A hollow sphere of mass `m` starting from rest rolls without slipping, on an inclined plane of inclination What is the total energy of the sphere after `10^(-5)s` if after `10^(-5) is v`?

Options

A. `5//6mv^(2)`
B. `2//3mv^(2)`
C. `10^-5mv^(2)`
D. `mv^(2)//10^(-5)`

Explanation

To solve the problem of finding the total energy of a hollow sphere rolling down an inclined plane after a time of \(10^{-5}\) seconds, we can follow these steps:

### Step 1: Understand the motion of the hollow sphere
The hollow sphere starts from rest and rolls without slipping down the inclined plane. This means that it has both translational and rotational motion.

**Hint:** Remember that rolling without slipping means that the linear velocity \(v\) and the angular velocity \(\omega\) are related by the equation \(v = r \omega\), where \(r\) is the radius of the sphere.

### Step 2: Write the expressions for kinetic energy
The total kinetic energy \(K\) of the sphere consists of translational kinetic energy and rotational kinetic energy:
\[ K = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
For a hollow sphere, the moment of inertia \(I\) is given by:
\[ I = \frac{2}{3} m r^2 \]

**Hint:** Recall the formulas for translational and rotational kinetic energy.

### Step 3: Substitute the moment of inertia into the kinetic energy equation
Substituting the moment of inertia into the kinetic energy equation:
\[ K = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{3} m r^2\right) \omega^2 \]

**Hint:** Make sure to express \(\omega\) in terms of \(v\) and \(r\) using the relationship \(v = r \omega\).

### Step 4: Replace \(\omega\) with \(v/r\)
Using the relationship \(v = r \omega\), we can express \(\omega\) as:
\[ \omega = \frac{v}{r} \]
Substituting this into the kinetic energy equation gives:
\[ K = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{3} m r^2\right) \left(\frac{v}{r}\right)^2 \]
\[ K = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{3} m r^2\right) \left(\frac{v^2}{r^2}\right) \]
\[ K = \frac{1}{2} mv^2 + \frac{1}{3} mv^2 \]

**Hint:** Combine the terms carefully to simplify the expression for total kinetic energy.

### Step 5: Simplify the total kinetic energy
Combining the terms:
\[ K = \left(\frac{1}{2} + \frac{1}{3}\right) mv^2 = \left(\frac{3}{6} + \frac{2}{6}\right) mv^2 = \frac{5}{6} mv^2 \]

**Hint:** Ensure that you correctly find a common denominator to combine the fractions.

### Step 6: Calculate the total energy
Since the sphere starts from rest and rolls down the incline, the potential energy at the top will convert into kinetic energy as it rolls down. After \(10^{-5}\) seconds, if the velocity is \(v\), the total energy \(E\) of the sphere is given by:
\[ E = K = \frac{5}{6} mv^2 \]

**Hint:** Remember that the total energy is conserved if we ignore friction and air resistance.

### Final Answer
The total energy of the hollow sphere after \(10^{-5}\) seconds is:
\[ E = \frac{5}{6} mv^2 \]