A body of mass M and radius R is rolling horizontally without slipping with speed v . It then rolls up a hill to a maximum height h . If ` h = (5v^(2))/(6g)` , what is the M.I of the body ?

Options

A. `(MR^(2))/2 `
B. `2/3 MR^(2)`
C. `3/4 Mr^(2)`
D. `2/5 MR^(2)`

Answer

To solve the problem, we will use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Understand the initial conditions
The body of mass \( M \) and radius \( R \) is rolling without slipping with an initial speed \( v \). The body has both translational and rotational kinetic energy.

### Step 2: Write the expression for initial kinetic energy
The total initial kinetic energy \( KE_{initial} \) of the body can be expressed as the sum of translational kinetic energy and rotational kinetic energy:
\[ KE_{initial} = KE_{translational} + KE_{rotational} = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2 \]
Where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

### Step 3: Relate angular velocity to linear velocity
For a body rolling without slipping, the relationship between linear velocity \( v \) and angular velocity \( \omega \) is given by:
\[ \omega = \frac{v}{R} \]
Substituting this into the rotational kinetic energy term:
\[ KE_{rotational} = \frac{1}{2} I \left(\frac{v}{R}\right)^2 = \frac{I v^2}{2R^2} \]

### Step 4: Substitute into the kinetic energy equation
Now, substituting \( KE_{rotational} \) back into the total kinetic energy:
\[ KE_{initial} = \frac{1}{2} M v^2 + \frac{I v^2}{2R^2} \]

### Step 5: Write the expression for potential energy at height \( h \)
When the body rolls up to a height \( h \), its kinetic energy is converted into gravitational potential energy \( PE \):
\[ PE = Mgh \]
Given that \( h = \frac{5v^2}{6g} \), we can substitute this into the potential energy equation:
\[ PE = Mg \left(\frac{5v^2}{6g}\right) = \frac{5Mv^2}{6} \]

### Step 6: Apply the conservation of energy principle
According to the conservation of energy:
\[ KE_{initial} = PE \]
Substituting the expressions we derived:
\[ \frac{1}{2} M v^2 + \frac{I v^2}{2R^2} = \frac{5Mv^2}{6} \]

### Step 7: Simplify the equation
To simplify, we can multiply through by \( 6R^2 \) to eliminate the fractions:
\[ 3M v^2 R^2 + 3I v^2 = 5M v^2 R^2 \]
Now, rearranging gives:
\[ 3I v^2 = 5M v^2 R^2 - 3M v^2 R^2 \]
\[ 3I v^2 = 2M v^2 R^2 \]
Dividing both sides by \( v^2 \) (assuming \( v \neq 0 \)):
\[ 3I = 2M R^2 \]

### Step 8: Solve for the moment of inertia \( I \)
Finally, solving for \( I \):
\[ I = \frac{2M R^2}{3} \]

### Final Answer
The moment of inertia of the body is:
\[ I = \frac{2M R^2}{3} \]

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