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In the above diagram, sec 1 (on the left side), an object of mass $m$, after releasing from rest from a slant track, continues into a vertical circular track. At a random position on the circular track, I have shown the forces acting on the object. Likewise in the sec 2 (on the right side), an object of mass $m$ climbs the inclined track with a initial velocity $v$. At a random position on the inclined plane, I have shown the forces acting on the object.All of the surfaces have no friction.

Hoping the forces which have shown above are correct, I have below questions:

  1. In sec 1, $N1 - mg\cos(a)$ acts as centripetal force which is required for changing of direction of velocity along the circular track. But what does $mg\sin(a)$ do? Does it acts as the decelerating force which changes the tangential velocity and hence leading this object into a non-uniform circular motion? If so, Is vertical uniform circular motion even possible?

  2. In sec 2, $N2 = mg\cos(a)$ since there is no acceleration in that direction. Also the object decelerates along the inclined plane, with a magnitude of $g\sin(a)$ until $t = v/g\sin(a)$, after that it accelerates back down the inclined plane. Is this analysis correct?

  3. Why are the normal reaction forces are different in the two scenarios? On what does the normal reaction force depend?

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  1. Is vertical uniform circular motion even possible?

No, it isn't. Because magnitude of velocity isn't constant and we know that in a uniform circular motion the object moves with constant speed. $\large{\frac {\mathrm d}{\mathrm dt}}v=g\sin\alpha\neq 0$ ($v$ is the speed (magnitude of the velocity vector $\vec v$) of the object)

  1. Is this analysis correct?

Yes, it is.

  1. Why are the normal reaction forces are different in the two scenarios?

Because the object experiences different motions in the two scenarios. Equation of motion for a particle with constant mass is $\Sigma\vec F=m\vec a$. If the right side of motion's equation is different for two scenarios; then, the left side of that will certainly be different. So, in the instant that angle $\alpha$ is same for two scenarios, the normal reaction forces will be different. Because in the first case, we have $N=mg\cos\alpha+m\large{\frac{v^2}R}$ and in the second case we have $N=mg\cos\alpha$

  1. On what does the normal reaction force depend?

Normal reaction force depends on the pressure that two surfaces exert on each other and area of contact surface $\mathrm dN=P\mathrm dA$

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  • $\begingroup$ +1 Nice and clear answer. To be fair, uniform circular motion is possible, but then the centripetal force must vary to make up for the loss in velocity caused by gravity. That's the reason for pulling harder in the string when the yo-yo is on it's way up, when you swing a yo-yo vertically around. $\endgroup$ – Steeven Jul 21 '16 at 7:01
  • $\begingroup$ @Steeven Thanks! To be honest, I couldn't get your point. Because in uniform circular motion, the magnitude of the velocity must be constant but in this case it varies due to variation of angle of $\alpha$. ($\large{\frac {\mathrm d}{\mathrm dt}}v=g\sin\alpha\neq 0$) And I cannot find out how centripetal force can effect on magnitude of the velocity when we use orthogonal coordinate system. Thanks again because of edit! $\endgroup$ – lucas Jul 21 '16 at 8:12
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1.mgsin(a) stands for the projection of the gravity on the x axis, if you take the axis like, y-axis is N1 and mgcos(a) and x-axis magsin(a). It is the force that stops the mass, so it can´t be uniform because if it has no acceleration and it is being dragged down it eventually stops. 2.I would write using Newton

$$m \cdot a = g \cdot sin (a)$$

then integrate on both sides using that a is the derivative of v

$$v-v_0=g \cdot sin (a) \cdot t \rightarrow t=\frac{v-v_0}{sin(a)}$$.

If it onlye has the downward force acting on it, it goes up with the starting velocity and then down. 3.So the projections $mgsin(a)$ and $mgcos(a)$ depend on a parameter called $a$ that is the angle between the vector $mg$ and $mgcos(a)$. in sec 1 that angle is changing along the curve. If your see the picture if the mass goes a but up that angle is inferior than before because the normal force is always perpendicular tu the surface. On the other hand in sec 2 that angle is always the same so $mgcos(a)$ is a constant.

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  • $\begingroup$ "because if it has no acceleration and it is being dragged down it eventually stops" This sounds contradictory $\endgroup$ – Steeven Jul 21 '16 at 6:59

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