A block of mass m is released from rest at a height R above a horizontal surface. The acceleration due to gravity is g. The block slides along the inside of a frictionless circular hoop of radius R.
What is the magnitude of the normal force exerted on the block by the hoop when the block has reached the bottom ?

Options

A. mg
B. 3mg
C. `(mg^(2))/(R)`
D. 2mg

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Initial Conditions
A block of mass \( m \) is released from rest at a height \( R \) above the bottom of a circular hoop of radius \( R \). The acceleration due to gravity is \( g \).

### Step 2: Apply Conservation of Energy
At the initial height \( R \), the block has potential energy and no kinetic energy since it is at rest. When it reaches the bottom of the hoop, all potential energy will have converted into kinetic energy.

- **Potential Energy at height \( R \)**:
\[ PE_A = mgh = mgR \]
- **Kinetic Energy at the bottom**:
\[ KE_B = \frac{1}{2} mv^2 \]

By conservation of energy:
\[ PE_A = KE_B \]
Thus,
\[ mgR = \frac{1}{2} mv^2 \]

### Step 3: Solve for Velocity \( v \)
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gR = \frac{1}{2} v^2 \]
Multiplying both sides by 2:
\[ 2gR = v^2 \]
Taking the square root:
\[ v = \sqrt{2gR} \]

### Step 4: Analyze Forces at the Bottom of the Hoop
At the bottom of the hoop, the block experiences two forces:
1. The gravitational force \( mg \) acting downward.
2. The normal force \( N \) exerted by the hoop acting upward.

For the block to maintain circular motion, the net force towards the center of the circle must equal the required centripetal force. The centripetal force \( F_c \) is given by:
\[ F_c = \frac{mv^2}{R} \]

### Step 5: Set Up the Equation for Forces
At the bottom of the hoop:
\[ N - mg = \frac{mv^2}{R} \]

### Step 6: Substitute for \( v^2 \)
From Step 3, we have \( v^2 = 2gR \). Substituting this into the force equation:
\[ N - mg = \frac{m(2gR)}{R} \]
This simplifies to:
\[ N - mg = 2mg \]

### Step 7: Solve for Normal Force \( N \)
Rearranging the equation gives:
\[ N = 2mg + mg = 3mg \]

### Conclusion
The magnitude of the normal force exerted on the block by the hoop when the block has reached the bottom is:
\[ \boxed{3mg} \]

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