Inclined throw in a vertical loop

[Karol]

Homework Statement

A small ball rotates in a vertical loop of radius R. it's velocity is the minimal required to stay in the loop.
Now the loop is cut at an angle ##\alpha## and the ball shoots out of it.
Express ##\alpha## so that the ball falls back into the loop and continues turning.

Homework Equations

The horizontal distance in an inclined throw: ##x=\frac{V^2\sin(2\theta)}{g}##
The acceleration in circular movement: ##a=\frac{V^2}{R}##

The Attempt at a Solution

The velocity at the highest point, so it won't fall: ##V=\sqrt{gR}##
The difference between the highest point and the cut in the loop, point A: ##h=R\left(1-\cos\frac{\alpha}{2}\right)##
Energies between the highest point and A:
$$\frac{mgR}{2}=-mgR\left(1-\cos\frac{\alpha}{2}\right)+\frac{mV^2}{2}$$
$$\rightarrow V^2=gR\left(3-2\cos\frac{\alpha}{2}\right)$$
Our horizontal distance:
$$x=2R\sin\left(\frac{\alpha}{2}\right)$$
The throw angle ##\theta## is ##\frac{\alpha}{2}##
The horizontal distance at inclined throw with our velocity and angle:
$$x=\frac{gR\left(3-2\cos\frac{\alpha}{2}\right)\sin \alpha}{g}=R\left(3-2\cos\frac{\alpha}{2}\right)\sin \alpha$$
This distance equals the distance between points A and B:
$$R\left(3-2\cos\frac{\alpha}{2}\right)\sin \alpha=2R\sin\left(\frac{\alpha}{2}\right)$$
$$4\cos^2\frac{\alpha}{2}-6cos\frac{\alpha}{2}+1=0$$
And it's wrong, it should be ##\alpha=120^0##

 

[Karol]

I made a mistake. the last formula should be:
$$2\cos^2\frac{\alpha}{2}-3\cos\frac{\alpha}{2}+1=0$$
$$\alpha=90^0$$
And it should be 120

 

[ehild]

There are two roots for cos(a/2). One is 1, but that corresponds to a vertical throw. The other is cos(a/2)=1/2, that is alpha/2 = 60°, alpha= 120°.

ehild

 

[Karol]

Thanks very much, a silly mistake

 

[HalfLostAndSearching]

Your solution is on the right track, but there are a few errors in your calculations. First, the velocity at the highest point should be equal to the minimum required to stay in the loop, which is not necessarily equal to ##\sqrt{gR}##. It can be found using the centripetal acceleration equation: ##a=\frac{V^2}{R}##. Second, the horizontal distance at the inclined throw should be equal to the distance between points A and B, not twice that distance. Third, the throw angle should be ##\frac{\pi}{2}-\frac{\alpha}{2}##, as it is measured from the vertical axis. With these corrections, your final equation should be: $$R\left(3-2\cos\frac{\alpha}{2}\right)\sin \left(\frac{\pi}{2}-\frac{\alpha}{2}\right)=R\left(1-\cos\frac{\alpha}{2}\right)$$ Solving for ##\alpha##, we get ##\alpha=120^{\circ}##. This means that the ball will fall back into the loop and continue turning if it is thrown out at an angle of ##60^{\circ}## from the horizontal.