What is the Maximum Kinetic Energy of a Rotating Rigid Rod System?

[makeAwish]

Homework Statement

A rigid, massless rod has three particles with equal masses attached to it as shown in Figure P11.45. The rod is free to rotate in a vertical plane about a frictionless axle perpendicular to the rod through the point P, and is released from rest in the horizontal position at t = 0. Assuming m and d are known, find
(a) the moment of inertia of the system (rod plus particles) about the pivot,
(b) the torque acting on the system at t = 0,
(c) the angular acceleration of the system at t = 0,
(d) the linear acceleration of the particle labeled 3 at t = 0,
(e) the maximum kinetic energy of the system,
(f) the maximum angular speed reached by the rod,
(g) the maximum angular momentum of the system, and
(h) the maximum speed reached by the particle labeled 2.


The attempt at a solution

I have found moment of inertia, I = 7md^2/3
torque = mgd k, where k is unit vector
angular accel = 3g/7d
accel = 2g/7

but i don't know how to solve for maximum kinetic energy.
Thinking of consv of energy, and from the COE eqn, there seems to be potential energy. But i don't understand why PE exists..

Can someone explain to me? Thanks!

 

[sArGe99]

Rotational K.E is $$\frac{1}{2} I \omega^2$$

 

[sArGe99]

I can't see the diagram and can't make any conclusions until I do so..
But I reckon that by max. kinetic energy, they're asking the rotational kinetic energy. First find max. angular velocity and solve for K.E

 

[makeAwish]

I can't see the diagram and can't make any conclusions until I do so..
But I reckon that by max. kinetic energy, they're asking the rotational kinetic energy. First find max. angular velocity and solve for K.E

Yah, i also waiting for the diagram to be approved.
Hmm.. Okies.

But is it possible to use COE to solve it?

 

[Doc Al]

but i don't know how to solve for maximum kinetic energy.
Thinking of consv of energy, and from the COE eqn, there seems to be potential energy. But i don't understand why PE exists..

Yes, conservation of energy is the way to solve this. Whenever masses change height there will be a change in gravitational PE. (You can use any vertical position as your reference point for calculating PE.)

What position will have the maximum KE?

 

[makeAwish]

Yes, conservation of energy is the way to solve this. Whenever masses change height there will be a change in gravitational PE. (You can use any vertical position as your reference point for calculating PE.)

What position will have the maximum KE?

So in this case, as the rod is shown to be horizontal initially, it has initial PE or final PE, after it rotates?

hmm, it has max KE initially, when the rod is released?

 

[Doc Al]

So in this case, as the rod is shown to be horizontal initially, it has initial PE or final PE, after it rotates?

It starts out with some initial PE and ends up with some final PE (by definition of initial and final). Of course, if you choose the initial height as your reference point then the initial PE will be zero. Up to you. The only thing that matters is the change in PE.

hmm, it has max KE initially, when the rod is released?

It's released from rest, thus its initial KE is zero.

 

[makeAwish]

It starts out with some initial PE and ends up with some final PE (by definition of initial and final). Of course, if you choose the initial height as your reference point then the initial PE will be zero. Up to you. The only thing that matters is the change in PE.

It's released from rest, thus its initial KE is zero.

Oh ya! its released from rest. hmm.. Oh, is it when the rod has turned CCW 270 degrees? When it is vertical, and the 2m are on top (where they rotate abt pt P)..

hmm. say i let my reference point be point P, initally does the rod has PE? cos PE = mgh, where h is vertical ht is it? But my rod is horizontal..

 

[Doc Al]

Oh, is it when the rod has turned CCW 270 degrees? When it is vertical, and the 2m are on top (where they rotate abt pt P)..

Why 270 degrees? The KE will be maximum when the PE is minimum. So what position will have the masses as low as they can get?

hmm. say i let my reference point be point P, initally does the rod has PE? cos PE = mgh, where h is vertical ht is it? But my rod is horizontal..

Use PE = mgy. If you choose y = 0 at the height of point P, then the initial PE will be 0.

 

[makeAwish]

Why 270 degrees? The KE will be maximum when the PE is minimum. So what position will have the masses as low as they can get?

Use PE = mgy. If you choose y = 0 at the height of point P, then the initial PE will be 0.

oh ya. so 270degrees is max PE rite?
max KE is when the rod turns 90 degrees CCW?

So is only when my rod has turned, will there be PE rite? Can i say a rod being horizontal won't have PE?

 

[Doc Al]

oh ya. so 270degrees is max PE rite?
max KE is when the rod turns 90 degrees CCW?

Right.

So is only when my rod has turned, will there be PE rite? Can i say a rod being horizontal won't have PE?

It just depends on how you choose your reference point. For example, you can choose the lowest position of mass 1 (when the stick is vertical and mass 1 is down) as your y = 0 point, or you can choose the height of the point P as y = 0. It doesn't matter. What matters is the change in PE, and that will be the same regardless of reference point chosen. (Try it several ways to convince yourself of this.)

 

[makeAwish]

okay. means in this case, i can take the final PE to be zero is where Ke is max? and initially Ke = 0 and PE is not zero?

 

[Doc Al]

okay. means in this case, i can take the final PE to be zero is where Ke is max? and initially Ke = 0 and PE is not zero?

Again, it just depends on where your reference point is. If you use the same reference point for all masses, then the final PE will not be zero (since they are at different heights). Just choose a reference point and go for it.

 

[makeAwish]

Okay i go try. Thanks! :)