How Does Mass Distribution Affect Rotational Dynamics of a Rod?

[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.