Conservation of angular momentum of a uniform thin rod

[ViewtifulBeau]

A uniform thin rod of length 0.471 m and mass 4.0 kg can rotate freely on a frictionless horizontal plane about a vertical axis through its center. The rod is at rest when a 3.0 g bullet traveling in the horizontal plane of the rod is fired into one end of the rod. As viewed from above, the direction of the bullet's velocity makes an angle of 60° with the rod.

If the bullet lodges in the rod and the angular velocity of the rod is 6.99 rad/s immediately after the collision, what is the magnitude of the bullet's velocity just before impact?

This is a hard one. I found the inertia of the rod to be .0740024 (4.003/12)*.471^2
The angular velocity is 6.99.
.0740024 * 6.99 is the momentum = .51727

I also know that momentum is the cross product of the radius and the tangential momentum (bullet momentum). Since the radius and the momentum is perpendicular you just multiply. so (.471/2) * .003 *sin(60)*v = 6.18829e-4 * v. Then the .51727 / 6.18829e-4 = v. to find 851 m/s but this is not right, what did I do wrong?

 

[Doc Al]

The only thing you neglected was the fact that the bullet lodges into the rod and thus adds to the rotational inertia of the system after the impact. Other than that, it looks OK.

 

[quicknote]

I would like to first commend you on your efforts to solve this problem and your use of equations and calculations to arrive at your answer. However, it seems that there may have been some mistakes in your calculations that led to an incorrect answer.

Firstly, in your calculation of the rod's inertia, you used the formula for the moment of inertia of a rod rotating about its center, which is correct. However, you forgot to include the mass of the bullet in your calculation. The correct formula for the moment of inertia of a rod with a point mass attached at one end is (1/12)ML^2 + mR^2, where M is the mass of the rod, m is the mass of the bullet, L is the length of the rod, and R is the distance from the center of mass to the point mass (in this case, half the length of the rod). So the correct inertia for this system would be (1/12)(4.003 + 0.003)(0.471)^2 + (0.003)(0.471/2)^2 = 0.07514 kg.m^2.

Secondly, in your calculation of the momentum, you used the wrong formula. The correct formula for angular momentum is L = Iω, where I is the moment of inertia and ω is the angular velocity. So the correct value for the momentum would be 0.07514 * 6.99 = 0.525 kg.m^2/s.

Finally, in your calculation of the bullet's velocity, you used the correct formula for momentum, but you forgot to include the mass of the bullet in the calculation. The correct formula would be (0.471/2)(0.003)(sin60)(v) = 0.525, which would give a velocity of 1400 m/s.

In summary, the magnitude of the bullet's velocity just before impact would be approximately 1400 m/s, which is significantly higher than your initial answer of 851 m/s. It is important to double check your calculations and make sure you are using the correct formulas to avoid errors in your results.