How Does Mass Distribution Affect Rotational Dynamics of a Rod?

[Panphobia]

Homework Statement

A rod of mass M and length l rotates in a vertical plane about its centre which is on a frictionless, horizontal pivot. On the ends of the rod are point-like masses m1 and m2, where m1 != m2.

a)moment of inerta about the center of the rod
b)Determine the angular momentum when the angular velocity is ω
c) Determine the angular acceleration when the rod makes an angle of θ with respect to the horizontal z-axis. For m1 > m2 what is the direction of the torque?
d) At what angle θ is the angular velocity ω the greatest?

The Attempt at a Solution

So I would know how to do this question without the point masses m1 and m2, what they do is change the centre of mass of the rod, so does this question have something to do with the change in the centre of mass? Also what does it mean when the rod makes an angle of θ with respect to the horizontal, does this mean it is slanted?

 

[Simon Bridge]

I'm reading this problem as implying that the rod is initially placed horizontal.
Ignoring the problem for a moment, just picture this situation - what will happen when the rod is let go?

 

[Panphobia]

What do you mean? Its already in motion.

 

[nil1996]

Just you need to find the moment of inertia first.The moment of inertia of a rod around its centre is ML^2/12.
Now find the moment of inertia when the two masses are attached.

 

[Panphobia]

What is the difference? I know there is one, but in class we have only gone so far as to calculate moments of inertia, not any of this. Would I add the moments of inertia of the points to the moment of inertia of the rod?

 

[nil1996]

Yes as they are point masses you can add them!

 

[Panphobia]

so (1/12)ML^2 + (1/2)m1L^2 + (1/2)m2L^2 = I ?
But since the moment of inertia for a point mass is mR^2 and R = L/2 would it actually be (1/4)mL^2?

 

[nil1996]

so (1/12)ML^2 + (1/2)m1L^2 + (1/2)m2L^2 = I ?
But since the moment of inertia for a point mass is mR^2 and R = L/2 would it actually be (1/4)mL^2?

moment of inertia is given by MR² where M==mass of the object;R==distance of the object from the axis around which we want moment of inertia.

 

[Panphobia]

ahhh so it is (1/4)m(1+2)L^2, thanks! I can figure out a) and b) with that...now for the others. What does c) mean by angle θ with respect to the horizontal? Is it slanted?

 

[nil1996]

ahhh so it is (1/4)m(1+2)L^2, thanks! I can figure out a) and b) with that...now for the others. What does c) mean by angle θ with respect to the horizontal? Is it slanted?

wait what is that (1/4)m(1+2)L^2 ?
why you have done m(1+2)

 

[Panphobia]

Because 1/4m1L^2 + 1/4m2L^2 = (1/4)m(1+2)L^2

 

[nil1996]

Because 1/4m1L^2 + 1/4m2L^2 = (1/4)m(1+2)L^2

It should like this-
1/4m₁L²+1/4m₂L²

those are only different masses indicated by m₁ and m₂

I=ML²/12 + 1/4m₁L² + 1/4m₂L²


I=ML²/12+1/4L²(m₁+m₂)

 

[Panphobia]

So about my question in post #9?

 

[nil1996]

when it is at an angle (theta ) then it is slanting to the z axis.

see the attachment

 

[Panphobia]

what significance does m1 > m2 have?

 

[nil1996]

it is necessary for rotating the rod, for giving it angular acceleration.

 

[Panphobia]

This has something to do with T = I\alpha correct?

 

[nil1996]

right:)

 

[Panphobia]

Hmm this may be a dumb question but it says it is making an angle with the horizontal axis, but I\alpha does not incorporate an angle, but T = r x F does, but then I am still stumped.

 

[nil1996]

T = r x F
yes it is a cross product and needs the angle

τ=r*F*sinθ

τ is the magnitude of the torque,
r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied),
F is the force vector,
θ is the angle between the force vector and the lever arm vector.

 

[Panphobia]

But that is difficult to say because there are two mass points. I was thinking r = L/2, then F = ma, and a = \alphar so T = m*r^2*\alpha*sinθ, but I can't help thinking I am missing something because there is another mass...

 

[nil1996]

But that is difficult to say because there are two mass points. I was thinking r = L/2, then F = ma, and a = \alphar so T = m*r^2*\alpha*sinθ, but I can't help thinking I am missing something because there is another mass...

you should first calculate the net torque at the center of the rod by both rods.Thn use T=I*(alpha)

 

[Panphobia]

soooo T1 = m1*r*a*sinθ and T2 = m2*r*a*sinθ, then Tnet = T1 + T2

 

[nil1996]

soooo T1 = m1*r*a*sinθ and T2 = m2*r*a*sinθ, then Tnet = T1 + T2

how it is sinθ?

 

[Panphobia]

What is the force vector in this question? I don't know that.

 

[nil1996]

the force due to point masses is the force vector here.It acts vertically downwards.

 

[Panphobia]

wait so that means that θ between the force vector and displacement vector is (90 - (horizontal angle))?

 

[nil1996]

wait here force vector is the component of mg that is perpendicular to the rod

 

[Panphobia]

oh right right, so then θ is the actual horizontal angle. I am still kind of iffy on this concept, it is in 3-d, so the rod has been tilted to the horizontal axis, and is not spinning towards this horizontal axis, so I am not completely sure how you can use the angle with respect to the horizontal to get the perpendicular component of mg.

