- #1
Karol
- 1,380
- 22
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##