The discussion revolves around the forces exerted by a rigid rod on a mass that is rotating in a vertical circle. Participants are analyzing the dynamics involved, particularly at various positions in the circular path, including the top and horizontal positions, and questioning the nature of tension in the rod compared to a string.
The discussion is ongoing, with various interpretations being explored regarding the forces at play. Some participants have offered insights into the conditions under which the rod may exert different types of forces, while others are seeking clarification on specific scenarios. There is no explicit consensus yet, as participants continue to question and analyze the problem.
Participants are considering the implications of the mass's weight and the role of external forces in maintaining constant speed. The discussion also touches on the differences between the forces exerted by a rod and those exerted by a string in similar scenarios.
Yes. All real forces are "action-reaction" forces. If the rod pushes up on the mass, the mass pushes down on the rod. That is Newton's third law.Father_Ing said:Is this an action-reaction force? Which means there will be a force downward (with the same in magnitude as the force you have stated) that applies on the rod?
Is it not possible for the end of the rod to be the one which keep the speed constant? As what @vcsharp2003 is trying to say.haruspex said:Not sure what you mean by "need" there.
We know it is radially inward because we are told the mass moves in a circle at uniform speed. Presumably this is arranged by a varying torque applied at the axle on which the rod rotates.
I understand that question even less.Rikudo said:Is it not possible for the end of the rod to be the one which keep the speed constant? As what @vcsharp2003 is trying to say.
I was just trying to decide whether ##a_t>0## or ##a_t=0##, where ##a_t## stands for tangential acceleration in horizontal position. I wasn't trying to understand why speed is constant, which you've explained through a varying torque mechanism.haruspex said:Not sure what you mean by "need" there
Tangential acceleration is the component of acceleration parallel to the velocity. If speed is constant, that must be zero.vcsharp2003 said:I was just trying to decide whether ##a_t>0## or ##a_t=0##, where ##a_t## stands for tangential acceleration in horizontal position. I wasn't trying to understand why speed is constant, which you've explained through a varying torque mechanism.
Yes, I get that part. If only direction of velocity is changing but not it's magnitude then tangential acceleration must be zero.haruspex said:Tangential acceleration is the component of acceleration parallel to the velocity. If speed is constant, that must be zero.
(In vectors, ##0=\frac{d}{dt}v^2=\frac{d}{dt}\vec v^2=\vec v.\dot{\vec v}=||\vec v|| a_t##.)
Why did you take derivative of ##v^2##?haruspex said:(In vectors, ##0=\frac{d}{dt}v^2=\frac{d}{dt}\vec v^2=\vec v.\dot{\vec v}=||\vec v|| a_t##.)
The speed, v, is constant if and only if ##v^2## is constant. Using the squared form let's me write it in vectors easily.vcsharp2003 said:Why did you take derivative of ##v^2##?
That's a very nice mathematical way of reasoning it, that I was unaware of. Thankyou for that great tip.haruspex said:The speed, v, is constant if and only if ##v^2## is constant. Using the squared form let's me write it in vectors easily.
But I did omit a factor 2, now corrected.