Conservation of Angular Momentum in Collision with Rotations

AI Thread Summary
A uniform rod of length 2.2 m and mass 2.8 kg is hinged and rotates in the vertical plane after being released from rest. A particle attached by a string of length 1.8 m sticks to the rod upon contact, with a maximum angle of 40° reached post-collision. The discussion emphasizes using conservation of angular momentum to analyze the collision, as kinetic energy is not conserved. Participants suggest setting the angular momentum before the collision equal to the angular momentum after the collision to solve for the unknown mass and energy dissipated. Understanding the conservation principles is crucial for progressing in the problem.
maniacp08
Messages
115
Reaction score
0
339698289.jpg


A uniform rod of length L1 = 2.2 m and mass M = 2.8 kg is supported by a hinge at one end and is free to rotate in the vertical plane. The rod is released from rest in the position shown. A particle of mass m is supported by a thin string of length L2 = 1.8 m from the hinge. The particle sticks to the rod on contact. After the collision, θmax = 40°.

(a) Find m.
kg
(b) How much energy is dissipated during the collision?
J

Relevant Equations:
Angular momentum = I * omega


I'm having trouble starting this problem. I should compare using energy of conservation before and after collision correct?

I for the thin rod = 1/3 ML^2

Energy of conservation for the rod before collision
Kf - Ki + Uf - Ui = 0
Ki = 0
Kf + Uf - Ui = 0
1/2(1/3ML^2) * omega ^2 + Uf - Ui = 0
What will Uf and Ui be?

Im not even sure if what I am doing is correct, can someone guide me?
 
Physics news on Phys.org
Anyone can help me get started with this problem please.
 
Hi maniacp08,

maniacp08 said:
339698289.jpg


A uniform rod of length L1 = 2.2 m and mass M = 2.8 kg is supported by a hinge at one end and is free to rotate in the vertical plane. The rod is released from rest in the position shown. A particle of mass m is supported by a thin string of length L2 = 1.8 m from the hinge. The particle sticks to the rod on contact. After the collision, θmax = 40°.

(a) Find m.
kg
(b) How much energy is dissipated during the collision?
J

Relevant Equations:
Angular momentum = I * omega


I'm having trouble starting this problem. I should compare using energy of conservation before and after collision correct?

You will compare them for part b, because kinetic energy is not conserved in this collision and the difference in energies will be equal to how much energy is dissipated.

You can use conservation of energy for other parts of this problem (you have to consider the swinging motion before the collision, and the swinging motion after the collision), but for the collision itself, what is conserved?
 

Attachments

  • 339698289.jpg
    339698289.jpg
    1.1 KB · Views: 375
Hi, alphysicist, thanks for responding.

The collision itself, the angular momentum is conserved.
Angular momentum = I * omega

Do I consider the before collision or after collision first?
Do I use conservation of energy on the particle or the rod?
Im just confuse on how to start this.
 
maniacp08 said:
Hi, alphysicist, thanks for responding.

The collision itself, the angular momentum is conserved.
Angular momentum = I * omega

Do I consider the before collision or after collision first?
Do I use conservation of energy on the particle or the rod?
Im just confuse on how to start this.

To use conservation of angular momentum for the collision, you don't do either one first, you set them equal to each other.

So the angular momentum of the system right before the collision (the instant the rod touches the particle) is equal to the total angular momentum of both of them right after the collision is over. Setting up that equation will then show you what else you need to find in the problem.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top