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The problem is as follows:

The figure from below shows a small sphere which is let to slide over a semicircular surface from rest on $A$ and exits the ramp on $B$ with a velocity $\vec{v}$. Find the maximum height from the floor when the sphere exits the surface. You may use $g=10\frac{m}{s^2}$.

Sketch of the problem

The alternatives are as follows:

$\begin{array}{ll} 1.&1.15\\ 2.&2.35\\ 3.&3.55\\ 4.&4.75\\ \end{array}$

In order to solve this problem I made a sketch which is seen in the figure from below:

Sketch of the problem

I assigned the values of $\phi$ and $R$ for radius.

Therefore to find the tangential speed which will be the one that the sphere will attain on exit will be as follows:

$E_ua=E_kb+E_ub$

$mgR=\frac{1}{2}mv^2+mg\left(R-R\cos\phi\right)$

From this expression masses do cancel on both sides and by multiplying by $2$ will give:

$2gR=v^2+2Rg-2Rg\cos\phi$

then:

$2Rg\cos\phi=v^2$

therefore:

$v=\sqrt{2Rg\cos\phi}$

Using this velocity the height can be calculated using the equations for vertical motion:

$v_{f}^2=v_{oy}^2-2g\Delta y$

Therefore:

$v_y=v_o\sin\phi=\sqrt{2Rg\cos\phi} \sin\phi$

$0=\left(\sqrt{2Rg\cos\phi} \sin\phi\right)^2 -2g\Delta y$

$\Delta y=\frac{\left(\sqrt{2Rg\cos\phi} \sin\phi\right)^2}{2g}$

$\Delta y=\frac{2Rg\cos\phi \sin^2\phi}{2g}=R\cos\phi \sin^2\phi$

Then to obtain the distance from the floor. I'm assuming the intended meaning is that the floor is the base where the disk is at its minimum, hence the height they're asking is given by:

$h=R-R\cos\phi+R\cos\phi \sin^2\phi$

$h=R\left(1-\cos\phi+\cos\phi \sin^2\phi\right)$

Then pluggin the values which were given it ends as follows:

$h=3\left(1-\cos53^{\circ}+\cos53^{\circ}\sin^{2}53^{\circ} \right )$

$h=3\left(1-\cos53^{\circ}+\cos53^{\circ}\sin53^{\circ} \right )$

$h=3\left(1-\frac{3}{5}+\frac{3}{5}\times\frac{4^2}{5^2} \right )$

$h=2.352\,m$

Which corresponds to the second answer.

However what would happen if I follow this logic?

Sketch of the solution second diagram

In the exit the sphere the tangential speed will be given as

$\frac{mv^2}{R}=mg\cos\phi$

Therefore:

$v^2=Rg\cos\phi$

$v=\sqrt{Rg\cos\phi}$

But why do I obtain a different value of the tangential velocity in the exit of the sphere compared to what I found using conservation of energy? Can somebody explain me the reason of this discrepancy?

The rest of the calculation would be the same for obtaining the height as follows:

$v_y=v_o\sin\phi=\sqrt{Rg\cos\phi} \sin\phi$

$0=\left(\sqrt{Rg\cos\phi} \sin\phi\right)^2 -2g\Delta y$

$\Delta y=\frac{\left(\sqrt{Rg\cos\phi} \sin\phi\right)^2}{2g}$

$\Delta y=\frac{Rg\cos\phi \sin^2\phi}{2g}=\frac{R\cos\phi \sin^2\phi}{2}$

Then for obtaining the required height will be:

$h=R-R\cos\phi+\frac{R\cos\phi \sin^2\phi}{2}$

$h=R\left( \frac{2-2\cos\phi+\cos\phi \sin^2\phi}{2}\right )$

Then pluggin the values given will be as follows:

$h=3\left( \frac{2-2\cos53^{\circ}+\cos53^{\circ} \sin^{2}53^{\circ}}{2}\right )$

$h=3\left( \frac{2-\frac{2\times 3}{5}+\frac{3}{5}\times \frac{4^2}{5^2}}{2}\right )$

$h=3\left( \frac{2-\frac{2\times 3}{5}+\frac{3}{5}\times \frac{4^2}{5^2}}{2}\right )$

$h=3.552\,m$

Which is also in the alternatives.

But according to my book the answer is the first one which I obtained and not this. Can someone explain to me why is this happening and why should I choose one over the other?. Did I overlooked something?.

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There are two forces acting on the ball, and you considered only one. The other force is the normal force to the surface $N$, acting inwards. Then $$N-mg\cos\phi=ma=\frac{mv^2}R$$Unfortunately, you must compute $v$ in order to get $N$, so you should use the first method.

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  • $\begingroup$ Thanks for that. I totally overlooked that fact. I am also wondering that had the sphere been tied to some wire I would had to account for that tension as well?. $\endgroup$ – Chris Steinbeck Bell Dec 13 '19 at 21:24
  • $\begingroup$ If possible I would like you could check this problem as well because I'm stuck with this problem. math.stackexchange.com/questions/3468517/… I'd like if possible an answer which would use an approach using conservation of momentum. $\endgroup$ – Chris Steinbeck Bell Dec 13 '19 at 21:28
  • $\begingroup$ Yes, the tension in the string would have the same role as the normal force to the surface. $\endgroup$ – Andrei Dec 13 '19 at 21:47
  • $\begingroup$ Wait.. does it mean that the existence of a string would had cancel the effect of the normal force? or is it additive? In other words $N+\textrm{Tension (going upwards)} -\textrm{weight} = \frac{mv^2}{R}$. Would this be the right equation in that fictional example that I put in my earlier comment?. $\endgroup$ – Chris Steinbeck Bell Dec 14 '19 at 5:32
  • $\begingroup$ It is additive. But remember that they are vectors. Tension is always along the string, the normal is perpendicular to the surface (in this case is also radial direction), while the weight is downwards. So you must only use the radial component. $\endgroup$ – Andrei Dec 14 '19 at 11:20
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The equation you use in your second version is based on the assumption that tangential velocity is constant, so it doesn’t apply here

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