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I'm considering the free body diagram for an object attached to a rigid rod undergoing uniform circular motion in a vertical plane where the net force acting on the object must always be acting towards the centre of the circle.

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The free body diagram at point B shows tension in the rod acting towards the centre of the circle and a weight force (mg) acting downwards. With just these 2 forces, the net force wouldn't be acting towards the centre of the circle, so there has to be some missing force on the free body diagram (that can counteract the weight force of the object). I'm just wondering what this force would be typically called (or what an example of this missing force would be)?

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  • $\begingroup$ Who says it has to be uniform circular motion? $\endgroup$
    – d_b
    Sep 14, 2022 at 5:08

5 Answers 5

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For positions $B$ and $B'$, The free body diagram at point B shows tension in the rod acting towards the centre of the circle,which provides the centripetal acceleration, and a weight force (mg) acting downwards, which provides the tangential acceleration thus changing the tangential speed.

So in all cases the forces $T$ and $mg$ produce a centripetal acceleration and a tangential acceleration, except at points $A$ and $C$ where there is no tangential force.

enter image description here

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Because the rod is rigid, it can apply an outward force and torque in addition to the tension force in your diagram depending in the position if the rod and object. At position B, the rod not only provides the centripetal tension force, but also a torque that pushes the object up against gravity--three forces in total. If you imagine that the system stops turning and holds its position at point B, then there must be an upward force that counteracts gravity to stop the object from falling. The stiffness of the rod provides this force. Saying that the rod is rigid means that any force that tries to bend, compress, or stretch the rod will result in a reaction force on the object applying the original force.

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Net force is not acting towards center the particle also has a component of force acting along the tangent and increasing its speed.

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We may need to think about the inertia in this way; we experienced an equal opposite force ( Reaction ) acting opposite to our action on an object because of the object's own inertia that resist to the motion (acceleration).

The same situation happened in the circular motion, the object experienced the centripetal acceleration towards the center but its own inertia is resisting to this acceleration and it experienced a force acting outwards.

If we are making a circular movement inside a car, are we experiencing the force towards the center or our body experienced the tendency flying outward ? That is what the inertia is.

If we account for this inertia force on the object during the motion, it makes the perfect balance, for example at

point C : $mg + T = mv^2/r $

point B : $T = mv^2/r $

point A : $T - mg = mv^2/r $

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To have uniform motion, in a gravitational field, you have to input an angular force that varies. Because you have to periodically either resist gravity or use gravity. So there's tension associated with $a=\frac{mv^2}{r}$, gravitational force and this extra periodically changing force that I tried to derive below.

The Lagrangian,

Take $r=1$.

$L=\text{Rotational Kinetic Energy}-\text{Gravitational Potential Energy} \\= \frac{m \dot{\theta}^2}{2}-\frac{mg}{2}((1+\sin(\theta ))$

From the Euler-Lagrange Equations,

$\frac{1}{dt}\frac{\partial L}{\partial \dot{\theta}} - \frac{\partial L}{\partial \theta}= 0 \rightarrow \ddot{\theta} = g \cos(\theta)$

$F=mg\cos(\theta)$.

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