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| ### Statement |
| ### Statement |
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| $2.1.53.$ [Insert the problem statement] |
| $2.1.53.$ A dumbbell (two balls of mass $m$ each connected by a weightless rod) was placed in a spherical cavity as shown in the figure. Determine the pressure force of the balls on the walls immediately after the dumbbell is released. The radius of the dumbbell balls is much smaller than the radius of the sphere. |
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| ### Solution |
| ### Solution |
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| If the cavity has radius $r$, then the center of mass (C.M.) moves on a circle of radius $r/\sqrt(2)$. Since it is initially at rest, the total force (shown in green) must be tangential to this circle. Consequently, |
| If the cavity has radius $r$, then the center of mass (C.M.) moves on a circle of radius $r/\sqrt2$. Since it is initially at rest, the total force (shown in green) must be tangential to this circle. Consequently, |
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| \begin{equation} |
| \[N=2mg-N',\] |
| N=2mg-N', |
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| \end{equation} |
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| and the total force has magnitude $\sqrt2N$. The C.M. has linear acceleration of $\sqrt2N/m$ and angular acceleration |
| and the total force has magnitude $\sqrt2N$. The C.M. has linear acceleration of $\sqrt2N/(2m)$ and angular acceleration |
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| \[\alpha=\frac{\sqrt2N/m}{r/\sqrt2}=\frac{2N}{mr}\] |
| \[\alpha=\frac{\sqrt2N/(2m)}{r/\sqrt2}=\frac{N}{mr}\] |
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| around the center of the cavity. This angular acceleration matches that of the rotation of the dumbbell around the C.M., which can be obtained by dividing the total torque by the moment of inertia around the C.M. |
| around the center of the cavity. This angular acceleration matches that of the rotation of the dumbbell around the C.M., which can be obtained by dividing the total torque by the moment of inertia around the C.M. |
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| \[\alpha=\frac{(N'-N)r/2}{2m(r/2)^2}=\frac{N'-N}{mr}.\] |
| \[\alpha=\frac{(N'-N)r/2}{2m(r/\sqrt2)^2}=\frac{N'-N}{2mr}.\] |
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| Equating the two expressions for $\alpha$, we have $N'=3N$. Using this result with (1), we have $N=mg/2$ and $N'=3mg/2$. |
| Equating the two expressions for $\alpha$, we have $N'=3N$. Using this result with the first equation, it follows that $N=mg/2$ and $N'=3mg/2$. |
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| #### Answer |
| #### Answer |
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| [Insert a concise answer or boxed result] |
| $N=mg/2$ and $N'=3mg/2$ |
