Kinetic Energy Ratio/Rotational Equilibrium

[Victorzaroni]

Homework Statement

Two equal masses, each m, are resting at the ends of a uniform rod of length 2a and negligible mass. The system is in equilibrium about the center C of the rod. A piece of clay of mass m is dropped down on the mass at the right end, hits it with velocity v as shown below and sticks to it.

The ratio of the kinetic energy E_f just after the collision to the kinetic energy E_i just before the collision, E_f/E_i, would be:

(A) 1
(B) 3/4
(C) 2/3
(D) 1/2
(E) 1/3

Homework Equations

Not really sure.

The Attempt at a Solution

If the system is initially in equilibrium, then it's not moving, and I don't understand how it could have kinetic energy if it's not moving. I want to write: E_f/0, but that's obviously wrong.

 

[tiny-tim]

Hi Victorzaroni!

… A piece of clay of mass m is dropped down on the mass at the right end, hits it with velocity v as shown below and sticks to it.
…
If the system is initially in equilibrium, then it's not moving, and I don't understand how it could have kinetic energy if it's not moving.

"The system" includes the clay, and that is moving!

 

[Victorzaroni]

ooohhh. wow i feel dumb. that makes sense. so its the initial kinetic energy of the system is the kinetic energy of the clay.

 

[Victorzaroni]

So E_i=(1/2)mv². Now what? I know torque is involved somehow.

 

[tiny-tim]

So E_i=(1/2)mv². Now what? I know torque is involved somehow.

torque has nothing to do with it

now find E_f (as a function of v_f)

 

[genericusrnme]

So E_i=(1/2)mv². Now what? I know torque is involved somehow.

Energy is a function of velocities (kinetic) and positions (potential), energy isn't a function of accelerations and torques

 

[Victorzaroni]

Alright... So the new kinetic energy after the clay has stuck to the mass would be KE=(1/2)(M+M)r²? How do I account for the mass on the other end of the board?

 

[tiny-tim]

add it on!

(it'll have the same speed, won't it? )​

 

[Victorzaroni]

Alright so I have: KEi=(1/2)Mv^2, and KEf=(1/2)(M+M+M)v^2. Dividing KEf by KEi yields (3/2)/(1/2), which is 3, which is not a choice? Dividing them the other way yields 1/3, choice E...

 

[tiny-tim]

i'm lost

what conservation equation are you using? ​

 

[Victorzaroni]

I'm not using conservation, KEf/KEi is what the question is asking for and the other stuff is what you've been saying. lol

 

[tiny-tim]

I'm not using conservation, KEf/KEi is what the question is asking for and the other stuff is what you've been saying. lol

how are you going to find the relation between before and after without a conservation law?

 

[Victorzaroni]

So how do i use a conservation law?

 

[tiny-tim]

what do you think will be conserved?

(and i'll repeat my previous question, since I'm still wondering …

how were you intending to find the relation between before and after without a conservation law?)​

 

[Victorzaroni]

The kinetic energy of the clay equals the rotational kinetic energy of the system?

 

[tiny-tim]

The kinetic energy of the clay equals the rotational kinetic energy of the system?

no, this is a perfectly inelastic collision …

A piece of clay … sticks to it.

… so energy is not conserved