Rotational energy & rotating rod

[Valerie Prowse]

Homework Statement

Consider a uniform rod of mass 12 kg and length l.0 m. At its end, the rod is attached to a fixed, friction-free pivot. Initially the rod is balanced vertically above the pivot and begins to fall (from rest) as shown in the diagram. Determine,

a. the angular acceleration of the rod as it passes through the horizontal at B.

b. the angular speed of the rod as it passes through the vertical at C.

Homework Equations

I = mr^2

KE(rotation)i = KE(rotation)f,

1/2Iω^2 + mgL = ...etc.

The Attempt at a Solution

I think I have it figured out, but I would just like to double check my answer as it's an assignment question.

I started with:

I (rod) = 1/3mL^2
L = 1.0m
m = 12kg
center of mass = L/2 = 0.5m

For a)

E₀ = E_f
mg(L/2) = 1/2 (1/3mL²)ω²
(mgL)/2 = 1/6mL²ω²
ω = sqrt[(3g/L)] = 5.4 rad/s

For b)
I wasn't entirely sure how to go about this part without including some form of potential energy.. since rotational energy at point A would just = E at point C and everything would be the same?? I thought maybe since point C is below the axis of rotation that the center of mass would be located at -0.5m?

1/2Iω² = 1/2Iω² - (mgL/2)
(mL²ω²)/6 + (mgL)/2 = (mL²ω²)/6
ω² (point C) = (mL²ω² + 3mgL)/mL²
ω² = (mL[Lω²+3g]/mL²) (L = 1)
ω = sqrt(ω²+3g)
ω = 7.65 rad/s

The answer does seem plausible to me, but I am not sure if my approach in adding mg(-L/2) to the KE in part B was correct. Any feedback is appreciated!

 

[haruspex]

a) asks for the angular acceleration, not the angular velocity

1/2Iω² = 1/2Iω² - (mgL/2)

Same ω on both sides?!

 

[Valerie Prowse]

a) asks for the angular acceleration, not the angular velocity

Same ω on both sides?!

Sorry, I didn't include subscripts. The angular velocity on the left in that equation is the angular velocity at point B, which I found in part A to be 5.4 rad/s, assuming that is the correct answer. The angular velocity on the right is unknown.

 

[haruspex]

Sorry, I didn't include subscripts. The angular velocity on the left in that equation is the angular velocity at point B, which I found in part A to be 5.4 rad/s, assuming that is the correct answer. The angular velocity on the right is unknown.

OK. I agree with your answer to part b. You could have got there directly using the method of part a, but not the result you got for part a. Please clarify whether a) asks for the angular velocity or angular acceleration.

 

[Valerie Prowse]

OK. I agree with your answer to part b. You could have got there directly using the method of part a, but not the result you got for part a. Please clarify whether a) asks for the angular velocity or angular acceleration.

Whoops, my bad. I misread part a.

Instead...

ω² = ω₀² + aα(θ-θ₀)
α = (ω²)/2θ
α = (5.4)²/ 2(1.57) = 9.28 rad/s²

I got the value 1.57 rad = 90° from the attached image.

 

[haruspex]

α = (ω²)/2θ

As with the linear SUVAT equations, that's only valid for constant acceleration. Look at the forces instead.

 

[Valerie Prowse]

So you mean torque?

If τ = rF⊥,
τ = (L/2)(mg) = 58.8

and

α = τ/I = τ/(1/3mL²) = 14.7 rad/s

 

[haruspex]

So you mean torque?

If τ = rF⊥,
τ = (L/2)(mg) = 58.8

and

α = τ/I = τ/(1/3mL²) = 14.7 rad/s

Looks right, except for the units.

 

[SteveS]

so for part a of this question, it's

t = r x F⊥ = 58.8 m*N

α = 58.8 / (1/3 (12kg) (1m)) = 14.7 rad/s^2

then for part b we can do

w^2 = w°^2 + 2αθ = (0) + 2 (14.7 rad/s^2)(3.14 rad)
w = 9.61 rad /s
Why couldn't you just do Ei = Ef in part a?

mgh = 1/2lw^2

with w = 5.4 rad/s

then find α with the relationship w^2 = w°^2 + 2αθ ?

