A solid cylinder rolls up an inclined plane of inclination `theta` with an initial velocity `v`. How far does the cylinder go up the plane ?

Options

A. `(3v^(2))/2g sintheta`
B. `v^(2)/4g sintheta`
C. `(3v^(2))/g sintheta`
D. `(3v^(2))/4g sintheta`

Explanation

To solve the problem of how far a solid cylinder rolls up an inclined plane with an initial velocity \( v \), we can use the principle of conservation of energy. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Energy Conservation
Initially, the cylinder has kinetic energy due to its translational and rotational motion. As it rolls up the incline, this kinetic energy is converted into gravitational potential energy.

### Step 2: Write the Expression for Initial Kinetic Energy
The total initial kinetic energy \( KE \) of the cylinder can be expressed as:
\[ KE = KE_{\text{translational}} + KE_{\text{rotational}} = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
For a solid cylinder, the moment of inertia \( I \) is given by \( \frac{1}{2} mr^2 \) and since it rolls without slipping, we have \( v = r\omega \), which implies \( \omega = \frac{v}{r} \).

### Step 3: Substitute the Values
Substituting \( I \) and \( \omega \) into the kinetic energy equation:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{1}{2} mr^2\right) \left(\frac{v}{r}\right)^2 \]
This simplifies to:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2 \]

### Step 4: Write the Expression for Potential Energy
As the cylinder rolls up the incline, it gains potential energy \( PE \):
\[ PE = mgh \]
where \( h \) is the height gained.

### Step 5: Apply Conservation of Energy
Setting the initial kinetic energy equal to the potential energy at the maximum height:
\[ \frac{3}{4} mv^2 = mgh \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{3}{4} v^2 = gh \]

### Step 6: Solve for Height \( h \)
Rearranging gives:
\[ h = \frac{3v^2}{4g} \]

### Step 7: Relate Height to Distance on the Incline
To find the distance \( s \) that the cylinder travels up the incline, we can use the relationship between height \( h \) and distance \( s \):
\[ h = s \sin \theta \]
Thus, we can express \( s \) as:
\[ s = \frac{h}{\sin \theta} \]

### Step 8: Substitute for \( h \)
Substituting the expression for \( h \):
\[ s = \frac{\frac{3v^2}{4g}}{\sin \theta} = \frac{3v^2}{4g \sin \theta} \]

### Final Answer
The maximum distance \( s \) that the cylinder rolls up the inclined plane is:
\[ s = \frac{3v^2}{4g \sin \theta} \]

---