A sphere of mass `m` and radius `r` rolls on a horizontal plane without slipping with a speed `u`. Now it rolls up vertically, then maximum height it would be attain will be

Options

A. `(3u^2)/(4g)`
B. `(5u^2)/(2g)`
C. `(7u^2)/(10 g)`
D. `(11u ^2)/(9g)`

Answer

To solve the problem of finding the maximum height attained by a sphere of mass `m` and radius `r` that rolls without slipping with an initial speed `u`, we can follow these steps:

### Step 1: Understand the Energy Conservation Principle
When the sphere rolls up vertically, its initial kinetic energy will be converted into potential energy at the maximum height. The total mechanical energy is conserved.

### Step 2: Write the Expression for Kinetic Energy
The total kinetic energy (KE) of the sphere consists of translational kinetic energy and rotational kinetic energy:
\[ KE = KE_{\text{translational}} + KE_{\text{rotational}} = \frac{1}{2} m u^2 + \frac{1}{2} I \omega^2 \]

### Step 3: Determine the Moment of Inertia
For a solid sphere, the moment of inertia \( I \) is given by:
\[ I = \frac{2}{5} m r^2 \]

### Step 4: Relate Angular Velocity to Linear Velocity
Since the sphere rolls without slipping, we have the relationship:
\[ u = \omega r \implies \omega = \frac{u}{r} \]

### Step 5: Substitute the Values into the Kinetic Energy Equation
Substituting the moment of inertia and angular velocity into the kinetic energy equation:
\[ KE = \frac{1}{2} m u^2 + \frac{1}{2} \left(\frac{2}{5} m r^2\right} \left(\frac{u}{r}\right)^2 \]
This simplifies to:
\[ KE = \frac{1}{2} m u^2 + \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{u^2}{r^2}\right) \]
\[ KE = \frac{1}{2} m u^2 + \frac{1}{5} m u^2 = \left(\frac{5}{10} + \frac{2}{10}\right) m u^2 = \frac{7}{10} m u^2 \]

### Step 6: Set the Kinetic Energy Equal to Potential Energy
At the maximum height \( h \), all kinetic energy is converted into potential energy:
\[ KE = PE \implies \frac{7}{10} m u^2 = mgh \]

### Step 7: Solve for Maximum Height \( h \)
Cancelling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{7}{10} u^2 = gh \]
Rearranging gives:
\[ h = \frac{7}{10} \frac{u^2}{g} \]

### Conclusion
The maximum height \( h \) that the sphere would attain is:
\[ h = \frac{7u^2}{10g} \]