This is calculated according to the definition of moment of inertia of rigid body. Because the thin rod has a mass of m and a length of l, the mass is evenly distributed, and its linear density is m/l, the moment of inertia of each segment varies with the distance from the axis, so it needs to be integrated, so that the moment of inertia of the thin rod to the rotating shaft is (upper integral limit L, lower integral limit 0):

I=∫(M/l)r？ dr=Ml？ /3 .

The moment of inertia of the elastic piece at one end of the thin rod relative to the rotating shaft is

i=ml？ .

The sum of angular momentum of thin rod and bullet is

J=(I+i)ω=ml？ ω+ 1/3Ml？ ω.

This is equal to the initial angular momentum mvl when the bullet hits the thin rod, so according to the conservation of angular momentum, mvl=ml? ω+ 1/3Ml？ ω.

On this basis, the initial angular velocity of the thin rod bullet system can be obtained: ω = 3mv/(3m+m) L. 。

And initial kinetic energy: Ek=(I+i)ω? /2=Jω/2= 1.5m？ v？ /(3m+M)。

When the system moves to the maximum swing angle, all the initial kinetic energy is converted into gravitational potential energy. Let the maximum swing angle be θ, then

H=0.5l*( 1-cosθ) of the center of gravity of the slender rod and EP1= mgh = 0.5mgl * (1-cos θ) of the potential energy;

The barycenter of the bullet increases h=l*( 1-cosθ), and the potential energy increases EP2 = mgh = mgl * (1-cos θ);

Therefore, according to the conservation of mechanical energy of the system, Ek=Ep 1+Ep2, that is

1.5m？ v？ /(3m+M)=(0.5M+M)GL *( 1-cosθ)。

Solvable:

cosθ= 1-3m？ v？ /[(3m+M)(M+2m)gl]。

You can also use the heavy moment to do negative work, so that all the initial kinetic energy is converted into gravitational potential energy, reaching the maximum swing angle θ. Let the swing angle at any moment be θ', then the weight moment of the system is

M = mgl * sinθ'+∫(M/l)gr * sinθ' dr = mgl * sinsθ'+0.5 mgl * sinsθ'。

The work done by the moment of gravity is (upper limit θ of integration, lower limit 0).

w =-∫MDθ=-∫(m+0.5M)GL * sinθ' dθ' =-(0.5M+m)GL *( 1-cosθ)。

This is equal to the change of kinetic energy of the system, that is

W=0-Ek=- 1.5m？ v？ /(3m+M)。

It can be solved from the above:

cosθ= 1-3m？ v？ /[(3m+M)(M+2m)gl]。

When cosθ= 1-3m? v？ /[(3m+M)(M+2m)GL]& gt； 0, θ

When cosθ= 1-3m? v？ /[(3m+M)(M+2m)gl]≤0, θ≥π/2, and the initial velocity satisfies V? ≥( 1+M/3m)(2+M/M)GL；

Further, when 1-3m? v？ /[(3m+M)(M+2m)gl]≤- 1，θ≥π。 At this time, the system still has kinetic energy when it rotates to the highest point, so it will continue to rotate and make a variable-speed circular rotation.