Question:
If a stone of mass \( 0.5~\text{kg} \) tied to one end of a wire is whirled in a circular path of radius \( 2~\text{m} \) with a speed \( 40~\text{rev/min} \) in a horizontal plane, then the tension in the wire is nearly
If a stone of mass \( 0.5~\text{kg} \) tied to one end of a wire is whirled in a circular path of radius \( 2~\text{m} \) with a speed \( 40~\text{rev/min} \) in a horizontal plane, then the tension in the wire is nearly
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Convert angular speed to rad/s when working with circular motion. Use \( T = mr\omega^2 \) for horizontal tension due to circular motion.
Updated On: Jun 6, 2025
- \( 14.8~\text{N} \)
- \( 12.4~\text{N} \)
- \( 17.5~\text{N} \)
- \( 20.8~\text{N} \)
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The Correct Option is C
Solution and Explanation
Step 1: Convert revolutions per minute to angular velocity in rad/s.
\[
n = 40~\text{rev/min} = \frac{40 \times 2\pi}{60} = \frac{4\pi}{3}~\text{rad/s}
\]
Step 2: Use centripetal force formula to calculate tension.
\[
T = mr\omega^2
\]
\[
m = 0.5~\text{kg},
r = 2~\text{m},
\omega = \frac{4\pi}{3} \] \[ T = 0.5 \times 2 \times \left(\frac{4\pi}{3}\right)^2 = 1 \times \frac{16\pi^2}{9} \approx 1 \times \frac{16 \times 9.87}{9} \approx 17.5~\text{N} \] % Final Answer \[ \boxed{17.5~\text{N}} \]
r = 2~\text{m},
\omega = \frac{4\pi}{3} \] \[ T = 0.5 \times 2 \times \left(\frac{4\pi}{3}\right)^2 = 1 \times \frac{16\pi^2}{9} \approx 1 \times \frac{16 \times 9.87}{9} \approx 17.5~\text{N} \] % Final Answer \[ \boxed{17.5~\text{N}} \]
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