Question

A uniform rod of length \[8a\] and mass \[6m\] lies on a smooth horizontal surface. Two point masses \[m\] and \[2m\] moving in the same plane with speed \[2v\] and \[v\] respectively strike the rod perpendicularly at distance \[a\] and \[2a\] from the midpoint of the rod in the opposite directions and stick to the rod. The angular velocity of the system immediately after the collision is:
a. $\dfrac{{6v}}{{32a}}$
b. $\dfrac{{6v}}{{33a}}$
c. $\dfrac{{6v}}{{40a}}$
d. $\dfrac{{6v}}{{41a}}$

Answer 1

Hint: First, you have to calculate moment of inertia for all the objects individually using their masses and lengths (distances) i.e for the rod, and the two point masses. And, also find the angular momentums corresponding to the moment of inertia.
- Calculate the initial angular momentums for the two point masses.
- Use the conservation of angular momentum about the centre of mass of the rod.

Formula used:
• The moment of inertia of the rod, ${I_r} = \dfrac{1}{{12}}{m_r}{l^2}$ , ${m_r}$ is the mass of the rod and ${l_r}$ is the length.
• Angular momentum of the rod = ${I_r}\omega $, $\omega $ is the angular velocity.
• The moment of inertia of point mass, ${I_p} = m{r^2}$ , $m$ is the point mass and $r$ is the distance from the midpoint of the rod.
• Angular momentum of the point mass = ${I_p}\omega $
• From the Conservation of angular momentum we get,
${m_1}{v_1}{r_1} + {m_2}{v_2}{r_2} = {I_r} + {I_{{p_1}}} + {I_{{p_2}}}$
${v_1}$ and ${v_2}$ are the speeds of the two point masses.

Complete step by step answer:
Let the masses of the two point masses ${m_1}$ and ${m_2}$ are at a distance of ${r_1}$ and ${r_2}$ from the midpoint of the rod. So the total angular momentum of the point masses will be, ${m_1}{v_1}{r_1} + {m_2}{v_2}{r_2}$ . ${v_1}$ and ${v_2}$ are the speeds of the two point masses.

The total moment of inertia of the point masses will be, ${I_{{p_1}}} + {I_{{p_2}}}$
Where, ${I_{{p_1}}} = {m_1}{r_1}^2$ and ${I_{{p_2}}} = {m_2}{r_2}^2$
Now if ${m_r}$ is the mass of the rod and ${l_r}$ is the length, The moment of inertia of the rod, ${I_r} = \dfrac{1}{{12}}{m_r}{l^2}$

So the total moment of inertia of the of the system will be, ${I_r} + {I_{{p_1}}} + {I_{{p_2}}} = \dfrac{1}{{12}}{m_r}{l^2} + {m_1}{r_1}^2 + {m_2}{r_2}^2$
The Angular momentum of the rod = ${I_r}\omega $,
The Angular momentum of the point mass = $({I_{{p_1}}} + {I_{{p_2}}})\omega $
$\omega $ is the angular velocity.

From the Conservation of angular momentum we get,
${m_1}{v_1}{r_1} + {m_2}{v_2}{r_2} = {I_r}\omega + ({I_{{p_1}}} + {I_{{p_2}}})\omega $
$ \Rightarrow {m_1}{v_1}{r_1} + {m_2}{v_2}{r_2} = \dfrac{1}{{12}}{m_r}{l^2}\omega + ({m_1}{r_1}^2 + {m_2}{r_2}^2)\omega $ ----(1)

Now we consider the given equation to find the angular velocity.
Where, ${m_r} = 6m$, ${m_1} = m$, ${m_2} = 2m$, ${l_r} = 8a$, ${r_1} = a$, ${r_2} = 2a$, ${v_1} = 2v$, ${v_2} = v$.

$\therefore$ Equation (1) becomes,
$m \times 2v \times a + 2mv \times 2a = \dfrac{1}{{12}}6m \times {(8a)^2} \times \omega + m{\left( a \right)^2} \times \omega + 2m{\left( {2a} \right)^2} \times \omega $
\[ \Rightarrow 2mva + 4mva = 32m{a^2}\omega + \omega (m{a^2} + 4m{a^2})\]
\[ \Rightarrow 6va = 41{a^2}\omega \]
\[ \Rightarrow \omega = \dfrac{{6v}}{{41a}}\]
So, the angular velocity , \[ \Rightarrow \omega = \dfrac{{6v}}{{41a}}\]

Hence, the correct answer is option (D).

Note:
• If there is no external torque acting on the object that is rotating about its axis, then the angular moment of that object always remains the same – this is called the conservation of law.
• If an object is rotating in a circular path of radius $r$ and the linear momentum of that object is $p$, the the angular momentum of the object $\overrightarrow L = \overrightarrow r \times \overrightarrow p $
• The change in angular moment of the object at that moment can be written as the product of the time and torque.