Question

A body of mass 1 kg is attached to one end of a string of 1 m length. It is rotated in a vertical circle with a constant speed of \( 4 \text{ ms}^{-1} \). When the object is at the highest point of the vertical circle, tension in the string is (\( g=10 \text{ ms}^{-2} \)):

• A. \( 6 \text{ N} \)
• B. \( 8 \text{ N} \)
• C. \( 10 \text{ N} \)
• D. \( 16 \text{ N} \)

Hint:

At the lowest point of a vertical circle, tension is at its maximum (\(T_{max} = \frac{mv^2}{r} + mg\)), whereas at the highest point, tension is at its minimum (\(T_{min} = \frac{mv^2}{r} - mg\)).

Answer 1

The Correct Option is A

Step 1: Look at the top of the circle.
At the highest point, the string tension $T$ and the weight $mg$ both point downward, toward the centre of the vertical circle.
Step 2: Apply Newton's second law for circular motion.
The net inward force equals the centripetal requirement: \[ T + mg = \frac{mv^2}{r} \]
Step 3: Write down the data.
$m = 1$ kg, $r = 1$ m, $v = 4$ ms$^{-1}$, $g = 10$ ms$^{-2}$.
Step 4: Centripetal force needed.
$\dfrac{mv^2}{r} = \dfrac{1\times 16}{1} = 16$ N.
Step 5: Weight contribution.
$mg = 1\times 10 = 10$ N, which already supplies part of the inward force.
Step 6: Solve for the tension.
$T = 16 - 10 = 6$ N.
\[ \boxed{6\ \text{N}} \]