A thin uniform bar of length L and mass 8 m lies on a smooth horizontal table. Two point masses m and 2 m moving in the same horizontal plane from opposite sides of the bar with speeds 2v and v respectively. The masses stick to the bar after collision at a distance $$\frac{L}{3}$$ and $$\frac{L}{6}$$ respectively from the centre of the bar. If the bar starts rotating about its center of mass as a result of collision, the angular speed of the bar will be:

Solution & Explanation

1. Find the Center of Mass of the Combined System After Collision

Let the thin uniform bar have a total length $$L$$ and mass $$M = 8m$$. Its initial center of mass sits at the geometric center (origin $$x = 0$$).

Two point masses ($$m$$ and $$2m$$) collide and stick to the bar at specific positions measured from the center:

• Mass $$m_1 = m$$ lands at $$x_1 = +\frac{L}{3}$$ with a velocity vector pointing in the negative direction ($$-2v$$).
• Mass $$m_2 = 2m$$ lands at $$x_2 = -\frac{L}{6}$$ with a velocity vector pointing in the positive direction ($$+v$$).

We calculate the new position of the center of mass ($$x_{\text{cm}}$$) for the fully joined system:

$$x_{\text{cm}} = \frac{M \cdot (0) + m_1 \cdot x_1 + m_2 \cdot x_2}{M + m_1 + m_2}$$

$$x_{\text{cm}} = \frac{8m(0) + m\left(\frac{L}{3}\right) + 2m\left(-\frac{L}{6}\right)}{8m + m + 2m}$$

$$x_{\text{cm}} = \frac{m\frac{L}{3} - m\frac{L}{3}}{11m} = 0$$

Because the net mass moments cancel out perfectly, the center of mass of the final combined system remains exactly at the original geometric center ($$x = 0$$) of the bar.

2. Calculate the Total Initial Angular Momentum ($$L_i$$)

We calculate the initial angular momentum about the system's stationary center of mass ($x = 0$) right before impact using $$\vec{L} = \vec{r} \times \vec{p}$$ (taking the counter-clockwise rotation direction as positive):

• For mass $$m$$: Moving with speed $$2v$$ from the opposite side, it hits at a position vector of $$+\frac{L}{3}$$ and tends to rotate the system counter-clockwise.

  $$L_1 = m \cdot (2v) \cdot \frac{L}{3} = \frac{2}{3}mvL$$

• For mass $$2m$$: Moving with speed $$v$$ from the bottom side, it hits at a position vector of $$-\frac{L}{6}$$ and also tends to rotate the system counter-clockwise.

  $$L_2 = 2m \cdot v \cdot \frac{L}{6} = \frac{1}{3}mvL$$

Summing them together gives the total initial angular momentum:

$$L_i = L_1 + L_2 = \frac{2}{3}mvL + \frac{1}{3}mvL = mvL$$

3. Calculate the Total Moment of Inertia ($$I_{\text{total}}$$) After Collision

After the completely inelastic collision, the total moment of inertia about the center point consists of the uniform rod plus both embedded point masses:

$$I_{\text{total}} = I_{\text{bar}} + I_1 + I_2$$

• Moment of Inertia of the bar: $$I_{\text{bar}} = \frac{1}{12}ML^2 = \frac{1}{12}(8m)L^2 = \frac{2}{3}mL^2$$
• Moment of Inertia of mass $$m$$: $$I_1 = m \left(\frac{L}{3}\right)^2 = \frac{1}{9}mL^2$$
• Moment of Inertia of mass $$2m$$: $$I_2 = 2m \left(-\frac{L}{6}\right)^2 = 2m\left(\frac{L^2}{36}\right) = \frac{1}{18}mL^2$$

Combine all terms by finding a common denominator (18):

$$I_{\text{total}} = \left(\frac{12}{18} + \frac{2}{18} + \frac{1}{18}\right)mL^2 = \frac{15}{18}mL^2 = \frac{5}{6}mL^2$$

4. Apply Conservation of Angular Momentum to Find $$\omega$$

Since no external torques act on the system during the horizontal plane collision, angular momentum is conserved ($$L_i = L_f$$):

$$L_i = I_{\text{total}} \cdot \omega$$

$$mvL = \left(\frac{5}{6}mL^2\right) \cdot \omega$$

Isolating the final angular speed ($$\omega$$):

$$\omega = \frac{mvL}{\frac{5}{6}mL^2} = \frac{6v}{5L}$$

Correct Option Key: Option B ($$\frac{6v}{5L}$$)