A circular ring of mass 3.14 kg and radius 0.05 m is rotating with a constant angular velocity 10 rad/s about its vertical axis on a smooth horizontal table. Then the tension in the ring is 10/n N, then n is ________.
Physics · Mixed Chapter
JEE Mains

Question

A circular ring of mass 3.14 kg and radius 0.05 m is rotating with a constant angular velocity 10 rad/s about its vertical axis on a smooth horizontal table. Then the tension in the ring is 10/n N, then n is ________.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: For a rotating thin ring, each small element needs inward centripetal force. For an element subtending dθ, two equal ring tensions contribute resultant ≈T dθ. Step 2: The element has mass dm=(m/2π)dθ. Centripetal force dm ω²R must equal T dθ. Step 3: Hence T=mω²R/(2π)=3.14×(10)²×0.05/(2π)≈2.5 N. Step 4: T=10/n N gives n=10/2.5=4. Answer: 4.