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.
Which one of the following expressions gives the speed of the mass at the bottom of the hoop ?

Options

A. zero`m//s^(2)`
B. `v=(mg)/(2R)`
C. `v^(2)=2gR`
D. `v=mgR`

Answer

To solve the problem of finding the speed of a block of mass \( m \) at the bottom of a frictionless circular hoop of radius \( R \), we can use the principle of conservation of mechanical energy. Here’s a step-by-step solution:

### Step 1: Understand the Initial and Final Conditions
- The block is released from rest at a height \( R \) above the horizontal surface.
- At the top (initial position), the block has potential energy and no kinetic energy since it starts from rest.
- At the bottom (final position), the block has kinetic energy and no potential energy (as it is at the reference height).

### Step 2: Write the Expression for Potential Energy at the Top
The potential energy (PE) at the top when the block is at height \( R \) is given by:
\[ PE_{\text{top}} = mgh = mgR \]
where \( h = R \) and \( g \) is the acceleration due to gravity.

### Step 3: Write the Expression for Kinetic Energy at the Bottom
The kinetic energy (KE) at the bottom when the block has speed \( v \) is given by:
\[ KE_{\text{bottom}} = \frac{1}{2} mv^2 \]

### Step 4: Apply the Conservation of Energy Principle
According to the conservation of mechanical energy, the total mechanical energy at the top must equal the total mechanical energy at the bottom:
\[ PE_{\text{top}} = KE_{\text{bottom}} \]
Substituting the expressions we derived:
\[ mgR = \frac{1}{2} mv^2 \]

### Step 5: Simplify the Equation
We can cancel the mass \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gR = \frac{1}{2} v^2 \]

### Step 6: Solve for \( v^2 \)
To isolate \( v^2 \), multiply both sides by 2:
\[ 2gR = v^2 \]

### Step 7: Take the Square Root to Find \( v \)
Taking the square root of both sides gives us the speed \( v \) at the bottom of the hoop:
\[ v = \sqrt{2gR} \]

### Final Expression
Thus, the speed of the mass at the bottom of the hoop is:
\[ v = \sqrt{2gR} \]