Question 88: Centre of Mass and Rotational Motion

Question

Two small balls with masses $m$ and 2 m are attached to both ends of a rigid rod of length $d$ and negligible mass. If angular momentum of this system is $L$ about an axis (A) passing through its centre of mass and perpendicular to the rod then angular velocity of the system about $A$ is :

Accepted Answer

Let the mass m be at position $\mathrm{x}=0$.
The mass 2 m is at position $\mathrm{x}=\mathrm{d}$.
The position of the centre of mass $\left(\mathrm{X}_{\mathrm{CM}}\right)$ is given by :
$$ \mathrm{x}_{\mathrm{CM}}=\frac{\mathrm{m}_1 \mathrm{x}_1+\mathrm{m}_2 \mathrm{x}_2}{\mathrm{~m}_1+\mathrm{m}_2} $$
$$ X_{C M}=\frac{m(0)+2 m(d)}{m+2 m}=\frac{2 m d}{3 m}=\frac{2 d}{3} $$
So, the distances of the masses from the centre of mass are :
Distance of mass m from $\mathrm{CM}\left(\mathrm{r}_1\right)$ :
$$ r_1=\frac{2 d}{3} $$
Distance of mass 2 m from CM ( $\mathrm{r}_2$ ) :
$$ \mathrm{r}_2=\mathrm{d}-\frac{2 \mathrm{~d}}{3}=\frac{\mathrm{d}}{3} $$
The moment of inertia of the system about the axis passing through the CM is the sum of the moments of inertia of the individual masses :
$$ \mathrm{I}=\mathrm{m}_1 \mathrm{r}_1^2+\mathrm{m}_2 \mathrm{r}_2^2 $$
$$\Rightarrow $$ $I=m\left(\frac{2 d}{3}\right)^2+2 m\left(\frac{d}{3}\right)^2$
$$\Rightarrow $$ $I=m\left(\frac{4 d^2}{9}\right)+2 m\left(\frac{d^2}{9}\right)$
$$\Rightarrow $$ $I=\frac{4 \mathrm{md}^2}{9}+\frac{2 \mathrm{md}^2}{9}$
$$\Rightarrow $$ $$ \mathrm{I}=\frac{6 \mathrm{md}^2}{9}=\frac{2}{3} \mathrm{md}^2 $$
The angular momentum L is related to the moment of inertia I and angular velocity $\omega$ by the formula:
$$ \mathrm{L}=\mathrm{I} \omega $$
$$\Rightarrow \omega=\frac{L}{I}=\frac{L}{\left(\frac{2}{3} \mathrm{md}^2\right)} $$
$$ \Rightarrow \omega=\frac{3}{2} \frac{\mathrm{~L}}{\mathrm{md}^2} $$
Therefore, the angular velocity of the system is $\frac{3}{2} \frac{\mathrm{~L}}{\mathrm{md}^2}$.
Hence, the correct option is (B).

Answer

$\frac{4}{3} \frac{L}{m d^2}$

Answer

$\frac{3}{2} \frac{L}{m d^2}$

Answer

$\frac{2 L}{5 m d^2}$

Answer

$\frac{2 L}{m d^2}$

Centre of Mass and Rotational Motion

Two small balls with masses m and 2 m are attached to both ends of a rigid rod of length d and negligible mass. If angular momentum of this system is L about an axis (A) passing through its centre of mass and perpendicular to the rod then angular velocity of the system about A is :

Choose the correct answer: