A uniform ball of radius `r` rolls without slipping down from the top of a sphere of radius `R` Find the angular velocity of the ball at the moment it breaks off the sphere. The initial velocity of the ball is negligible.

Options

A. `sqrt((5g(R+r))(17r^(2)))`
B. `((10g(R+r))/(17r^(2)))`
C. `sqrt((5g(R-r))/(10r^(2)))`
D. `sqrt((10g(R+r))/(7r^(2)))`

Explanation

To find the angular velocity of a uniform ball of radius \( r \) as it rolls down a sphere of radius \( R \) and breaks off, we can follow these steps:

### Step 1: Understand the Problem
The ball rolls down the sphere without slipping. At the moment it breaks off, we need to find its angular velocity. The initial velocity of the ball is negligible.

### Step 2: Define the Variables
Let:
- \( m \) = mass of the ball
- \( r \) = radius of the ball
- \( R \) = radius of the sphere
- \( v \) = linear velocity of the ball at the moment it breaks off
- \( \omega \) = angular velocity of the ball at the moment it breaks off
- \( \theta \) = angle at which the ball breaks off

### Step 3: Apply Conservation of Energy
The potential energy lost by the ball as it rolls down is converted into kinetic energy (both translational and rotational). The height \( h \) from which the ball falls can be expressed as:
\[ h = R - (R - r) \cos \theta = R - r \cos \theta \]
The potential energy lost is:
\[ \Delta PE = mg(R - r \cos \theta) \]
The total kinetic energy at the point of breaking off is:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
For a uniform ball, the moment of inertia \( I \) is:
\[ I = \frac{2}{5} m r^2 \]
Since the ball rolls without slipping, we have the relation:
\[ v = r \omega \]
Substituting \( \omega \) gives:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v}{r}\right)^2 = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \frac{7}{10} mv^2 \]

### Step 4: Set Up the Energy Equation
Equating potential energy lost to kinetic energy gained:
\[ mg(R - r \cos \theta) = \frac{7}{10} mv^2 \]
Cancelling \( m \) from both sides:
\[ g(R - r \cos \theta) = \frac{7}{10} v^2 \]

### Step 5: Analyze Forces at the Break-off Point
At the break-off point, the centripetal force required for circular motion is provided by the component of gravitational force:
\[ mg \cos \theta = \frac{mv^2}{R + r} \]
Cancelling \( m \):
\[ g \cos \theta = \frac{v^2}{R + r} \]

### Step 6: Solve for \( v^2 \)
From the centripetal force equation:
\[ v^2 = g \cos \theta (R + r) \]
Substituting this into the energy equation:
\[ g(R - r \cos \theta) = \frac{7}{10} g \cos \theta (R + r) \]
Cancelling \( g \):
\[ R - r \cos \theta = \frac{7}{10} \cos \theta (R + r) \]

### Step 7: Rearranging and Solving for \( \cos \theta \)
Rearranging gives:
\[ 10(R - r \cos \theta) = 7 \cos \theta (R + r) \]
\[ 10R - 10r \cos \theta = 7R \cos \theta + 7r \cos \theta \]
\[ 10R = (10r + 7R) \cos \theta \]
\[ \cos \theta = \frac{10R}{10r + 7R} \]

### Step 8: Substitute Back to Find \( v^2 \)
Substituting \( \cos \theta \) back into the equation for \( v^2 \):
\[ v^2 = g \left(\frac{10R}{10r + 7R}\right)(R + r) \]

### Step 9: Find Angular Velocity \( \omega \)
Using \( v = r \omega \):
\[ \omega = \frac{v}{r} = \frac{1}{r} \sqrt{g \left(\frac{10R}{10r + 7R}\right)(R + r)} \]

### Final Result
Thus, the angular velocity \( \omega \) of the ball at the moment it breaks off is:
\[ \omega = \sqrt{\frac{10g(R + r)}{r(10r + 7R)}} \]