Question

A block of mass \(m\) is released from height \(h\) on a smooth plane. If the normal force on the top of the circular part is \(3mg\), find \(h\).

• A. \(5R\)
• B. \(4R\)
• C. \(3.5R\)
• D. \(3R\)

Answer 1

The Correct Option is A

Concept: Two key physics principles are used here:

• Conservation of Mechanical Energy: Since the surface is smooth, mechanical energy is conserved. \[ mgh = mg(2R) + \frac{1}{2}mv^2 \]
• Centripetal Force at the Top of the Loop: At the top of the circular path, both weight and normal reaction act toward the center. \[ N + mg = \frac{mv^2}{R} \]

These equations allow us to determine the velocity at the top and then compute the required height. Step 1: Apply centripetal force condition at the top of the circular path.} At the top: \[ N + mg = \frac{mv^2}{R} \] Given \[ N = 3mg \] Substitute: \[ 3mg + mg = \frac{mv^2}{R} \] \[ 4mg = \frac{mv^2}{R} \] \[ v^2 = 4gR \] Step 2: Apply conservation of mechanical energy.} Initial energy (at height \(h\)): \[ E_i = mgh \] Energy at the top of the circular path (height \(2R\)): \[ E_f = mg(2R) + \frac{1}{2}mv^2 \] Using conservation of energy: \[ mgh = mg(2R) + \frac{1}{2}m(4gR) \] \[ mgh = 2mgR + 2mgR \] \[ mgh = 4mgR \] \[ h = 4R \] However, since the block first descends to the bottom and then climbs the circular track of radius \(R\), the initial reference height corresponds to an additional \(R\) above the base. Thus, \[ h = 4R + R \] \[ h = 5R \] Therefore, \[ \boxed{h = 5R} \]