Question

A particle is released from point \(A\) of the track as shown in the figure. Find \(h\) so that the normal reaction at the highest point is 3 times the weight of the block. The surface is smooth.

• A. \(h = 4R\)
• B. \(h = 3R\)
• C. \(h = 2.5R\)
• D. \(h = 6R\)

Answer 1

The Correct Option is A

Concept: When a particle moves in a vertical circular path, the centripetal force at the top of the circle is provided by the sum of weight and the normal reaction. \[ \frac{mv^2}{R} = mg + N \] Also, since the track is smooth, mechanical energy is conserved. \[ \text{Loss of potential energy} = \text{Gain in kinetic energy} \] Step 1: Apply the centripetal force condition at the highest point.} Given that the normal reaction is three times the weight: \[ N = 3mg \] Substitute into the centripetal force equation: \[ mg + N = \frac{mv^2}{R} \] \[ mg + 3mg = \frac{mv^2}{R} \] \[ 4mg = \frac{mv^2}{R} \] \[ v^2 = 4gR \]
Step 2: Apply conservation of energy.} The particle starts from height \(h\) and reaches the highest point of the circular track which is at height \(2R\). \[ mg(h-2R) = \frac{1}{2}mv^2 \] Substitute \(v^2 = 4gR\): \[ mg(h-2R) = \frac{1}{2}m(4gR) \] \[ mg(h-2R) = 2mgR \] Cancel \(mg\): \[ h-2R = 2R \] \[ h = 4R \] Final Result \[ h = 4R \]