A ring starts to roll down the inclined plane of height h without slipping . The velocity when it reaches the ground is

Options

A. `sqrt((10 gh)/7)`
B. `sqrt((4gh)/7)`
C. `sqrt((10 gh)/3)`
D. `sqrt(gh)`

Answer

To solve the problem of a ring rolling down an inclined plane of height \( h \) without slipping, we can use the principles of energy conservation and the properties of rotational motion. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the System
We have a ring that starts at a height \( h \) on an inclined plane and rolls down without slipping. The ring has both translational and rotational motion as it rolls.

### Step 2: Apply the Conservation of Energy
The potential energy (PE) at the top of the incline will convert into kinetic energy (KE) at the bottom. The total energy at the top is given by:
\[ PE = mgh \]
where \( m \) is the mass of the ring, \( g \) is the acceleration due to gravity, and \( h \) is the height.

At the bottom, the ring has both translational kinetic energy and rotational kinetic energy:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
where \( v \) is the linear velocity of the ring, \( I \) is the moment of inertia of the ring, and \( \omega \) is the angular velocity.

### Step 3: Moment of Inertia and Relationship Between Linear and Angular Velocity
For a ring, the moment of inertia \( I \) is given by:
\[ I = mr^2 \]
where \( r \) is the radius of the ring. Since the ring rolls without slipping, we have the relationship:
\[ v = r\omega \quad \Rightarrow \quad \omega = \frac{v}{r} \]

### Step 4: Substitute \( \omega \) in the Kinetic Energy Equation
Substituting \( \omega \) into the kinetic energy equation gives:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} (mr^2) \left(\frac{v}{r}\right)^2 \]
This simplifies to:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} mv^2 = mv^2 \]

### Step 5: Set Potential Energy Equal to Kinetic Energy
Now, equate the potential energy at the top to the kinetic energy at the bottom:
\[ mgh = mv^2 \]
Canceling \( m \) from both sides gives:
\[ gh = v^2 \]

### Step 6: Solve for \( v \)
Taking the square root of both sides, we find:
\[ v = \sqrt{gh} \]

### Conclusion
Thus, the velocity of the ring when it reaches the ground is:
\[ \boxed{v = \sqrt{gh}} \]