The discussion revolves around the rotational dynamics of a rod with point masses attached to its ends, focusing on how mass distribution affects its motion. The problem involves calculating the moment of inertia, angular momentum, and angular acceleration of the system as it rotates in a vertical plane about a pivot.
The discussion is active, with participants sharing insights on calculating moment of inertia and torque. Some guidance has been provided regarding the addition of moments of inertia for point masses, and there is ongoing exploration of the relationship between angles and forces in the context of torque.
Participants are navigating the complexities of a multi-mass system and its rotational dynamics, with some uncertainty about the definitions and implications of angles in the problem setup. There is a focus on understanding the gravitational forces acting on the masses and their contributions to torque.
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 said: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.
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).Panphobia said:wait let me change that T1 = (L/2)*m1gcosθ'*sinθ, T2 = (L/2)*m2gcosθ'*sinθ where θ' = angle with respect to horizontal
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.Panphobia said: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.
The direction they're moving does not affect the torque. They are at opposite ends of the rod, so the torques must oppose.Panphobia said: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?
Panphobia said:Ohhhh yes, since the mg's are all point downwards, the direction of torque is opposite, good good. So it is Tnet = T1 - T2?
Panphobia said:Is the mass of the rod irrelevant to the torque?
Panphobia said: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.