A uniform disc of mass M and radius R is rotating about a horizontal axis passing through its centre with angular velocity `omega` . A piece of mass m breaks from the disc and flies off vertically upwards. The angular speed of the disc will be

Options

A. `((M-2m)omega)/((M-m))`
B. `((M+2m)omega)/((M+m))`
C. `((M-2m)omega)/((M+m))`
D. `((M+2m)omega)/((M-m))`

Answer

To solve the problem of finding the new angular speed of the disc after a piece of mass \( m \) breaks off, we will use the principle of conservation of angular momentum. Here’s a step-by-step solution:

### Step 1: Understand the Initial Conditions
The disc has a mass \( M \) and radius \( R \), and it is rotating about a horizontal axis with an initial angular velocity \( \omega \).

### Step 2: Calculate the Initial Angular Momentum
The moment of inertia \( I \) of a uniform disc about an axis through its center is given by:
\[ I = \frac{1}{2} M R^2 \]
The initial angular momentum \( L_i \) of the disc is:
\[ L_i = I \cdot \omega = \frac{1}{2} M R^2 \cdot \omega \]

### Step 3: Analyze the Situation After the Mass Breaks Off
When the piece of mass \( m \) breaks off and flies vertically upwards, it retains its tangential velocity \( V \) at the edge of the disc, which is given by:
\[ V = \omega R \]
The angular momentum of the piece of mass \( m \) about the axis of rotation (using the perpendicular distance \( R \)) is:
\[ L_m = m \cdot V \cdot R = m \cdot (\omega R) \cdot R = m \omega R^2 \]

### Step 4: Calculate the Moment of Inertia of the Remaining Disc
After the mass \( m \) breaks off, the remaining mass of the disc is \( M - m \). The new moment of inertia \( I' \) of the remaining disc is:
\[ I' = \frac{1}{2} (M - m) R^2 \]

### Step 5: Apply Conservation of Angular Momentum
According to the conservation of angular momentum, the initial angular momentum must equal the final angular momentum:
\[ L_i = L_f \]
Thus,
\[ \frac{1}{2} M R^2 \cdot \omega = \frac{1}{2} (M - m) R^2 \cdot \omega' + m \omega R^2 \]

### Step 6: Simplify the Equation
Dividing through by \( R^2 \) (assuming \( R \neq 0 \)):
\[ \frac{1}{2} M \omega = \frac{1}{2} (M - m) \omega' + m \omega \]

### Step 7: Rearrange to Solve for \( \omega' \)
Rearranging gives:
\[ \frac{1}{2} M \omega - m \omega = \frac{1}{2} (M - m) \omega' \]
\[ \left(\frac{1}{2} M - m\right) \omega = \frac{1}{2} (M - m) \omega' \]
Now, isolating \( \omega' \):
\[ \omega' = \frac{(M/2 - m) \cdot \omega}{(M - m)/2} \]
\[ \omega' = \frac{M - 2m}{M - m} \cdot \omega \]

### Final Answer
The new angular speed of the disc after the mass \( m \) flies off is:
\[ \omega' = \frac{M - 2m}{M - m} \cdot \omega \]