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NEET PHYSICSMedium

What is the minimum velocity with which a body of mass mm must enter a vertical loop of radius RR so that it can complete the loop?

A

2gR\sqrt{2gR}

B

3gR\sqrt{3gR}

C

5gR\sqrt{5gR}

D

gR\sqrt{gR}

Step-by-Step Solution

  1. Critical Condition at the Top: To successfully complete the vertical loop, the body must not lose contact (or the string must not go slack) at the highest point. The minimum condition is when the tension (or normal force) is zero. At the highest point (C), the forces are gravity and centripetal force: mg=mvC2R  ⟹  vC=gRmg = \frac{mv_C^2}{R} \implies v_C = \sqrt{gR} (Refer to NCERT Class 11, Chapter 6, Eq. 5.14 ).
  2. Conservation of Energy: Apply the conservation of mechanical energy between the lowest point (A) and the highest point (C). Let the velocity at the lowest point be vAv_A. EA=ECE_A = E_C 12mvA2=12mvC2+mg(2R)\frac{1}{2}mv_A^2 = \frac{1}{2}mv_C^2 + mg(2R)
  3. Calculation: Substitute vC=gRv_C = \sqrt{gR}: 12mvA2=12m(gR)+2mgR\frac{1}{2}mv_A^2 = \frac{1}{2}m(gR) + 2mgR 12vA2=12gR+2gR=52gR\frac{1}{2}v_A^2 = \frac{1}{2}gR + 2gR = \frac{5}{2}gR vA=5gRv_A = \sqrt{5gR} This is the minimum velocity required at the lowest point to just complete the circle.
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