To solve this problem, we need to analyze the forces acting on both masses $m_1$ and $m_2$.

1. For the hanging mass $m_2$:

   • The weight of $m_2$ exerts a force downwards, which is equal to $m_2 g$, where $g$ is the acceleration due to gravity.
   • The tension $T$ in the string is the force holding $m_2$ at rest, so $T = m_2 g$.

2. For the mass $m_1$ on the table:

   • Since $m_1$ is rotating in a circle, there is a centripetal force acting on it that keeps it in circular motion. This centripetal force is provided by the tension in the string.
   • The formula for the centripetal force is: $F_{\text{centripetal}} = \frac{m_1 v^2}{r}$
   • This force is equal to the tension in the string, so: $\frac{m_1 v^2}{r} = T$

3. Equating the forces: Since $T = m_2 g$, we substitute this into the equation for the centripetal force: $\frac{m_1 v^2}{r} = m_2 g$

4. Solving for $v$ (the speed of $m_1$): Multiply both sides of the equation by $r$ to isolate $v^2$: $m_1 v^2 = m_2 g r$ Divide by $m_1$: $v^2 = \frac{m_2 g r}{m_1}$ Take the square root to solve for $v$: $v = \sqrt{\frac{m_2 g r}{m_1}}$

Final Answer:

The speed at which $m_1$ must rotate for $m_2$ to remain hanging at rest is: $v = \sqrt{\frac{m_2 g r}{m_1}}$

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Would you like further details or have any specific questions?

Here are 5 related questions for further exploration:

1. How would the speed change if $m_1$ were increased?
2. What happens if the radius $r$ is doubled?
3. How does the free-fall acceleration $g$ affect the system if it's not constant?
4. What would happen if there was friction on the table affecting $m_1$'s motion?
5. Can the tension in the string ever be greater than $m_2 g$?

Tip: Always check the units of your variables in problems like this to ensure dimensional consistency when solving equations.