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I've been going in circles (no pun intended) with this problem for several hours and yet I can't seem to find what am I doing wrong or misunderstanding, hence I hope somebody can help me with this problem.

It states as follows:

A 1500-kg roller coaster (including passengers) passes point A at $3\,\frac{m}{s}$(see figure 1 as a reference). Due to safety concerns, you must design the track so that at point B the passengers do not experience an upward force that exceeds 4g. If the arc at B is circular with radius 15 m,(a) determine the minimum value of h that satisfies this requirement,and (b) determine the speed of the coaster at C.

Sketch of the problem

Figure 1. Illustrates the problem as it appears in my book

Part a. In my attempt to solve the problem. I thought that the key was to use the conservation of mechanical energy in both points. In A and B. As this is described in Figure 2. I also thought that because the Radius of the circle is given then there is an implicit hint that the upward acceleration they're talking about is the centripetal acceleration (the one which is pulling to the center of the circle, hence upwards), so that this must not exceed 4g. As a consequence I equated both expressions in order to obtain the speed v_b so with that all there is left to solve is the height which is given by the potential energy U reached at that point.

For better visualizing of the situation I made an sketch of all the forces which I could find, seen in Figure 2.

Sketch of the solution

The above paragraph is summarized in the following equations:

$$E_a=E_b$$

$$\frac{1}{2}mv^2_a+mgh_a=\frac{1}{2}mv^2_b+mgh_b$$

Mass cancels in both expressions and multiplied by $2$:

$$v^2_a+2gh_a=v^2_b+2gh_b$$

$$h_b=\frac{v^2_a+2gh_a-v^2_b}{2g}$$

Edit:

Apparently there is an acting force which is going outwards the circle, this summed up with Newton's second law would yield:

$$\frac{mv^2_b}{R}+mg=m\left(a_{max}\right)$$

(Here's where I use the fact of the "threshold of $4g$)

$$\frac{mv^2_b}{R}+mg=m\left(4g\right)$$

$$v^2_b=3gR$$

Inserting this final expression into the one obtained earlier:

$$h_b=\frac{v^2_a+2gh_a-3gR}{2g}$$

Then all is left is to insert the values given and make the computation of the height:

$$h_b=\frac{\left(3\,\frac{m}{s}\right)^2+2\times 9.8\,\frac{m}{s^2}\times 40\,m-3\times9.8\,\frac{m}{s^2}\times 15\,m}{2\times 9.8\,\frac{m}{s^2}}$$

$$h_b=\frac{\left(3\,\frac{m}{s}\right)^2+2\times 9.8\,\frac{m}{s^2}\times 40\,m-3\times9.8\,\frac{m}{s^2}\times 15\,m}{2\times 9.8\,\frac{m}{s^2}}$$

Computing this expression yields:

$h_{b}=17.95918367\,m$

Which results with what my book describes as $h_b\approx 17.96\,m$

Part b. Is very straightforward. It was just comparing both Energies from the top in A and lower in C as stated below.

This process is illustrated in Figure 3.

Sketch of the second solution

$$E_a=E_c$$

$$\frac{1}{2}mv^2_a+mgh_a=\frac{1}{2}mv^2_c+mgh_c$$

Again masses cancels and multiplying by $2$:

$$v^2_a+2gh_a=v^2_c+2gh_c$$

$$v^2_c=v^2_a+2gh_a-2gh_c$$

$$v_c=\sqrt{v^2_a+2gh_a-2gh_c}$$

$$v_c=\sqrt{v^2_a+2g\left(h_a-h_c\right)}$$

Finally inserting the given values into the above equation:

$$v_c=\sqrt{\left(3.0\,\frac{m}{s}\right)^2+2\times 9.8\,\frac{m}{s^2}\left(40\,m-20\,m\right)}$$

Which results into:

$$v_c= 20.02498439\,\frac{m}{s}$$

This answer does check with what my book lists as the correct answer, $20.02\,\frac{m}{s}$ rounded to two decimals.

Although the procedure which I intended to do initially seems reasonable, I'm still dumbfounded on why part a doesn't check, so far I believe I haven't overlooked anything, but if I did, please explain me where is the mistake and some conceptual basis as maybe I could have misinterpreted something.

