The rod is hinged and the projectile embeds itself in it. Before calculating anything, the decisive question is: what is conserved and what isn’t?

  • Angular momentum about the hinge is conserved: the hinge’s constraint reaction has zero arm about the hinge itself, so it produces no moment.
  • Momentum is not conserved: the hinge’s impulsive constraint reaction transmits an external impulse to the system.
  • Kinetic energy is not conserved: the collision is inelastic, part of the energy turns into heat.

Pole: the hinge

The hinge is the only sensible choice. About any other pole, the hinge’s reaction would have a non-zero arm and would contribute an external angular impulse, making angular momentum non-conserved.

One unknown (ω\omega, the angular velocity after the collision with the embedded projectile) and one equation (conservation of LL about the hinge):

mpvALLA=ItotωLB\underbrace{m_p v_A L}_{L^A} = \underbrace{I_\text{tot}\,\omega}_{L^B}

The system’s moment of inertia (rod + point-like projectile at the end) about the hinge is

Itot=13ML2+mpL2=1344+0,34=5,33+1,2=6,53  kgm2I_\text{tot} = \tfrac{1}{3}ML^2 + m_p L^2 = \tfrac{1}{3}\cdot 4\cdot 4 + 0{,}3\cdot 4 = 5{,}33 + 1{,}2 = 6{,}53\;\mathrm{kg\,m^2}

from which

ω=mpvALItot=0,390026,5382,7  rad/s\omega = \frac{m_p v_A L}{I_\text{tot}} = \frac{0{,}3\cdot 900\cdot 2}{6{,}53} \approx 82{,}7\;\mathrm{rad/s}

Energy tells us how violent the collision was:

Ecin,A=12mpvA2=120,3810000=121500  JE_\text{cin,A} = \tfrac{1}{2}m_p v_A^2 = \tfrac{1}{2}\cdot 0{,}3\cdot 810\,000 = 121\,500\;\mathrm{J}

Ecin,B=12Itotω2=126,5382,7222320  JE_\text{cin,B} = \tfrac{1}{2}I_\text{tot}\omega^2 = \tfrac{1}{2}\cdot 6{,}53\cdot 82{,}7^2 \approx 22\,320\;\mathrm{J}

Etermica=1215002232099180  JE_\text{termica} = 121\,500 - 22\,320 \approx 99\,180\;\mathrm{J}

About 82% of the initial kinetic energy went into heat: the hallmark of a strongly inelastic collision.

A body pinned at a fixed point

For a rigid body hinged at a fixed point PP you can use directly L=IPωL = I_P\,\omega and Ecin=12IPω2E_\text{cin} = \tfrac{1}{2}I_P\,\omega^2, where IPI_P is the moment of inertia about PP (not about the centre of mass).

Collegamenti

Argomenti: Dinamica rotazionale Concetti: Conservazione del momento angolare · Urto anelastico Competenze: Scelta del polo Oggetti: Asta

Esercizi collegati: Esercizio svolto — proiettile contro asta incernierata · Problema — Disco rotante colpito tangenzialmente · Problema — La pattinatrice che si stringe