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How can we prove that a particle (ball) moving along a circular path with the only forces acting being its weight and the normal force (forming the centripetal force) will result in uniform (constant speed) circular motion. Furthermore why is the normal force present in point D and why is it bigger in magnitude in point B compared to the y-axis component of the weight?

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  • $\begingroup$ Please define "Natural force." I've not heard of it. $\endgroup$
    – Bill N
    Mar 9, 2017 at 20:00
  • $\begingroup$ If for example the ball is moving along a circular path (road) the natural force is the reaction of the weight exerted from the ground to our ball. If we imagine the ball being tied by a string the role of the natural force would be the tension of the string. $\endgroup$
    – TheNotoriousSc
    Mar 9, 2017 at 20:03
  • $\begingroup$ @BillN The force of nature, I presume ;) $\endgroup$
    – SchrodingersCat
    Mar 9, 2017 at 20:08
  • $\begingroup$ Oh, you mean the "normal force." It's called normal because it is perpendicular to the tangent of the contact point/surface, not because it's "natural." $\endgroup$
    – Bill N
    Mar 9, 2017 at 20:08
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    $\begingroup$ You can't prove uniform circular motion happens for those two forces. Uniform circular motion is usually a constraint which the author of a problem imposes on a system. In fact, for a vertical circle with only weight and normal force (from some track or rail), uniform circular motion isn't possible. $\endgroup$
    – Bill N
    Mar 9, 2017 at 20:13

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You can't prove uniform circular motion happens for those two forces. Uniform circular motion is usually a constraint which the author of a problem imposes on a system. In fact, for a vertical circle with only weight and normal force (from some track or rail), uniform circular motion isn't possible.

The normal force acts perpendicular to the circular motion of the particle, so it does zero mechanical work on the ball. Only the weight does work, and we can account for that by considering a constant mechanical energy problem including the gravitational potential energy of the ball/Earth system: $$ E_{\mathrm{top}}=E_{\mathrm{bottom}}$$ $$ \frac{1}{2}mv_{\mathrm{top}}^2+mg(2R)=\frac{1}{2}v_{\mathrm{bottom}}^2$$ where $R$ is the radius of the circle, $m$ is the ball mass, $g$ is the gravitational field strength, and the reference point for the gravitational potential energy is chosen to be a the bottom of the circle.

From this it's clear that the speed at the top is different from the speed at the bottom. We could generalize this to angular positions (you can do the maths if you're curious), but we clearly see this is not, nor can it be, uniform circular motion.

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  • $\begingroup$ So it is not necessary for an object that has a circular trajectory to have $\sum_{}^{}F=F_k=\frac{u^2}{R}$ $\endgroup$
    – TheNotoriousSc
    Mar 9, 2017 at 22:07
  • $\begingroup$ That is still necessary for curved motion. That is not restricted to uniform circular motion. $v^2/R$ is the required perpendicular acceleration for any curvilinear motion. $\endgroup$
    – Bill N
    Mar 9, 2017 at 22:54
  • $\begingroup$ Does it also have to be constant the accelaration or can we have an object moving alonga a circular path with different speed at different times $\endgroup$
    – TheNotoriousSc
    Mar 9, 2017 at 22:57
  • $\begingroup$ Does not have to be constant. $\endgroup$
    – Bill N
    Mar 10, 2017 at 3:23
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Consider circle of radius $R$ with the origin at its center, where $\vec{r} = R(cos\theta \hat{i} + sin\theta \hat{j})$ is the radius vector of a particle moving on this circle.

If you differentiate this vector twice wrt time, you will find that the centripetal acceleration has to, at every instant be equal to $v^2/R$

Now if you can show that the net force on a particle is always directed towards the center and has a magnitude of $v^2/r$(which won't change because the net force is perpendicular to the disp) , it means that the particle will travel in a circle of radius $R$ with constant velocity $v$

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  • $\begingroup$ How do you prove that $v$ is constant if it's a vertical circle as the OP originally assumes (see the diagram)? Velocity isn't constant because the direction is changing, and the speed isn't constant because the tangential acceleration is not zero. $\endgroup$
    – Bill N
    Mar 10, 2017 at 13:12
  • $\begingroup$ If it is a vertical circle, then the magnitude of velocity simply isn't constant and it isn't uniform circular motion which the OP is trying to prove. I've simply shown him how to prove that a circular motion is uniform. He can apply this 'test' and easily prove that in a vertical circle, a particle doesn't undergo uniform circular motion $\endgroup$
    – xasthor
    Mar 10, 2017 at 14:31

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