Rotational Energy and momentum help

[mistasong]

Homework Statement

A uniform rod of length L1 and mass M = 0.75 kg is supported by a hinge at one end and is free to rotate in the vertical plane (Figure). The rod is released from rest in the position shown. A particle of mass m = 0.5 kg is supported by a thin string of length L2 from the hinge. The particle sticks to the rod on contact. What should be the ratio L2/L1 so that θmax = 60o after the collision?

Homework Equations

Kf − Ki +Uf −Ui = 0

The Attempt at a Solution

I used energy concept to get Kf +Uf −Ui = 1/2(1/3ML)ω^2 +MgL1/2 − MgL1 = 0
I do not know where to go from here though and the teacher said there would be a cubic equation.

 

[SammyS]

Homework Statement

A uniform rod of length L₁ and mass M = 0.75 kg is supported by a hinge at one end and is free to rotate in the vertical plane (Figure). The rod is released from rest in the position shown. A particle of mass m = 0.5 kg is supported by a thin string of length L₂ from the hinge. The particle sticks to the rod on contact. What should be the ratio L₂/L₁ so that θmax = 60° after the collision?

Homework Equations

Kf − Ki +Uf −Ui = 0

The Attempt at a Solution

I used energy concept to get Kf +Uf −Ui = 1/2(1/3ML)ω^2 +MgL1/2 − MgL₁ = 0
I do not know where to go from here though and the teacher said there would be a cubic equation.

There is no figure.

From what angle is the rod released?

Also, we've provided icons to facilitate the use of subscripts, X₂, and superscripts X² .

 

[mistasong]

Ops sorry
http://www.solutioninn.com/physics/mechanics/angular-momentum/a-uniform-rod-of-length-l1-and-mass-m-
This was the only link with a diagram

 

[haruspex]

You have to take into account inelastic collision. You can use conservation of work to find the speed of the rod just before collision, but you have to do something else to find the speeds after collision... Any ideas?

 

[Nickie]

To solve this problem, you will need to use both conservation of energy and conservation of momentum. First, let's consider the conservation of energy equation you have written:

Kf + Uf - Ui = 1/2(1/3ML)ω^2 + MgL1/2 - MgL1 = 0

This equation is correct, but it only accounts for the energy before and after the collision. We also need to consider the energy during the collision, which is where the particle sticks to the rod. This energy will be in the form of rotational kinetic energy.

To account for this, we can add a term for the kinetic energy of the particle before the collision, which is 1/2mv^2, where v is the velocity of the particle. Since the particle is initially at rest, this term will be 0.

We can also add a term for the kinetic energy of the rod after the collision, which will be in the form of rotational kinetic energy. This term will be 1/2Iω^2, where I is the moment of inertia of the rod. We can calculate the moment of inertia of the rod using the formula I = 1/3ML^2.

Now, let's consider the conservation of momentum equation. We know that momentum is conserved before and after the collision, so we can write:

Mv + mv = Mv'

where v is the velocity of the rod before the collision, m is the mass of the particle, and v' is the velocity of the combined system after the collision.

We can also relate the velocity of the particle to the angular velocity of the rod using the relationship v = ωL1.

Now, we have two equations and two unknowns (v and ω), so we can solve for both of them. Once we have the values for v and ω, we can use the relationship between angular velocity and angle (ω = θ/t) to solve for the ratio L2/L1 that will give us an angle of 60 degrees after the collision.

This problem may indeed result in a cubic equation, but by using both conservation of energy and momentum, you should be able to solve for the unknowns and find the correct ratio.