A 1 kg stone at the end of 1 m long string is whirled in a vertical circle at a constant speed of 4 m/s. The t

Question

A 1 kg stone at the end of 1 m long string is whirled in a vertical circle at a constant speed of 4 m/s. The tension in the string is 6 N, when the stone is at (Take g = 10 m/s²):

Accepted Answer

1. **Identify the Dynamics:** For a body moving in a vertical circle, the net radial force provides the centripetal acceleration. The forces involved are the tension ($T$) in the string and the component of weight ($mg$) along the radius [Source 73, 82].
2. **Calculate Centripetal Force:**
   $$ F_c = \frac{mv^2}{r} $$
   Given $m=1	ext{ kg}$, $v=4	ext{ m/s}$, $r=1	ext{ m}$.
   $$ F_c = \frac{1 	imes (4)^2}{1} = 16	ext{ N} $$
3. **Analyze Tension at Key Positions:**
   - **At the Top:** Both tension and gravity point downwards (towards the center).
     $$ T_{top} + mg = \frac{mv^2}{r} \implies T_{top} = \frac{mv^2}{r} - mg $$
     $$ T_{top} = 16 - (1 	imes 10) = 6	ext{ N} $$
   - **At the Bottom:** Tension points upwards (towards center), gravity downwards.
     $$ T_{bottom} - mg = \frac{mv^2}{r} \implies T_{bottom} = \frac{mv^2}{r} + mg $$
     $$ T_{bottom} = 16 + 10 = 26	ext{ N} $$
4. **Conclusion:** The given tension is 6 N, which matches the calculated tension at the **top of the circle**.

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