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In a fixed quarter circular track of radius R which lies in a vertical plane, a block is released from point A and it leaves the path at point B . The radius of curvature of its trajectory when it just leaves the path will be

Options

  1. AR
  2. BR 4
  3. CR 2
  4. DNone of these

Correct answer

C. R 2

Step-by-step solution

By energy conservation between A and B ⇒ Mg 2 R 5 + 0 = MgR 5 + 1 2 Mv 2 v = 2 gR 5 Now, radius of curvature after just leaving B is r = v ⊥ 2 a r = 2 gR / 5 g cos 37 = R 2

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