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Q.A motor cyclist loops a vertical loop of diameter 50m, without dropping down even at uppermost point. What is the minimum speed at lowest and highest points of the loop ?

Manipur CohsemCouncil of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2024Subjective· 2mImportance★★★★★
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Using v_top = √(gR) at the top and energy conservation to get v_bottom = √(5gR), with R = 25 m: v_top ≈ 15.65 m/s and v_bottom = 35 m/s.

The diameter of the vertical loop is 50 m, so its radius is R = 25 m.

Step 1 — Minimum speed at the top:

At the topmost point of the loop, for the motorcyclist to just maintain contact with the track (without falling), gravity alone must be just enough to provide the necessary centripetal force (normal reaction N = 0 at the minimum condition):

mg = m v_top^2 / R

v_top = √(gR) = √(9.8 × 25) = √245 ≈ 15.65 m/s

Step 2 — Minimum speed at the bottom:

Using conservation of mechanical energy between the bottom and top of the loop (height gained = 2R = diameter = 50 m):

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