How to finish part (a) and (b) of this dynamics problem?
There are no solution or answer in the book...
My professor say that this problem is very tricky, don't even try by yourself.
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1$\begingroup$ Tell us what you tried? How did you attempt to solve the problem. $\endgroup$– papabicepsFeb 12, 2017 at 9:46
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$\begingroup$ I will give you a hint. The speed of the ball is maximum when the components of forces acting on the ball that are parallel to track are equal. $\endgroup$– papabicepsFeb 12, 2017 at 9:57
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5$\begingroup$ @papabiceps Forget about "forces." Use energy methods instead, and there is nothing particularly tricky about the question. $\endgroup$– alephzeroFeb 12, 2017 at 13:33
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1 Answer
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Starting with equating the total energies at the starting and max velocity point. The energy at the starting point comes from the potential energies of the mass and spring. The energy at the other point also has the kinetic energy component of the mass.
$$ m g (R \tan (\alpha )+R)+\frac{1}{2} k \left(\sqrt{R^2+(R \tan (\alpha )+R)^2}-\ell \right)^2 = \\ m g (R-R \sin (\beta ))+\frac{1}{2} k \left(\sqrt{(R-R \sin (\beta ))^2+(2 R-R \cos (\beta ))^2}-\ell \right)^2+\frac{1}{2}m v^2 $$
($\ell$ is the unstretched spring length)
Now we can solve for $v^2$ (or $v$).
$$ \frac{1}{m}(-4. \sin ^2(\beta )-4. \cos ^2(\beta )+16. \cos (\beta )+40. \sqrt{-0.08 \sin (\beta )-0.16 \cos (\beta )+0.24}+(3.92 m+8.) \sin (\beta )-12 +\\ 4. \tan ^2(\alpha )-40. \sqrt{0.04 \tan ^2(\alpha )+0.08 \tan (\alpha )+0.08}+8. \tan (\alpha )+3.92 m \tan (\alpha )) $$
Looking at this expression we see that part of it depends on $\beta$ and the other part on $\alpha$, and there are no cross variables. The other variable is the mass $m$. Given the information that the max velocity is at $\beta=34 {}^{\circ}$, we can equate the derivative of $v^2$ to 0 and solve for $m$. This gives $m = 0.143 kg$.
I think this is as far as we can get with the information provided. The actual velocity is a function of $\alpha$ and as expected higher values produce larger velocities.
$$ 2.64447 \sqrt{4. \tan ^2(\alpha )-40. \sqrt{0.04 \tan ^2(\alpha )+0.08 \tan (\alpha )+0.08}+8.56054 \tan (\alpha )+12.0611} $$
[Update] If $\alpha=45{}^{\circ}$ then the velocity is $6.86192\ m/s$.
(All the computations were done in Mathematica and I am attaching the screenshot).
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$\begingroup$ The answer should be mass= 143 g and the speed is 6.86 m/s when I email my professor $\endgroup$– VictorFeb 15, 2017 at 21:50
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$\begingroup$ This jives with what I got. The mass is $142.99\ g$. The velocity is $6.86192\ m/s$ if $\alpha=45{}^{\circ}$. The wording of the question, however, seems to convey that $\alpha$ could be any value higher than $45{}^{\circ}$. I will add this update. $\endgroup$ Feb 16, 2017 at 14:20