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Q4.21: NCERT Physics - Part 1 LAWS OF MOTION 4.21 Solution: A stone of mass 0.25 kg tied to the end of a string is whirled…
⚠️ Common Exam Traps & Conceptual Pitfalls
Students frequently forget to convert the rotational speed from revolutions per minute (rev/min) to radians per second (rad/s) before substituting it into the centripetal force formula $T = m\omega^2 r$. You must multiply the frequency in Hz by $2\pi$ to obtain the correct angular velocity $\omega$ in SI units.
A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?
Step-by-Step Verified Solution
**Answer:** T = 6.57 N, v_max = 34.6 m/s
**Step-by-Step Explanation:**
f = 40 rev/min = 40/60 = 0.667 Hz. _ = 2�f = 4.19 rad/s.
Tension: T = m__r = 0.25 _ (4.19)_ _ 1.5 = 6.57 N.
For T_max = 200 N:
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