The discussion centers on the forces exerted by a rigid rod on a mass moving in a vertical circle. Participants agree that at the top of the circle, the tension in the rod is at its minimum, as it must counterbalance the weight of the mass and provide the necessary centripetal force. The net force exerted by the rod is a combination of gravitational and tension forces, which varies depending on the position of the mass in the circle. Key insights include the distinction between the behavior of a rod and a string, particularly in how they can exert forces in different directions based on their structural properties.
PREREQUISITESPhysics students, mechanical engineers, and anyone interested in dynamics and the forces involved in circular motion will benefit from this discussion.
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.