An object of radius R and mass M is rolling horizontally without slipping with speed v . It then rolls up the hill to a maximum height `h = (3v^(2))/(4g)` . The moment of inertia of the object is ( g = acceleration due to gravity)

Answer

To find the moment of inertia of the object rolling up the hill, we will use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Identify the initial kinetic energy
The object is rolling without slipping, so it has both translational and rotational kinetic energy. The total initial kinetic energy (KE_initial) can be expressed as:
\[ KE_{\text{initial}} = KE_{\text{translational}} + KE_{\text{rotational}} = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2 \]
where:
- \( M \) is the mass of the object,
- \( v \) is the linear speed,
- \( I \) is the moment of inertia,
- \( \omega \) is the angular velocity.

### Step 2: Relate angular velocity to linear speed
For rolling without slipping, the relationship between linear speed \( v \) and angular velocity \( \omega \) is given by:
\[ \omega = \frac{v}{R} \]
where \( R \) is the radius of the object.

### Step 3: Substitute \( \omega \) in the kinetic energy equation
Substituting \( \omega \) in the kinetic energy equation gives:
\[ KE_{\text{initial}} = \frac{1}{2} M v^2 + \frac{1}{2} I \left(\frac{v}{R}\right)^2 \]
This simplifies to:
\[ KE_{\text{initial}} = \frac{1}{2} M v^2 + \frac{1}{2} \frac{I v^2}{R^2} \]

### Step 4: Identify the potential energy at maximum height
When the object reaches its maximum height \( h \), all the kinetic energy is converted into potential energy (PE):
\[ PE = Mgh \]
Given that \( h = \frac{3v^2}{4g} \), we can substitute this into the potential energy equation:
\[ PE = Mg \left(\frac{3v^2}{4g}\right) = \frac{3Mv^2}{4} \]

### Step 5: Set the initial kinetic energy equal to potential energy
By conservation of energy:
\[ KE_{\text{initial}} = PE \]
Substituting the expressions we derived:
\[ \frac{1}{2} M v^2 + \frac{1}{2} \frac{I v^2}{R^2} = \frac{3Mv^2}{4} \]

### Step 6: Rearrange the equation to solve for \( I \)
Multiply through by 2 to eliminate the fractions:
\[ M v^2 + \frac{I v^2}{R^2} = \frac{3Mv^2}{2} \]
Now, isolate the term with \( I \):
\[ \frac{I v^2}{R^2} = \frac{3Mv^2}{2} - M v^2 \]
This simplifies to:
\[ \frac{I v^2}{R^2} = \frac{3Mv^2}{2} - \frac{2Mv^2}{2} = \frac{Mv^2}{2} \]

### Step 7: Solve for \( I \)
Now, we can solve for \( I \):
\[ I = \frac{Mv^2}{2} \cdot \frac{R^2}{v^2} = \frac{1}{2} M R^2 \]

### Final Answer
The moment of inertia \( I \) of the object is:
\[ I = \frac{1}{2} M R^2 \]