What is the Angular Speed of a Rod Pivoted at One End?

[fight_club_alum]

Homework Statement

A uniform rod (mass = 1.5 kg) is 2.0 m long. The rod is pivoted about a horizontal, frictionless pin through one end. The rod is released from rest in a horizontal position. What is the angular speed of the rod when the rod makes an angle of 30 degrees with the horizontal? (The moment of inertia of the rod about the pin is 2 kg m^2)

Homework Equations

mg(H intial - H final) =1/2 I w^2

The Attempt at a Solution

My attempt:
M = 1.5 kg
L = 2 m
Moment of inertia (I) = 2 kg m^2
mg(H intial - H final) =1/2 I w^2
1.5 x 9.8 x 2sin(30) = 1/2 x 2 x w^2
w^2 = 14.7
w = 3.64

 

[Merlin3189]

A uniform rod (mass = 1.5 kg) is 2.0 m long. The rod is pivoted about a horizontal, frictionless pin through one end. The rod is released from rest in a horizontal position. What is the angular speed of the rod when the rod makes an angle of 30 degrees with the horizontal? (The moment of inertia of the rod about the pin is 2 kg m^2)

Homework Equations

mg(H intial - H final) =1/2 I w^2

The Attempt at a Solution

My attempt:
M = 1.5 kg
L = 2 m
Moment of inertia (I) = 2 kg m^2
mg(H intial - H final) =1/2 I w^2
1.5 x 9.8 x 2sin(30) = 1/2 x 2 x w^2
w^2 = 14.7
w = 3.64[/B]

So you seem to think that (H intial - H final) = 2sin(30)
How about a diagram?
Especially to show the movement of the centre of mass.

 

[fight_club_alum]

So you seem to think that (H intial - H final) = 2sin(30)
How about a diagram?
Especially to show the movement of the centre of mass.

I see my mistake so I should always use the center of mass to determine the angular acceleration or speed?
Thank you

 

[Merlin3189]

Well, I don't know about 'always'. But here, mgh represents the change in gravitational PE as the rod changes position.
Although the free end of the rod falls 1 m, the other end does not fall at all. So you can't say all the mass of the rod has fallen 1 m.
We simplify the calculation by using the centre of mass concept, which sums up the vertical movement of all the bits of rod.

This will not work for the angular momentum: you can't treat the rod as having all its mass at the centre of mass when calculating angular momentum. This is obvious if you think about it rotating about its own CoM: it still has angular momentum even though the product of mass x distance from axis is zero.

In fact the angular momentum of an object about a particular axis is thought of in two parts: the angular momentum of the object about its own CoM, which is given by the moment of inertia, plus the angular momentum of the CoM about the given axis. (If the given axis is through the CoG, then the second part is zero, but not the first.)

So for the other part of your calculation (which you glossed over and didn't mention in the "Relevant Formulae" part of the template) is, what is the formula for rotational KE?

 

[fight_club_alum]

Well, I don't know about 'always'. But here, mgh represents the change in gravitational PE as the rod changes position.
Although the free end of the rod falls 1 m, the other end does not fall at all. So you can't say all the mass of the rod has fallen 1 m.
We simplify the calculation by using the centre of mass concept, which sums up the vertical movement of all the bits of rod.

This will not work for the angular momentum: you can't treat the rod as having all its mass at the centre of mass when calculating angular momentum. This is obvious if you think about it rotating about its own CoM: it still has angular momentum even though the product of mass x distance from axis is zero.

In fact the angular momentum of an object about a particular axis is thought of in two parts: the angular momentum of the object about its own CoM, which is given by the moment of inertia, plus the angular momentum of the CoM about the given axis. (If the given axis is through the CoG, then the second part is zero, but not the first.)

So for the other part of your calculation (which you glossed over and didn't mention in the "Relevant Formulae" part of the template) is, what is the formula for rotational KE?

I think I mentioned
mg(H intial - H final) =1/2 I w^2
So, KE = 1/2 x I x w^2

 

[Merlin3189]

Sorry, so you did.
However, it might help you to step back from the "Equation" and thnk about the formulae you've used.

mg(H intial - H final) is the formula for change in gravitational PE (I'll guess that m is mass, g is accn due to gravity and H is height)

1/2 I w^2 is the formula for the rotational KE of an object rotating about its own CoM (assuming that I is the moment of inertia and w the angular speed)

Now that it is stated clearly, can you see the flaw?

 

[fight_club_alum]

Sorry, so you did.
However, it might help you to step back from the "Equation" and thnk about the formulae you've used.

mg(H intial - H final) is the formula for change in gravitational PE (I'll guess that m is mass, g is accn due to gravity and H is height)

1/2 I w^2 is the formula for the rotational KE of an object rotating about its own CoM (assuming that I is the moment of inertia and w the angular speed)

Now that it is stated clearly, can you see the flaw?

Yes I see thank you so much I will pay more attention to the meaning in the future

 

[Merlin3189]

Aaarghh! I've just noticed, I'm wrong about the moment of inertia.

Your question specified that it is the moment of inertia.about the pin at the end of the rod. So they've taken that into account and that part of your formula is correct.

Sorry about that. I must have been in a mental rut about CoM and pivot points from looking at the first bit.

 

[fight_club_alum]

Aaarghh! I've just noticed, I'm wrong about the moment of inertia.

Your question specified that it is the moment of inertia.about the pin at the end of the rod. So they've taken that into account and that part of your formula is correct.

Sorry about that. I must have been in a mental rut about CoM and pivot points from looking at the first bit.

It's ok no problem