \mathrm{K}_{\mathrm{f}}=\mathrm{K}_{	ext {trans }(\mathrm{m})}+\mathrm{K}_{	ext {trans }(2 \mathrm{~m})}+\mathrm{K}_{	ext {rot }(	ext { pulley })} $$
$$\Rightarrow $$ $$ K_f=\frac{1}{2} m v^2+\frac{1}{2}(2 m) v^2+\frac{1}{2} I \omega^2 $$
The pulley is given as a solid disc. The moment of inertia (I) of a uniform solid disc about its central axis is $\frac{1}{2} \mathrm{Mr}^2$, where M is the mass of the disc.
$$ \mathrm{I}=\frac{1}{2}(30 \mathrm{~m}) \mathrm{r}^2=15 \mathrm{mr}^2 $$
So, the rotational kinetic energy of pulley is, $\mathrm{K}_{\mathrm{rot}}=\frac{1}{2} \mathrm{I} \omega^2=\frac{1}{2}\left(15 \mathrm{mr}^2ight)\left(\frac{\mathrm{v}}{\mathrm{r}}ight)^2$
$\mathrm{K}_{\mathrm{rot}}=\frac{1}{2}\left(15 \mathrm{mr}^2ight)\left(\frac{\mathrm{v}^2}{\mathrm{r}^2}ight)$
$$\Rightarrow $$ $$ \mathrm{K}_{\mathrm{rot}}=7.5 \mathrm{mv}^2 $$
So, the final kinetic energy of the system is
$$ \mathrm{K}_{\mathrm{f}}=\frac{1}{2} m v^2+m v^2+7.5 m v^2 $$
$$\Rightarrow $$ $$ \mathrm{K}_{\mathrm{f}}=0.5 m v^2+1 m v^2+7.5 m v^2=9 m v^2 $$
By the principle of conservation of energy, the magnitude of the potential energy lost equals the kinetic energy gained:
$$ \left|\Delta U_{	ext {net }}ight|=K_f $$
$$\Rightarrow $$ $\mathrm{mgh}=9 \mathrm{mv}^2$
$$\Rightarrow $$ $v^2=\frac{g h}{9}$
$$\Rightarrow $$ $v^2=\frac{10 	imes 3.6}{9}$
$$\Rightarrow $$ $$ \mathrm{v}^2=4 \Rightarrow \mathrm{v}=2 \mathrm{~m} / \mathrm{s} $$
Therefore, the speed of the 2 m mass when it has descended through a height of 3.6 m is $2 \mathrm{~m} / \mathrm{s}$. Hence, the correct answer is 2 .

Centre of Mass and Rotational Motion

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