A uniform ball of radius r is placed on the top of a sphere of radius R = 10 r. It is given a slight push due to which it starts rolling down the sphere without slipping. The spin angular velocity of the ball when it breaks off from the sphere is `omega=sqrt((p)/(q)((g)/(r)))`, where g is the acceleration due to gravity and p and q are the smallest integers. What is the value of `p+q`?

Answer

To solve the problem, we will analyze the motion of a uniform ball rolling down a larger sphere. We will derive the spin angular velocity of the ball when it breaks off from the sphere.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a uniform ball of radius \( r \) placed on top of a sphere of radius \( R = 10r \).
- The ball is given a slight push and starts rolling down the sphere without slipping.

2. **Conservation of Energy**:
- When the ball rolls down, it loses potential energy and gains kinetic energy.
- The potential energy lost when the ball descends from height \( h \) is given by:
\[ \Delta PE = mgH \]
- The height \( H \) can be expressed in terms of the angle \( \theta \) as:
\[ H = R - R \cos \theta = R(1 - \cos \theta) = 10r(1 - \cos \theta) \]
- Thus, the potential energy lost becomes:
\[ \Delta PE = mg \cdot 10r(1 - \cos \theta) \]

3. **Kinetic Energy**:
- The total kinetic energy when the ball is rolling down consists of translational and rotational components:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
- For a solid sphere, the moment of inertia \( I \) is given by:
\[ I = \frac{2}{5} m r^2 \]
- Using the rolling condition \( v = r \omega \), we can express \( \omega \) in terms of \( v \):
\[ \omega = \frac{v}{r} \]
- Substituting \( \omega \) into the kinetic energy equation gives:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{5} m r^2\right \left(\frac{v}{r}\right)^2 = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \frac{7}{10} mv^2 \]

4. **Setting Up the Energy Equation**:
- Equating the potential energy lost to the kinetic energy gained:
\[ mg \cdot 10r(1 - \cos \theta) = \frac{7}{10} mv^2 \]
- Canceling \( m \) from both sides:
\[ g \cdot 10r(1 - \cos \theta) = \frac{7}{10} v^2 \]

5. **Centripetal Force Condition**:
- At the point of breaking off, the normal force \( N \) becomes zero. The gravitational force component providing centripetal acceleration is:
\[ mg \cos \theta = \frac{mv^2}{R} \]
- Substituting \( R = 11r \):
\[ mg \cos \theta = \frac{mv^2}{11r} \]
- Canceling \( m \):
\[ g \cos \theta = \frac{v^2}{11r} \]

6. **Substituting for \( v^2 \)**:
- From the energy equation, we can express \( v^2 \):
\[ v^2 = \frac{10g(1 - \cos \theta)}{7} \cdot 10r \]
- Substituting this into the centripetal force equation:
\[ g \cos \theta = \frac{\frac{10g(1 - \cos \theta)}{7}}{11r} \]

7. **Solving for \( \cos \theta \)**:
- Rearranging gives:
\[ 11g \cos \theta = \frac{10g(1 - \cos \theta)}{7} \]
- Multiplying through by 7:
\[ 77g \cos \theta = 10g - 10g \cos \theta \]
- Combining terms:
\[ 87g \cos \theta = 10g \]
- Thus:
\[ \cos \theta = \frac{10}{87} \]

8. **Finding \( \omega \)**:
- Substitute \( \cos \theta \) back into the expression for \( v^2 \):
\[ v^2 = \frac{10g(1 - \frac{10}{87})}{7} \cdot 10r \]
- Simplifying gives:
\[ v^2 = \frac{10g \cdot \frac{77}{87}}{7} \cdot 10r \]
- Finally, substituting \( v = r \omega \):
\[ r^2 \omega^2 = \frac{10g \cdot \frac{77}{87}}{7} \cdot 10r \]
- Solving for \( \omega \):
\[ \omega^2 = \frac{1100g}{609r} \]

9. **Identifying \( p \) and \( q \)**:
- Comparing with the given form \( \omega = \sqrt{\frac{p}{q}\frac{g}{r}} \):
\[ p = 1100, \quad q = 609 \]
- Therefore, \( p + q = 1100 + 609 = 1709 \).

### Final Answer:
The value of \( p + q \) is \( 1709 \).