Why do I obtain two value for tangential velocity when a ball slides in a semicircular ramp?

Jun 2017
Lima, Peru
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 $g=10\frac{m}{s^2}$.

The alternatives are as follows:


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

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:



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






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

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


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

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

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

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

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:

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

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

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


Which corresponds to the second answer.

However what would happen if I follow this logic?

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





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:

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

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

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


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?.
Jun 2019
There seems to be some information missing from your problem statement.

Purely from conservation of energy, if there is no friction (therefore no work except by gravity) and the ball slides instead of rolling (therefore no rotational kinetic energy), shouldn't it reach a height of R?
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Jun 2019
shouldn't it reach a height of R?
Actually, nix that part. I forgot it will still have horizontal velocity.

For the second method, your free-body diagram is incorrect. The centripetal force (mv^2/R) should equal the normal force minus the centrifugal component of the weight. In other words, the normal force from the ramp will be greater than mgcosΦ.

I would try it this way:
1) From conservation of energy, calculate the velocity at exit.
2) From the angle of exit, calculate the horizontal component of velocity, $v_x$.
3) At the top of its trajectory, $v_y=0$. Apply conservation of energy between the initial release ($v_0 = 0, h_0 = mgR$) and the top of the trajectory to solve for $h_f$.
This might be essentially what you did in the first method. I don't have enough active brain cells to follow that many sines and cosines at the moment.


Math Team
Jul 2011
Given R=3 (had to look for that essential piece of info), the ball launches at a height of 1.2 m

At the launch point, $v_0 = 6 \text{ m/s } \implies v_{y0} = 4.8 \text{ m/s}$.

$0 = v_{y0} - gt \implies t = 0.48 \text{ s}$

$h = 1.2 + 4.8t - 5t^2 = 2.35 \text{ m}$