Question 88: Centre of Mass and Rotational Motion

Question

A uniform bar of length 12 cm and mass 20 m lies on a smooth horizontal table. Two point masses $m$ and $2 m$ are moving in opposite directions with same speed of $v$ and in the same plane as the bar, as shown in figure. These masses strike the bar simultaneously and get stuck to it. After collision the entire system is rotating with angular frequency $\omega$. The ratio of $v$ and $\omega$ is :

Accepted Answer

The mass of the rod is $M=20 m$
The length of the rod is $\mathrm{L}=12 \mathrm{~cm}$
Distance of mass m from center, $\mathrm{r}_1=4 \mathrm{~cm}$
Distance of mass 2 m from center, $\mathrm{r}_2=2 \mathrm{~cm}$
Let the geometrical center of the rod be the origin $(0,0)$.
The position of the new center of mass $\left(\mathrm{x}_{\mathrm{cm}}\right)$ after the collision :
$$ \mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{M}(0)+\mathrm{m}\left(\mathrm{r}_1\right)-2 \mathrm{~m}\left(\mathrm{r}_2\right)}{\mathrm{M}+\mathrm{m}+2 \mathrm{~m}} $$
$$\Rightarrow $$ $\mathrm{x}_{\mathrm{cm}}=\frac{20 \mathrm{~m}(0)+\mathrm{m}(4)-2 \mathrm{~m}(2)}{23 \mathrm{~m}}$
$$\Rightarrow $$ $ \mathrm{x}_{\mathrm{cm}}=\frac{4 \mathrm{~m}-4 \mathrm{~m}}{23 \mathrm{~m}}=0 $
Since $\mathrm{x}_{\mathrm{cm}}=0$, the center of mass of the system does not shift. The entire system will only rotate about the geometrical center of the rod.
The initial angular momentum $\left(\mathrm{L}_{\mathrm{i}}\right)$ of the system about the center must equal the final angular momentum $\left(\mathrm{L}_{\mathrm{f}}\right)$.
The initial angular momentum of the system is,
$$ \mathrm{L}_{\mathrm{i}}=\left(\mathrm{m} \cdot \mathrm{v} \cdot \mathrm{r}_1\right)+\left(2 \mathrm{~m} \cdot \mathrm{v} \cdot \mathrm{r}_2\right) $$
$$\Rightarrow $$ $\mathrm{L}_{\mathrm{i}}=\mathrm{m} \cdot \mathrm{v} \cdot 4+2 \mathrm{~m} \cdot \mathrm{v} \cdot 2$
$$\Rightarrow $$ $$ \mathrm{L}_{\mathrm{i}}=4 \mathrm{mv}+4 \mathrm{mv}=8 \mathrm{mv} $$
The final moment of inertia ( $\mathrm{I}_{\mathrm{f}}$ ) about the center is the sum of the moment of inertia of the rod and the two point masses.
$$ \mathrm{I}_{\mathrm{f}}=\mathrm{I}_{\mathrm{rod}}+\mathrm{I}_{\mathrm{m}}+\mathrm{I}_{2 \mathrm{~m}} $$
$$\Rightarrow $$ $I_f=\frac{M L^2}{12}+m r_1^2+2 m r_2^2$
$$\Rightarrow $$ $\mathrm{I}_{\mathrm{f}}=\frac{(20 \mathrm{~m})(12)^2}{12}+\mathrm{m}(4)^2+2 \mathrm{~m}(2)^2$
$$\Rightarrow $$ $$ \mathrm{I}_{\mathrm{f}}=20 \mathrm{~m}(12)+16 \mathrm{~m}+2 \mathrm{~m}(4)=264 \mathrm{~m} $$
The angular momentum of the system is given as $\mathrm{L}=\mathrm{I} \omega$, where $\omega$ is the angular speed and I is the moment of inertia.
Using conservation of angular momentum,
$$ \mathrm{L}_{\mathrm{i}}=\mathrm{L}_{\mathrm{f}} \Rightarrow 8 \mathrm{mv}=264 \mathrm{~m} \omega $$
$\Rightarrow \omega=\frac{8 \mathrm{mv}}{264 \mathrm{~m}}$
$$\Rightarrow $$ $\frac{\omega}{v}=\frac{1}{33} \Rightarrow \frac{v}{\omega}=33$

Answer

33

Answer

$2 \sqrt{88}$

Answer

32

Answer

66

Centre of Mass and Rotational Motion

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