 

[haruspex]

oh right right, so then θ is the actual horizontal angle. I am still kind of iffy on this concept, it is in 3-d, so the rod has been tilted to the horizontal axis, and is not spinning towards this horizontal axis, so I am not completely sure how you can use the angle with respect to the horizontal to get the perpendicular component of mg.

No. it's all in 2D, a vertical plane.
It doesn't matter whether the rod is rotating or not at this point. We just want to calculate the acceleration due to gravity when it's at angle theta to the horizontal. What torques do the gravitational forces on the different masses exert about the axis? How do they combine to a net torque?

 

[Panphobia]

Oh my I read this question totally wrong, I am so sorry, silly me. The torques are T1 = (L/2)*m1gcosθ*sinθ, T2 = (L/2)*m2gcosθ*sinθ where sinθ = 1 because the angle between the two vectors is 90.

 

[haruspex]

Oh my I read this question totally wrong, I am so sorry, silly me. The torques are T1 = (L/2)*m1gcosθ*sinθ, T2 = (L/2)*m2gcosθ*sinθ where sinθ = 1 because the angle between the two vectors is 90.

The two vectors, surely, are the vertical gravitational force and the radius from the axis to the mass. These are only at right angles when the rod is horizontal. But even then, your formula would make the torque zero there because cos θ would be zero. That is clearly not right.

 

[Panphobia]

wait let me change that T1 = (L/2)*m1gcosθ'*sinθ, T2 = (L/2)*m2gcosθ'*sinθ where θ' = angle with respect to horizontal

 

[haruspex]

wait let me change that T1 = (L/2)*m1gcosθ'*sinθ, T2 = (L/2)*m2gcosθ'*sinθ where θ' = angle with respect to horizontal

Now you have two angles, θ and θ'. Are you taking them as adding to 90 degrees, so θ is the angle to the vertical? If so, that's giving you sin²(θ).
You have these two vectors at each mass: the vertical gravitational force and the radius vector to the mass from the axis. The torque is given by the cross product. What is the magnitude of the cross product in terms of the magnitudes of the vectors and the angle between them?
Or, if you don't want to deal in terms of cross products, what is the horizontal distance from the axis to one of the point masses?

 

[Panphobia]

Wait look, the angle between m1gcosθ' and L/2 is 90, so that means that sinθ = 1, and then θ' = the original thing given in c) (the angle θ with respect to the horizontal). So I do not know what you are talking about.

 

[haruspex]

Wait look, the angle between m1gcosθ' and L/2 is 90, so that means that sinθ = 1, and then θ' = the original thing given in c) (the angle θ with respect to the horizontal). So I do not know what you are talking about.

Things may have become confused because at one point I deduced from your equations that theta was the angle to the vertical, and I didn't go back and check the OP. When you introduced θ' you didn't define it, and I guessed wrongly.
I think you are now using θ' for what was given in the OP as θ (the angle to the horizontal) and fixing θ at 90 degrees. (Not sure why you introduce it in that case.)
If so, your equation T1 = (L/2)*m1gcosθ'*sinθ reduces to T1 = (L/2)*m1gcosθ', and in terms of the original θ that's T1 = (L/2)*m1gcosθ. With which I agree!
The next step is to get the net torque. Watch the signs.

 

[Panphobia]

So I have to figure out the positive direction of Torque. But then since both of the masses will be moving either clockwise or counter clockwise at the same time, that is kind of pointless, so just adding T1 and T2 should be enough, or am I missing something?

 

[haruspex]

So I have to figure out the positive direction of Torque. But then since both of the masses will be moving either clockwise or counter clockwise at the same time, that is kind of pointless, so just adding T1 and T2 should be enough, or am I missing something?

The direction they're moving does not affect the torque. They are at opposite ends of the rod, so the torques must oppose.

 

[Panphobia]

Ohhhh yes, since the mg's are all point downwards, the direction of torque is opposite, good good. So it is Tnet = T1 - T2?

 

[haruspex]

Ohhhh yes, since the mg's are all point downwards, the direction of torque is opposite, good good. So it is Tnet = T1 - T2?

Yes.

 

[Panphobia]

Is the mass of the rod irrelevant to the torque?

 

[haruspex]

Is the mass of the rod irrelevant to the torque?

Yes, it balances out. Would there be any acceleration if there point masses were not there?

 

[Panphobia]

I guess not, yea I see what you are talking about, because the rotation axis is at the centre of mass of the rod. So if I wanted the angle at which the ω was maximum, would torque = 0 then? if torque = 0 then the angle would be 90 to the horizontal correct? By the way my thinking process is, L = Iω and the derivative of L is T, when velocity is a maximum, acceleration is 0, soo that's what I think.

 

[haruspex]

I guess not, yea I see what you are talking about, because the rotation axis is at the centre of mass of the rod. So if I wanted the angle at which the ω was maximum, would torque = 0 then? if torque = 0 then the angle would be 90 to the horizontal correct? By the way my thinking process is, L = Iω and the derivative of L is T, when velocity is a maximum, acceleration is 0, soo that's what I think.

Yes, that works.