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

Answer

To find the angular velocity of a uniform ball of radius \( r \) rolling without slipping down from the top of a sphere of radius \( R \) at the moment it breaks off, we can follow these steps:

### Step 1: Understand the Geometry of the Problem
The ball rolls down the sphere and breaks off at an angle \( \theta \) with the vertical. The distance from the center of the sphere to the center of the ball when it breaks off is \( R + r \).

### Step 2: Apply Energy Conservation
Initially, the ball is at rest at height \( h = R + r \). As it rolls down, it gains kinetic energy. The potential energy lost is converted into kinetic energy. The height \( h \) at the moment it breaks off can be expressed as:
\[ h = (R + r) - (R + r) \cos \theta = (R + r)(1 - \cos \theta) \]

### Step 3: Write the Energy Conservation Equation
Using the work-energy theorem, the work done by gravity equals the change in kinetic energy:
\[ mgh = \text{K.E.} \]
The kinetic energy consists of translational and rotational parts:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
For a uniform ball, the moment of inertia \( I \) is given by \( \frac{2}{5}mr^2 \). The relationship between linear velocity \( v \) and angular velocity \( \omega \) is \( v = r\omega \).

### Step 4: Substitute and Simplify
Substituting \( I \) and \( \omega \):
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{5}mr^2\right) \left(\frac{v^2}{r^2}\right) \]
This simplifies to:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \frac{7}{10} mv^2 \]
Cancelling \( m \) from both sides gives:
\[ gh = \frac{7}{10} v^2 \]

### Step 5: Substitute for \( h \)
Substituting for \( h \):
\[ g(R + r)(1 - \cos \theta) = \frac{7}{10} v^2 \]
From this, we can express \( v^2 \):
\[ v^2 = \frac{10g(R + r)(1 - \cos \theta)}{7} \]

### Step 6: Centripetal Force Condition
At the point of breaking off, the centripetal force required for circular motion is provided by the component of the gravitational force acting on the ball. The equation is:
\[ mg \cos \theta = \frac{mv^2}{R + r} \]
Cancelling \( m \) gives:
\[ g \cos \theta = \frac{v^2}{R + r} \]
Substituting for \( v^2 \):
\[ g \cos \theta = \frac{10g(R + r)(1 - \cos \theta)}{7(R + r)} \]
This simplifies to:
\[ \cos \theta = \frac{10(1 - \cos \theta)}{7} \]
Solving for \( \cos \theta \):
\[ 7 \cos \theta = 10 - 10 \cos \theta \]
\[ 17 \cos \theta = 10 \implies \cos \theta = \frac{10}{17} \]

### Step 7: Calculate \( v \)
Substituting \( \cos \theta \) back into the equation for \( v^2 \):
\[ v^2 = \frac{10g(R + r)(1 - \frac{10}{17})}{7} = \frac{10g(R + r)(\frac{7}{17})}{7} = \frac{10g(R + r)}{17} \]

### Step 8: Find Angular Velocity \( \omega \)
Using \( v = r \omega \):
\[ \omega = \frac{v}{r} = \frac{\sqrt{\frac{10g(R + r)}{17}}}{r} = \sqrt{\frac{10g(R + r)}{17r^2}} \]

### Final Answer
The angular velocity \( \omega \) of the ball at the moment it breaks off the sphere is:
\[ \omega = \sqrt{\frac{10g(R + r)}{17r^2}} \]