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.
Step 1: Understanding the Concept:
When a particle moves inside a vertical circular track, the forces acting on it at the highest point are gravity (downwards) and the normal reaction (downwards). These forces provide the necessary centripetal acceleration. We use energy conservation to link the release height to the velocity at the top.
Step 2: Key Formula or Approach:
Equation for circular motion at the highest point:
\[ mg + N = \frac{mv^2}{R} \]
Conservation of Mechanical Energy between release point and top of the loop:
\[ mgh = mg(2R) + \frac{1}{2}mv^2 \]
Step 3: Detailed Explanation:
We are given that the normal reaction \(N = 3mg\).
Substitute this into the force equation:
\[ mg + 3mg = \frac{mv^2}{R} \]
\[ 4mg = \frac{mv^2}{R} \implies v^2 = 4gR \]
Now apply the energy conservation equation. The particle falls from height \(h\) and reaches a height of \(2R\) at the top of the circular track:
\[ mgh = mg(2R) + \frac{1}{2}mv^2 \]
Substitute \(v^2 = 4gR\):
\[ mgh = 2mgR + \frac{1}{2}m(4gR) \]
\[ mgh = 2mgR + 2mgR = 4mgR \]
Divide by \(mg\):
\[ h = 4R \]
Step 4: Final Answer:
The required height is \(h = 4R\).
