A solid sphere rolls down a parabolic path , whose vertical dimension is given by `y = kx^(2)` and base of the parabola is at x = 0 . During the fall the sphere does not slip, but beyond the base the climb is frictionless. If the sphere falls through a height h, the height above the base that it would climb is

Options

A. 54
B. 7h
C. `(7)/( 5) h`
D. `( 5)/( 7)h`

Answer

To solve the problem of a solid sphere rolling down a parabolic path and then climbing back up, we can follow these steps:

### Step 1: Understand the Energy Conservation
When the sphere rolls down the parabolic path, it converts potential energy into kinetic energy. The potential energy lost by the sphere when it falls through a height \( h \) is given by:
\[ PE = mgh \]

### Step 2: Calculate the Kinetic Energy
The kinetic energy of the sphere at the bottom of the parabolic path consists of both translational and rotational kinetic energy. The total kinetic energy \( KE \) 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 sphere, the moment of inertia \( I \) is given by \( I = \frac{2}{5} m r^2 \) and the relationship between linear velocity \( v \) and angular velocity \( \omega \) is \( v = r \omega \). Thus, we can substitute \( \omega \):
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v^2}{r^2}\right) = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \frac{7}{10} mv^2 \]

### Step 3: Apply Energy Conservation
By the principle of conservation of energy, the potential energy lost equals the kinetic energy gained:
\[ mgh = \frac{7}{10} mv^2 \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh = \frac{7}{10} v^2 \]

### Step 4: Solve for \( v^2 \)
Rearranging the equation gives:
\[ v^2 = \frac{10}{7} gh \]

### Step 5: Analyze the Climb
When the sphere climbs back up, it will convert its kinetic energy back into potential energy. The kinetic energy at the base is:
\[ KE = \frac{7}{10} mv^2 = \frac{7}{10} m \left(\frac{10}{7} gh\right) = mgh \]
As the sphere climbs, it will reach a maximum height \( h_0 \) where all the kinetic energy is converted into potential energy:
\[ mgh = mgh_0 \]
Thus, we can write:
\[ h_0 = h \]

### Step 6: Find the Height Above the Base
However, since the sphere rolls down a height \( h \) and then climbs back up, we need to consider the energy loss due to the climb. The energy available for climbing is:
\[ KE = mgh - mgh_0 \]
Setting the kinetic energy equal to the potential energy at height \( h_0 \):
\[ mgh_0 = \frac{7}{10} mgh \]
Cancelling \( m \) gives:
\[ h_0 = \frac{7}{10} h \]

### Final Result
The height above the base that the sphere would climb is:
\[ h_0 = \frac{5}{7} h \]