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January 2nd, 2019, 01:41 AM  #1 
Member Joined: Jun 2017 From: Lima, Peru Posts: 63 Thanks: 1 Math Focus: Calculus  How to determine the minimum height of a roller coaster given an acceleration?
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: 1500kg 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.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. 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 $2$: $v^2_a+2gh_a=v^2_b+2gh_b$ $h_b=\frac{v^2_a+2gh_av^2_b}{2g}$ For the second part, all there is left is to replace with the centripetal acceleration: $\frac{mv^2_b}{R}=m\left(a_{max}\right)$ $\frac{mv^2_b}{R}=m\left(4g\right)$ $v^2_b=4gR$ Inserting this final expression into the one obtained earlier: $h_b=\frac{v^2_a+2gh_a4gR}{2g}$ Then all is left is to insert the values given and make the computation of the height: $h_b=\frac{\left(3\,\frac{m}{s}\right)^2+2\times 9.8\,\frac{m}{s^2}\times 40\,m4\times9.8\,\frac{m}{s^2}\times 15\,m}{2\times 9.8\,\frac{m}{s^2}}$ $h_b=\frac{\left(3\,\frac{m}{s}\right)^2+2\times 9.8\,\frac{m}{s^2}\times 40\,m4\times9.8\,\frac{m}{s^2}\times 15\,m}{2\times 9.8\,\frac{m}{s^2}}$ Computing this expression yields: $h_b=\,10.4591\,m\,\textrm{why ??}$ However this answer is not right as my book answers sheet states. (The supposed answer is $17.96\,m$) Could I be doing something wrong?. This is the part where I'm stuck at and really need somebody could lend me a hand with this matter. 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. $E_a=E_c$ $\frac{1}{2}mv^2_a+mgh_a=\frac{1}{2}mv^2_c+mgh_c$ Again masses cancels and multiplying by $2$: $v^2_a+2gh_a=v^2_c+2gh_c$ $v^2_c=v^2_a+2gh_a2gh_c$ $v_c=\sqrt{v^2_a+2gh_a2gh_c}$ $v_c=\sqrt{v^2_a+2g\left(h_ah_c\right)}$ Finally inserting the given values into the above equation: $v_c=\sqrt{\left(3.0\,\frac{m}{s}\right)^2+2\times 9.8\,\frac{m}{s^2}\left(40\,m20\,m\right)}$ 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. Last edited by Chemist116; January 2nd, 2019 at 02:14 AM. Reason: fixed typo 
January 2nd, 2019, 09:44 AM  #2  
Math Team Joined: Jul 2011 From: Texas Posts: 2,818 Thanks: 1462  Quote:
I agree with your solution. $\Delta KE = \Delta U_g$ $\dfrac{m}{2} (v_f^2  v_0^2) = mg(h_f  h_0)$ solving for $h_f$ ... $h_f = h_0  \dfrac{v_f^2v_0^2}{2g}$ note $a_c = \dfrac{v_f^2}{r} \le 4g \implies v_f^2 \le 4gr \implies h_f \ge h_0  \dfrac{4grv_0^2}{2g} \approx 10.5 \, m$ btw ... recommend you post physics problems in the physics forum to benefit those users searching for help on similar problems.  
January 2nd, 2019, 07:36 PM  #3  
Member Joined: Jun 2017 From: Lima, Peru Posts: 63 Thanks: 1 Math Focus: Calculus  Some thoughts into the answer... Quote:
Apparently there must be a reason why it isn't right the way as indicated being the proposed answer not checking with what the author says. Upon on inspecting what could had been wrong. I found that what was intended to be "computed" is the reactive force or the normal which is pointing to the center of the radius. Hence this added with the centripetal force would be what must not exceed the $4g$. By translating this into equations would yield: $f_c+mg=ma$ $\frac{mv^2_b}{R}+mg=m\left(4g\right)$ Masses cancel: $\frac{v^{2}_{b}}{R}+g=4g$ $v^{2}_{b}=3gR$ This inserted into the above equation: $v^2_a+2gh_a=3gR+2gh_b$ $h_{b}=\frac{v^2_a+2gh_a3gR}{2g}$ By replacing the values given in the problem translates into: $h_{b}=\frac{\left(3\,\frac{m}{s}\right)^2+2\times 9.8\,\frac{m}{s^2}\times 40\,m3\times9.8\,\frac{m}{s^2}\times 15\,m}{2\times 9.8\,\frac{m}{s^2}}$ Therefore the answer obtained is: $h_{b}=17.95918367\,m$ Which is $h_{b}\approx 17.96\,m$ which is what my book lists as the answer. Now the question which arises is, the "effect" that experiences the passengers in the coaster is the normal. But exactly why?. I'd like to know what's exactly this kind of force. Quote:
 
January 3rd, 2019, 06:14 AM  #4  
Math Team Joined: Jul 2011 From: Texas Posts: 2,818 Thanks: 1462  Quote:
$F_c = F_N  mg = ma$, where $F_N$ is the normal force of the track pushing upward on the cart. If the author's intention was to have $F_N \le m(4g)$, then he should have stated such in his problem statement. As I stated earlier, $4g$ is not a force but an acceleration. "Due to safety concerns, you must design the track so that at point B the passengers do not experience an upward force that exceeds $4mg$." ... would be much clearer as to his intent, making the centripetal force, $F_c = 3mg$.  
January 3rd, 2019, 09:45 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,298 Thanks: 1971 
I've moved this to the Physics forum.


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acceleration, coaster, determine, height, minimum, roller 
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