A circular disc has a mass of 1kg and radius 40 cm. It is rotating about an axis passing through its centre and perpendicular to its plane with a speed of 10rev/s. The work done in joules in stopping it would be-

Options

A. `4`
B. `47.5`
C. `79`
D. `158`

Answer

To solve the problem of finding the work done in stopping a rotating circular disc, we can follow these steps:

### Step 1: Identify the given values
- Mass of the disc (m) = 1 kg
- Radius of the disc (r) = 40 cm = 0.4 m (convert to meters)
- Rotational speed (N) = 10 revolutions per second (rev/s)

### Step 2: Calculate the moment of inertia (I) of the disc
The moment of inertia (I) for a circular disc rotating about an axis through its center is given by the formula:
\[ I = \frac{1}{2} m r^2 \]
Substituting the values:
\[ I = \frac{1}{2} \times 1 \, \text{kg} \times (0.4 \, \text{m})^2 \]
\[ I = \frac{1}{2} \times 1 \times 0.16 \]
\[ I = 0.08 \, \text{kg m}^2 \]

### Step 3: Convert rotational speed from revolutions per second to radians per second
To convert revolutions per second (rev/s) to radians per second (rad/s), we use the conversion factor \( 2\pi \) radians per revolution:
\[ \omega = N \times 2\pi \]
Substituting the values:
\[ \omega = 10 \, \text{rev/s} \times 2\pi \]
\[ \omega = 20\pi \, \text{rad/s} \]

### Step 4: Calculate the initial rotational kinetic energy (KE)
The rotational kinetic energy (KE) is given by the formula:
\[ KE = \frac{1}{2} I \omega^2 \]
Substituting the values we calculated:
\[ KE = \frac{1}{2} \times 0.08 \, \text{kg m}^2 \times (20\pi)^2 \]
\[ KE = \frac{1}{2} \times 0.08 \times 400\pi^2 \]
\[ KE = 0.04 \times 400\pi^2 \]
\[ KE = 16\pi^2 \]

### Step 5: Calculate the numerical value of the kinetic energy
Using \( \pi \approx 3.14 \):
\[ KE \approx 16 \times (3.14)^2 \]
\[ KE \approx 16 \times 9.8596 \]
\[ KE \approx 157.74 \, \text{J} \]

### Step 6: Determine the work done to stop the disc
The work done (W) in stopping the disc is equal to the negative of the initial kinetic energy (since we are removing energy):
\[ W = -KE \]
Thus:
\[ W \approx -157.74 \, \text{J} \]

### Step 7: Final answer
The work done in stopping the disc is approximately \( 158 \, \text{J} \) (taking the magnitude).

### Summary
The work done in stopping the disc is approximately 158 J.

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