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 find the velocity of a ring rolling down an inclined plane of height \( h \) without slipping, we can use the principle of conservation of energy. Here’s the step-by-step solution:

### Step 1: Identify the energies involved
When the ring is at the height \( h \), it possesses gravitational potential energy given by:
\[ PE = mgh \]
where \( m \) is the mass of the ring, \( g \) is the acceleration due to gravity, and \( h \) is the height of the inclined plane.

### Step 2: Determine the kinetic energy at the bottom
As the ring rolls down the incline, this potential energy is converted into kinetic energy. The total kinetic energy (\( KE \)) of the rolling ring consists of translational kinetic energy and rotational kinetic energy:
\[ KE = KE_{translational} + KE_{rotational} = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
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 3: Substitute the moment of inertia and angular velocity
Substituting \( I \) and \( \omega \) 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 4: Apply the conservation of energy principle
According to the conservation of energy, the potential energy at the top equals the kinetic energy at the bottom:
\[ mgh = mv^2 \]

### Step 5: Solve for \( v \)
Cancelling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh = v^2 \]
Taking the square root of both sides gives:
\[ v = \sqrt{gh} \]

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

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