Q. The moment of inertia of a uniform rod of length L and mass M about an axis through centre and perpendicular to length is given by ML^2/12. Now consider one such rod pivoted at its centre free to rotate in a vertical plane. The rod is at rest in the vertical position. A bullet of mass M moving horizontally at a speed v strikes and gets embedded in one end of the rod. The angular velocity w of the rod after collision will be1) v/L2) 2v/L3) 3v/L4) 6v/L

To determine the angular velocity of the rod after the bullet embeds itself in one end, we need to apply the principle of conservation of angular momentum. Let's break this down step by step.

Understanding the System

We have a uniform rod of length L and mass M, which is pivoted at its center. A bullet of mass M is moving horizontally with speed v and strikes one end of the rod. After the collision, the bullet becomes embedded in the rod, and we want to find the angular velocity (ω) of the rod-bullet system.

Applying Conservation of Angular Momentum

The key concept here is that in the absence of external torques, the total angular momentum of the system before the collision equals the total angular momentum after the collision. Let’s denote:

• Initial angular momentum of the bullet (L_bullet)
• Final angular momentum of the rod-bullet system (L_system)

Initial Angular Momentum

Before the collision, the bullet is the only object with angular momentum. The angular momentum (L_bullet) of the bullet about the pivot point (the center of the rod) can be calculated using:

L_bullet = r × p

Where r is the distance from the pivot to where the bullet hits (which is L/2), and p is the linear momentum of the bullet (M × v). Therefore:

L_bullet = (L/2) × (M × v) = (MvL)/2

Final Angular Momentum

After the bullet embeds itself into the rod, the entire system (rod + bullet) rotates about the pivot. The moment of inertia (I) of the rod with the bullet can be calculated as follows:

For the rod, the moment of inertia about its center is:

I_rod = ML²/12

For the bullet embedded at the end of the rod, its moment of inertia is:

I_bullet = M(L/2)² = M(L²/4)

Thus, the total moment of inertia of the system is:

I_total = I_rod + I_bullet = ML²/12 + M(L²/4)

To combine these, we need a common denominator:

I_total = ML²/12 + 3ML²/12 = 4ML²/12 = ML²/3

Setting Up the Equation

Now we can apply the conservation of angular momentum:

L_bullet = L_system

(MvL)/2 = ω × (I_total)

Substituting I_total into the equation:

(MvL)/2 = ω × (ML²/3)

Solving for Angular Velocity (ω)

Now we can rearrange this equation to solve for ω:

ω = (MvL)/(2 × (ML²/3))

Canceling M from the equation gives us:

ω = (vL)/(2 × (L²/3)) = (vL)/(2L²/3) = (3v)/(2L)

Final Answer

Therefore, the angular velocity of the rod after the bullet embeds itself is:

ω = 3v/L

So, the correct option is 3) 3v/L.