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 need to find the spin angular velocity of a uniform ball when it breaks off from the surface of a larger sphere. Here's a step-by-step solution:

### Step 1: Understand the Setup
We have a uniform ball of radius \( r \) placed on top of a sphere of radius \( R = 10r \). The ball rolls down the sphere without slipping.

### Step 2: Determine the Height
When the ball is at the top of the sphere, its height from the reference point (the bottom of the sphere) is \( h = R + r = 10r + r = 11r \).

### Step 3: Apply Conservation of Energy
The total mechanical energy at the top (potential energy only) is equal to the total mechanical energy at the point just before it breaks off (potential energy + kinetic energy).

- At the top:
\[ PE_{\text{top}} = mg(11r) \]
(where \( m \) is the mass of the ball and \( g \) is the acceleration due to gravity)

- At the point of breaking off:
- The height at this point is \( h = 11r \cos \theta \).
- The potential energy at this point:
\[ PE = mg(11r \cos \theta) \]
- The kinetic energy consists of translational and rotational components:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
where \( I = \frac{2}{5}mr^2 \) (moment of inertia of a solid sphere) and \( \omega = \frac{v}{r} \).

### Step 4: Set Up the Energy Conservation Equation
Setting the total energy at the top equal to the total energy at the point of breaking off:
\[ mg(11r) = mg(11r \cos \theta) + \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{5}mr^2\right) \left(\frac{v^2}{r^2}\right) \]

### Step 5: Simplify the Equation
Cancelling \( m \) from all terms and simplifying:
\[ g(11r) = g(11r \cos \theta) + \frac{1}{2} v^2 + \frac{1}{5} v^2 \]
\[ g(11r) = g(11r \cos \theta) + \frac{7}{10} v^2 \]

### Step 6: Rearranging the Equation
Rearranging gives:
\[ g(11r)(1 - \cos \theta) = \frac{7}{10} v^2 \]

### Step 7: Analyze Forces at the Point of Breaking Off
At the point of breaking off, the centripetal force is provided by the component of gravitational force:
\[ mg \cos \theta = \frac{mv^2}{R} \]
Substituting \( R = 11r \):
\[ g \cos \theta = \frac{v^2}{11r} \]
Thus,
\[ v^2 = 11rg \cos \theta \]

### Step 8: Substitute \( v^2 \) Back into Energy Equation
Substituting \( v^2 \) into the energy equation:
\[ g(11r)(1 - \cos \theta) = \frac{7}{10} (11rg \cos \theta) \]
Cancelling \( g \) and \( r \):
\[ 11(1 - \cos \theta) = \frac{7}{10} \cdot 11 \cos \theta \]
\[ 1 - \cos \theta = \frac{7}{10} \cos \theta \]
\[ 1 = \frac{17}{10} \cos \theta \]
\[ \cos \theta = \frac{10}{17} \]

### Step 9: Find Angular Velocity \( \omega \)
Now substituting \( \cos \theta \) back into the expression for \( v^2 \):
\[ v^2 = 11rg \cos \theta = 11rg \cdot \frac{10}{17} \]
Thus,
\[ v^2 = \frac{110rg}{17} \]

Using \( v = r \omega \):
\[ (r \omega)^2 = \frac{110rg}{17} \]
\[ \omega^2 = \frac{110g}{17r} \]
Taking the square root:
\[ \omega = \sqrt{\frac{110g}{17r}} \]

### Step 10: Identify \( p \) and \( q \)
Comparing with the given form \( \omega = \sqrt{\frac{p}{q} \frac{g}{r}} \):
- \( p = 110 \)
- \( q = 17 \)

### Final Answer
Thus, \( p + q = 110 + 17 = 127 \).