Circular Motion: Tricky Questions Solved (6)

Question

A small block of mass $m = 0.5 \text{ kg}$ is attached to a string of length $L = 1 \text{ m}$ and whirled in a vertical circle. The minimum speed at the top of the circle is just enough so that the string remains taut. Find (a) the speed at the top, (b) the speed at the bottom, (c) the tension in the string at the bottom. Take $g = 10 \text{ m/s}^2$.

Solution — Step by Step

Final answers: $v_{top} \approx \mathbf{3.16 \text{ m/s}}$, $v_{bot} \approx \mathbf{7.07 \text{ m/s}}$, $T_{bot} = \mathbf{30 \text{ N}}$.

Why This Works

Two distinct ideas merge here. First, the minimum-speed condition at the top isn’t $v = 0$ — it’s the minimum that keeps the string taut, which means gravity alone provides centripetal acceleration. Second, between top and bottom, energy is conserved (string tension does no work — it’s perpendicular to motion).

The factor of 5 in $v_{bot}^2 = 5gL$ is worth memorizing — it shows up in every vertical circular motion problem with the minimum-speed condition.

Alternative Method

Tension at the bottom can be derived in one line: $T_{bot} - T_{top} = 6mg$ for vertical circular motion at minimum speed (a standard result from combining centripetal + energy conservation). Since $T_{top} = 0$ here, $T_{bot} = 6mg = 6 \times 0.5 \times 10 = 30 \text{ N}$.

Common Mistake