A uniform rope with length `L` and mass `m` is held at one end and whirled in a horizontal circle with angular velocity `omega`. You can ignor the force of gravity on the rope. Find the time required for a transverse wave to travel from one end of the rope to the other.
Hint: `int (dx)/(sqrt(a^(2) - (x^(2)))) = sin^(-1)((x)/(a))`

Explanation

To solve the problem of finding the time required for a transverse wave to travel from one end of a uniform rope to the other while being whirled in a horizontal circle, we can follow these steps:

### Step 1: Understand the System
We have a uniform rope of length \( L \) and mass \( m \) being whirled in a horizontal circle with angular velocity \( \omega \). The rope is held at one end. We need to find the time it takes for a transverse wave to travel from one end of the rope to the other.

### Step 2: Determine the Tension in the Rope
As the rope is whirled, the tension \( T \) in the rope varies along its length. The tension at a point \( x \) from the fixed end can be expressed as:
\[ T(x) = m \cdot \omega^2 \cdot x \]
where \( m \) is the mass of the rope and \( \omega \) is the angular velocity.

### Step 3: Relate Tension to Wave Velocity
The velocity \( v \) of a wave traveling through the rope is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( \mu \) is the mass per unit length of the rope. For a uniform rope, \( \mu = \frac{m}{L} \).

Substituting the expression for tension:
\[ v = \sqrt{\frac{m \cdot \omega^2 \cdot x}{\mu}} = \sqrt{\frac{m \cdot \omega^2 \cdot x}{\frac{m}{L}}} = \sqrt{\omega^2 \cdot x \cdot L} = \omega \sqrt{x \cdot L} \]

### Step 4: Set Up the Differential Equation
The relationship between wave velocity and the distance traveled can be expressed as:
\[ \frac{dx}{dt} = v = \omega \sqrt{x \cdot L} \]

### Step 5: Rearrange and Integrate
Rearranging gives:
\[ \frac{dx}{\sqrt{x}} = \omega \sqrt{L} \, dt \]

Integrating both sides:
\[ \int_0^L \frac{dx}{\sqrt{x}} = \omega \sqrt{L} \int_0^t dt \]

### Step 6: Solve the Integrals
The left side integral evaluates to:
\[ 2\sqrt{x} \bigg|_0^L = 2\sqrt{L} \]
The right side evaluates to:
\[ \omega \sqrt{L} \cdot t \]

Setting the two sides equal gives:
\[ 2\sqrt{L} = \omega \sqrt{L} \cdot t \]

### Step 7: Solve for Time \( t \)
Dividing both sides by \( \omega \sqrt{L} \):
\[ t = \frac{2\sqrt{L}}{\omega} \]

### Step 8: Final Result
Thus, the time required for a transverse wave to travel from one end of the rope to the other is:
\[ t = \frac{2\sqrt{L}}{\omega} \]