Step 1: Understanding the Question:
We need to find the critical minimum speed for an object moving in a vertical circle (attached to a string) at the highest point of its path, such that the string does not go slack.
Step 2: Key Formula or Approach:
For an object to move in a circle, there must be a net centripetal force directed towards the center. At the topmost point of a vertical circle, the forces acting on the object along the radial direction are the tension in the string (\(T\)) and the force of gravity (\(mg\)), both pointing downwards (towards the center).
The equation for centripetal force is:
\[ F_{net, center} = \frac{mv^2}{r} \]
At the top, this becomes:
\[ T + mg = \frac{mv^2}{L} \]
The condition for the string to "remain taut" means the tension \(T\) must be greater than or equal to zero (\(T \geq 0\)). The minimum speed occurs at the critical point where \(T = 0\).
Step 3: Detailed Explanation:
Let \(v_{top}\) be the speed at the topmost point. The force equation is:
\[ T + mg = \frac{mv_{top}^2}{L} \]
To find the minimum possible speed, we set the tension \(T\) to its minimum possible value, which is zero. If the speed were any lower, the string would go slack.
Setting \(T = 0\), the equation becomes:
\[ 0 + mg = \frac{mv_{min}^2}{L} \]
We can cancel the mass \(m\) from both sides:
\[ g = \frac{v_{min}^2}{L} \]
Now, solve for \(v_{min}\):
\[ v_{min}^2 = gL \]
\[ v_{min} = \sqrt{gL} \]
Step 4: Final Answer:
The minimum speed required at the topmost point for the string to remain taut is \(\sqrt{gL}\).