Find the max edge length $a$ for which cube will remain in stable equilibrium on hemispherical surface. Consider no sliding of the base of cube and hemisphere.

Please help to begin.

1 vote

Find the max edge length $a$ for which cube will remain in stable equilibrium on hemispherical surface. Consider no sliding of the base of cube and hemisphere.

Please help to begin.

I'm getting $a=2R$ but not aware of the correct answer.

Draw a diagram of the cube after it has rocked (not slid) through some small angle. Determine whether the position of the cube is stable - ie whether the moment of the weight of the cube about the contact point is turning the cube back towards its equilibrium position or turning it further away from equilibrium.

The difficult part is to calculate the position of the COM of the cube relative to the contact point when the cube has rocked through an angle.

The difficult part is to calculate the position of the COM of the cube relative to the contact point when the cube has rocked through an angle.

1 vote

Best answer

In the diagram I have drawn only one quarter of the cube, which has centre C and side $a=2b$.

Initially the cube is balanced in equilibrium with point O below C in contact with the hemisphere. After it has rolled through some angle $\theta$ contact has shifted to point P. If C is to the left of the vertical through P ($b \lt h$) then the cube will roll back to the equilibrium position shown on the left, whereas if C is to the right of the vertical through P ($b \gt h$) then the cube will continue rolling and will topple.

$\tan\theta=\frac{s}{h}=\frac{R\theta}{h}$

For small angles $\tan\theta \approx \theta$ so for stable equilibrium

$b \le h \approx R$

$a = 2b \le 2 R$

**Comment :**

As $\theta$ increases the limit for $a$ gets smaller :

$h=R\theta \cot\theta=R\theta (\frac{1}{\theta}-\frac{\theta}{3}-\frac{\theta^3}{45}-...)$

$a \le 2R(1-\frac{\theta^2}{3}-\frac{\theta^4}{45}-...)$

For a cube the maximum angle $\theta$, beyond which it can no longer roll, is given by $R\theta=b$ so the maximum angle is $\theta_m=\frac{b}{R}=\frac{a}{2R}$ hence

$\frac{a}{2R} = \theta_m \approx 1-\frac13 \theta_m^2-\frac{1}{45} \theta_m^4-...$

$\theta_m^2+3\theta_m -3 \approx 0$

$\theta_m \approx \frac12 (-3 + \sqrt{21}) = 0.791^c \approx 45^{\circ}$

The problem assumes that the coefficient of friction is sufficient to avoid sliding. The condition for no sliding on an inclined plane applies :

$\mu \ge \tan\theta$

...