Question

A thin uniform bar of length $L$ and mass $8m$ lies on a smooth horizontal table. Two point masses $m$ and $2m$ are 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 :

• A. $\frac{v}{5L}$
• B. $\frac{6 v}{5L}$
• C. $\frac{3 v}{5L}$
• D. $\frac{ v}{6L}$

Answer 1

The Correct Option is B

To find the angular speed of the bar about its center of mass after the collision, we need to consider the conservation of angular momentum. Initially, the bar is at rest, so the total initial angular momentum is due solely to the moving point masses.

The initial angular momentum \(L_i\) about the center of the bar is given by:

\(L_i = m \cdot 2v \cdot \frac{L}{3} + 2m \cdot v \cdot \frac{L}{6}\)

Plugging in values, we get:

\(L_i = \frac{2mvL}{3} + \frac{2mvL}{6}\)

Finding a common denominator, this becomes:

\(L_i = \frac{4mvL}{6} + \frac{2mvL}{6} = \frac{6mvL}{6} = mvL\)

Next, we calculate the final angular momentum \(L_f\). After the collision, the bar and the masses rotate together. The final angular momentum is given by:

\(L_f = I \omega\)

Where \(I\) is the moment of inertia of the system about the center of mass, and \(\omega\) is the angular speed we want to find. The total moment of inertia \(I\) is the sum of the moment of inertia of the bar and the point masses:

• The moment of inertia of the bar about its center: \(I_{\text{bar}} = \frac{1}{12} \times 8m \times L^2 = \frac{2mL^2}{3}\)
• For point mass \(m\) at \(L/3\) from the center, its moment of inertia: \(I_1 = m \left(\frac{L}{3}\right)^2 = \frac{mL^2}{9}\)
• For point mass \(2m\) at \(L/6\) from the center, its moment of inertia: \(I_2 = 2m \left(\frac{L}{6}\right)^2 = \frac{mL^2}{18}\)

Thus, the total moment of inertia \(I\) is:

\(I = \frac{2mL^2}{3} + \frac{mL^2}{9} + \frac{mL^2}{18}\)

Simplifying further, we find:

\(I = \frac{2mL^2}{3} + \frac{2mL^2}{18} = \frac{2mL^2}{3} + \frac{mL^2}{9}\)

Putting it all together, it simplifies to:

\(I = \frac{12mL^2 + mL^2 + 2mL^2}{18} = \frac{15mL^2}{18} = \frac{5mL^2}{6}\)

Equating initial and final angular momentum, we have:

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

Simplifying for \(\omega\):

\(\omega = \frac{6v}{5L}\)

Thus, the angular speed of the bar is \(\frac{6v}{5L}\), which corresponds to option: \(\frac{6v}{5L}\).