Rotation of Rod Homework: Max Angle, Motion Analysis

[rbwang1225]

Homework Statement

A uniform thin rod with mass M and length L nailed by frictionless pivot can swing freely on the wall as shown in Fig. The pivot locates at the distance L/4 from the bottom and stops inside. The velocity of the bullet before hitting the rod is v. (a) Compute the max. angle that the rod can reach after the shot. (b) If the swinging angle is small, describe the motion of the whole system (rod+bullet) in detail.

The Attempt at a Solution

I used the conservation of energy and concept of mass center.
The result of (a) is $cos\theta = \frac{2mv^2-(M+2m)gL}{L(1+\frac{m}M)}$.
I don't really understand how to get (b).

 

[tiny-tim]

hi rbwang1225!

I used the conservation of energy …

nooo …

energy is never conserved in a collision unless the question says so

in this case, energy is obviously not conserved, since the bullet becomes part of the rod, and the collision is perfectly inelastic

however, momentum or angular momentum is always conserved in a collision ​

(for b, the system becomes a pendulum, with shm)

 

[rbwang1225]

hi rbwang1225!


nooo …

energy is never conserved in a collision unless the question says so

in this case, energy is obviously not conserved, since the bullet becomes part of the rod, and the collision is perfectly inelastic

however, momentum or angular momentum is always conserved in a collision ​

(for b, the system becomes a pendulum, with shm)

Oh...right! I forgot the reason why the bullet stuck in the rod!
Now that means I have to calculate the moments of the inertia...
I know it would be a SHM but how do I describe "in detail"?
Thanks a lot!

 

[tiny-tim]

I know it would be a SHM but how do I describe "in detail"?

τ = Iα

 

[asteriatic]

I would like to commend you on your use of the conservation of energy and concept of mass center in solving part (a) of the problem. Your solution demonstrates a solid understanding of these concepts and their application to rotational motion.

In order to solve part (b) of the problem, we can use the small angle approximation, which states that for small angles, the sine of the angle is approximately equal to the angle itself (in radians). This approximation is valid for angles less than 0.1 radians.

Using this approximation, we can describe the motion of the system as simple harmonic motion. This means that the rod and bullet will oscillate back and forth around the pivot point, with a period that can be calculated using the formula T = 2π√(I/mgd), where I is the moment of inertia of the system, m is the mass of the bullet, g is the acceleration due to gravity, and d is the distance between the pivot point and the center of mass of the system.

The amplitude of the oscillation can be determined using the maximum angle calculated in part (a). The bullet will also impart a small amount of angular momentum to the system, causing a slight precession in the motion.

Overall, the motion of the system will be a combination of rotational and oscillatory motion, with the bullet acting as a small perturbation on the system. Further analysis can be done by considering the forces and torques acting on the system, but for small angles, the simple harmonic motion approximation is a good starting point.