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  • $\begingroup$ Are you certain that this is the right site for your problem? Have you tried physics.SE? It might be off-topic there, though, on account of their stance on homework (and homework-like) problems. I don't know. $\endgroup$ – Arthur Jan 2 at 9:13
  • $\begingroup$ @Arthur I thought that maths was an appropiate site as my question is about if my use of mathematics was right given the conditions of the problem. I am not very experienced in the field so, verification of my reasoning is needed. $\endgroup$ – Chris Steinbeck Bell Jan 2 at 9:22
  • $\begingroup$ Where did you get $v_a = 3$ m/s from? $\endgroup$ – caverac Jan 2 at 9:25
  • $\begingroup$ @caverac Sorry it was a typo. It was a given data from the problem, now I've corrected that mistake. $\endgroup$ – Chris Steinbeck Bell Jan 2 at 9:29
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    $\begingroup$ When you calculate $h_b$ you use $R = 10$ m, is that also a typo? $\endgroup$ – caverac Jan 2 at 9:32
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The downward force felt by a passenger of mass $m$ at point $B$ has two components:

  • The weight of the passenger, $mg\downarrow$
  • The radial force, $\frac{mv_b^2}R\downarrow$

Your equation gets modified as $mg+\frac{mv_b^2}R=4mg$, solving which gets you $h_b=17.96$m.

Figure1Figure2

Edit. Let us first discuss the answer in ground frame. As you have correctly pointed out, a radially upward acceleration with magnitude $\frac{mv_b^2}R$ is required to maintain circular motion of the roller coaster at point $B$. The required 'centripetal force' is delivered by the normal reaction at point $B$. This is because there is no agent exclusively exerting the centripetal force; the normal reaction and gravitational force are the only forces acting on the coaster.

As seen from the free body diagram of the coaster, $$N-mg=\frac{mv_b^2}R$$ Since the upward force on the coaster is $N$, you require $$N\le4mg\implies\frac{mv_b^2}R+mg\le4mg$$

Now, for the solution in the frame of reference of the roller coaster. Imagine being seated on the coaster. You would perceive yourself to be at rest without acceleration while the world around would be moving circularly. You would also feel more compressed at $B$: this suggests that in your reference frame, a certain phantom force other than your weight acts downward. This force is called the pseudo-force; 'pseudo', because it doesn't actually exist. It is merely an arrangement to solve in accelerated frames of reference.

Recall that the centripetal acceleration has direction opposite to that of its corresponding pseudo-force. In general too, for an object with acceleration $\vec a$ in ground frame, the pseudo-force acting on the object in its frame of reference is $-m\vec a$; that is, it has magnitude $ma$ and direction opposite to $\vec a$. This gives $$N=mg+\frac{mv_b^2}R$$ since in this frame, the coaster is at rest so the forces cancel. Once again, the upward force is $N$, which gives you the same result as the ground frame.

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  • $\begingroup$ Shubham.Yes, mv^2/R = Reaction(upward) -mg.+) $\endgroup$ – Peter Szilas Jan 2 at 9:45
  • $\begingroup$ @Shubham John Why the radial force goes downwards and not upwards?. Can you explain to me this part?. I thought that the radial force goes into the circle not the other way around. Perhaps am I mistaking this? $\endgroup$ – Chris Steinbeck Bell Jan 2 at 9:50
  • $\begingroup$ @ChrisSteinbeckBell The centrifugal force is a pseudo-force and has the direction opposite to that of the centripetal acceleration. $\endgroup$ – Shubham Johri Jan 2 at 9:55
  • $\begingroup$ Think about it: at the bottom of the coaster, doesn't one feel compressed? $\endgroup$ – Shubham Johri Jan 2 at 10:00
  • $\begingroup$ @Shubham John I am not familiar with the concept of pseudo-force. Checking with my book it states that "centripetal acceleration" is inward seeking. Yes indeed when you're at the bottom you feel compressed but I don't know to which force is due this. Do you know any link where I can read more about it? $\endgroup$ – Chris Steinbeck Bell Jan 2 at 10:04

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