Sphere rolling in a semisphere

Shyan

Consider a sphere and a semicircle with radii r and R respectively.(R>>r)
The sphere has mass m.Imagine we place it in the semicircle and let it rotate in it.
Let's take the z axis the line which passing through the center of the semicircle and the bottom of the semicircle with the origin at the center of the semicircle.
Now we write the Lagrangian with coordinates [itex] \alpha [/itex] and [itex] \beta [/itex].
[itex] \alpha [/itex] is the angle that the radius passing the sphere makes with the z axis and [itex] \beta [/itex] is the angle which shows the rotation of the sphere around its axis.

[itex]
L=m(\frac{1}{2} R^2 \dot{\alpha}^2+\frac{1}{5} r^2 \dot{\beta}^2)-mgR (1-\cos{\alpha})
[/itex]

With this Lagrangian,the equations of motion become

[itex]
\ddot{\alpha}-\frac{g}{R} \sin{\alpha}=0
[/itex]

[itex]
\frac{2}{5} m r^2 \dot{\beta}=constant=h
[/itex]

Also we assume that the sphere is pure rolling so

[itex]
v_{cm}=R\dot{\alpha}=r\dot{\beta}
[/itex]

Solving for [itex]\dot{\beta}[/itex] and putting it in the second equation we get

[itex]
\frac{2}{5}mr^2\frac{R}{r}\dot{\alpha}=h \rightarrow \frac{2}{5} m r R \dot{\alpha}=h
[/itex]

Solving the above equation for [itex] \dot{\alpha} [/itex], we have

[itex]
\dot{\alpha}=\frac{5h}{2mrR}
[/itex]

But this means that [itex]\dot{\alpha}[/itex] is constant which means [itex] \ddot{\alpha} [/itex] is zero.Taking this and the differential equation of motion for [itex] \alpha [/itex] we get

[itex]
-\frac{g}{R}\sin{\alpha}=0
[/itex]

Which means that [itex] \alpha=k\pi[/itex]!!!This seems to be a quantization which is completely nonsense.

Does this mean that in such a condition,pure roll is impossible?
If not,what's wrong?
Thanks

mfb

You cannot extract your equations of motion like this, if alpha and beta are not independent. Use your rolling constraint first and let the Lagrangian be a function of alpha only. This will give you a proper equation of motion.

DaleSpam

I agree with mfb. You need to put the constraints in before calculating the Euler-Lagrange equations. You should get [itex]\frac{7}{5}R^2 \alpha'' = -g R sin(\alpha)[/itex]