Let's break down the problem from the image:

The scenario involves a block of mass $m$ moving from rest at a height $h$ at point $A$ and going through a vertical loop of radius $R$. There are some key questions related to forces and the energy conservation involved.

Part a: Vector Force Diagram at Points B and C

For Point B (at the lowest point of the loop):

• Gravitational force ($mg$): This always points downward.
• Normal force ($F_n$): This will point upward because it's perpendicular to the surface of the loop, supporting the block against gravity.

For Point C (at the top of the loop):

• Gravitational force ($mg$): This again points downward.
• Normal force ($F_n$): This also points downward, as it acts towards the center of the circular loop to keep the block in contact with the track.

Part b: Minimum Height $h_{\text{min}}$

To find the minimum height $h_{\text{min}}$ necessary for the block to maintain contact at point $C$ (the top of the loop), we can use energy conservation and centripetal force concepts.

1. Energy conservation from point $A$ to point $C$:

   • At point $A$, the block starts with only potential energy $mgh$.
   • At point $C$, it has potential energy $mg(2R)$ (since it is $2R$ above the ground) and kinetic energy $\frac{1}{2} mv_C^2$.

   Using energy conservation: $mgh = mg(2R) + \frac{1}{2} mv_C^2$

   Simplifying, we get: $gh = 2gR + \frac{1}{2} v_C^2$

2. Centripetal force at the top of the loop:

   • For the block to stay in contact at point $C$, the centripetal force requirement must be met. At point $C$, the gravitational force and the normal force together provide the centripetal force: $mg + F_n = \frac{mv_C^2}{R}$ For minimum height, the normal force $F_n$ should be zero (since we want the block to just stay in contact), so: $mg = \frac{mv_C^2}{R}$ This simplifies to: $v_C^2 = gR$

3. Substituting $v_C^2 = gR$ into the energy conservation equation: $gh = 2gR + \frac{1}{2} gR$ $gh = \frac{5}{2} gR$

   Therefore, the minimum height $h_{\text{min}}$ is: $h_{\text{min}} = \frac{5R}{2}$

This concludes the solution to the problem.

Relative Questions:

1. How does the normal force at point B compare to the gravitational force?
2. What happens if the height $h$ is less than $\frac{5R}{2}$?
3. Why is the normal force zero at point C for the minimum height?
4. How would friction affect the minimum height calculation?
5. Could you calculate the velocity at point B, given $h = \frac{5R}{2}$?

Tip:

When dealing with energy conservation, always check if there are forces doing work that could add or remove mechanical energy (e.g., friction).