 

[haruspex]

then for part b we can do

w^2 = w°^2 + 2αθ = (0) + 2 (14.7 rad/s^2)(3.14 rad)

That equation is only valid for a constant acceleration.

 

[SteveS]

I've been doing this stuff all day and my mind is really fried.

There was a similar example my professor did before though slightly different situation. Instead of continuing to position C in the diagram, the ground was position B. The example goes like this,



Ei = ½mgL.

Ef = ½Iω2

I = ⅓mL2.

Ef = 1/6 ML^2w^2

then put them together

1/2mgl = 1/6 ML^2 w^2

w = √(3g/L)

but then there is this massive jump to

v = Lw

which I do not understand.. All this is beyond me for some reason today and my final is friday :(

 

[haruspex]

v = Lw
which I do not understand.. All this is beyond me for some reason today and my final is friday :(

Where v, I assume, is the velocity at the tip of the rod.
If the rod rotates through an angle $\theta$, the tip of the rod moves through a distance (along the arc) $r\theta$, yes? If you differentiate that you should get the above equation.
Note that these equations only work if we measure angles in radians (which is the whole reason radians were invented).

 

[SteveS]

ok I'm lost lol...

So back to the question originally posted,

This is correct for part a?

t = r x F⊥ = 58.8 m*N

α = 58.8 / (1/3 (12kg) (1m)) = 14.7 rad/s^2

From there then I'm not sure where to go? Am I using the conservation of angular momentum?

 

[SteveS]

Ok so I went back at it,

I used the conservation of energy..

mgh = 1/2Iw^2

2mgh = 1/3ML^2 * w^2

works down to

w^2 = 6g/L

w = 7.67 rad/s

which is just the square root of the torque found in part a... is that just a coincidence or a short cut?

 

[haruspex]

ok I'm lost lol...

So back to the question originally posted,

This is correct for part a?

t = r x F⊥ = 58.8 m*N

α = 58.8 / (1/3 (12kg) (1m)) = 14.7 rad/s^2

From there then I'm not sure where to go? Am I using the conservation of angular momentum?

Yes, that's right for part a.
Angular momentum will not be conserved. (Can you see why?)
What will be conserved?

 

[SteveS]

Is angular momentum not conserved because the F⊥ changes as it moves through the arc?

 

[haruspex]

Is angular momentum not conserved because the F⊥ changes as it moves through the arc?

First, you have to specify what point you are taking as the reference axis for angular momentum. But in the present case, you will find that whatever point you choose there will be a force (gravity or the force from the pivot) which has a moment about that axis. Hence, angular momentum will not be conserved.

 

[SteveS]

Ah, gravity provides a constant external force so momentum can't be conserved, am I correct?

 

[haruspex]

Ah, gravity provides a constant external force so momentum can't be conserved, am I correct?

Well, if you could find a reference axis that was always in the same vertical line as mg then mg would have no torque about it so would not affect the angular momentum. The mass centre of the rod is such a point. (It is ok to take the mass centre as such an axis, but in general you should not take an accelerating axis.). But the force from the pivot will have moment about the mass centre, except when the rod is vertical, so still angular momentum will not be conserved.

 

[Jam51]

If we know angular acceleration of the first half of the rotation (to horizontal), how could I use this value to figure out the angular velocity at the bottom? I'm assuming angular acceleration would be constant throughout? I looked at some equations, but nothing that gave me the answer I had using conservation of energy.

 

[haruspex]

If we know angular acceleration of the first half of the rotation (to horizontal), how could I use this value to figure out the angular velocity at the bottom? I'm assuming angular acceleration would be constant throughout? I looked at some equations, but nothing that gave me the answer I had using conservation of energy.

No, the angular acceleration is certainly not constant. The question does not ask for the angular acceleration "for" rotation to the horizontal, it asks for it at the horizontal. For the purposes of this question, you do not need to find the acceleration at any other position.

 

[Jam51]

Ok, thanks for clearing that up! I do understand the difference now.