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Normally, if an object of mass $m$ is inclined to the horizontal at an angle $b$, we set the reaction force of the object on the inclined plane as $R = mg\cos{b}$ (if we resolve the force of gravity so the line of action coming out of the plane is perpendicular to it).

However in circular motion*. it's assumed that $R\cos{b} = mg$. In the example above, one would have to do this in order to arrive to the correct answer, instead of $R = mg\cos{b}$. Using $R = mg\cos{b}$ seems natural enough, as I am resolving vertically, however, both equations would produce two different values for $R$. Why is this?

To show what I mean: If we set the reaction force in this question as $mg\cos{a}$, then the centripetal force will be $mg\cos{b}\cos(\pi/2-b) = mg\cos{b}\sin{b} = \frac{1}{2}mg\sin(2b)$

Whereas If we use $R\cos{b} = mg$, $R = mg\sec{b}$ and the centripetal force will be $mg\sec{b}\sin{b} = mg\tan{b}$. This will end up with two different values for the radius of the circular motion, and hence two different final answers.

*In the circular motion questions I've seen in my mechanics module

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  • $\begingroup$ I understand there's obviously something I'm not understanding, but I can't figure it out. $\endgroup$ – Logan545 May 3 '17 at 15:57
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    $\begingroup$ "we set the reaction force of the object on the inclined plane as R=mgcosa" you don't set it: you decompose the vector along its components. When doing so on a case by case scenario the equation changes according to the geometry. $\endgroup$ – gented May 3 '17 at 16:05
  • $\begingroup$ @GennaroTedesco I see where I went wrong. By the way, after decomposing a vector into components, does it make any sense to decompose the components as well? $\endgroup$ – Logan545 May 3 '17 at 16:36
  • $\begingroup$ Decomposing the components...into what? $\endgroup$ – gented May 3 '17 at 16:39
  • $\begingroup$ @GennaroTedesco Into other components $\endgroup$ – Logan545 May 3 '17 at 16:48
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Never, ever, just blindly memorize formulae.

What you need to do is draw a free-body diagram of your particle, which will have an angled normal force, and a downward gravitational force, and you know that the net acceleration is inward with magnitude $v^{2}/r$. You can either rotate your reference frame so that the normal force is upward, and the gravitational force is angled, or work out the two equations, eliminating the normal force.

Either way, you'll arrive at an answer. But the text of the question presupposes that you can just memorize a formula for a situation. Never do this, look at a situation, and work out the answer. You will end up wrong as often as you don't if you try and solve problems the way you seem to be -- because all it takes to be wrong is someone labeling an angle in a funny way, or using a slightly different convention.

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  • $\begingroup$ I never blindly memorise formuale. I simply cut to the chase and assumed readers would know what I mean when I said R can be expressed as mgcosa . My point is, is that shouldn't the reaction force (R) have the same value no matter what method (as long as it's correct) I use to arrive to it? If I use the two different values I stated above, my centripetal force will be different and so I'll arrive at two different answers (by the way, the circular motion doesn't have a radius of 3a). $\endgroup$ – Logan545 May 3 '17 at 15:44
  • $\begingroup$ You never specify what is "a" under sin or cosine. From the given data, a seems to be a length and not an angle. This is an examples of using formulas "blindly". $\endgroup$ – nasu May 3 '17 at 16:04
  • $\begingroup$ @nasu My bad, amended. $\endgroup$ – Logan545 May 3 '17 at 16:06
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    $\begingroup$ The fact that you change its name does not help to identify it. The use of sin or cos depends of what angle are you talking about. $\endgroup$ – nasu May 3 '17 at 16:08
  • $\begingroup$ @nasu b is the angle of the cone to the horizontal. $\endgroup$ – Logan545 May 3 '17 at 16:14
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@JerrySchirmer's advice is generally good, and worth heeding. If you actually construct the free-body diagrams for a particle on an inclined plane and your particle on a cone, you will note the following important difference:

  • A particle at rest on an incline (or sliding down an incline) has an acceleration vector that is parallel to the surface.
  • Your particle moving around the inside of a cone, on the other hand, does not have an acceleration vector parallel to the surface: it is accelerating horizontally towards the center of the circle instead.

In both cases, you can then use the fact that the particle is not accelerating in the "other" direction (perpendicular to the plane for the incline; vertically for the cone) to write down a relation between the particle's weight and the reaction force. But these respective equations deal with the components of those forces in different directions, and so they turn out differently from each other.

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  • $\begingroup$ Parellel to the surface is same as tangential to the surface......in this case it does have tangential acceleration but not vertical acceleration.....so how come it is not accelerating in parallel to the surface. $\endgroup$ – Abhinav Jul 20 '18 at 0:24

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