A circuit disc of mass 49 kg and of radius 50 cm is mounted axially and made to rotate. Calculate the K.E.. It possesses, when executing 120 rpm

Answer

To calculate the kinetic energy of a rotating disc, we can follow these steps:

### Step 1: Convert the rotational speed from RPM to radians per second.
The formula to convert revolutions per minute (RPM) to radians per second is:
\[ \omega = \frac{2\pi \times \text{RPM}}{60} \]
Given that the disc rotates at 120 RPM:
\[ \omega = \frac{2\pi \times 120}{60} = 4\pi \, \text{rad/s} \]

### Step 2: Calculate the moment of inertia (I) of the disc.
The moment of inertia for a solid disc rotating about its axis is given by:
\[ I = \frac{1}{2} M R^2 \]
Where:
- \( M = 49 \, \text{kg} \) (mass of the disc)
- \( R = 0.5 \, \text{m} \) (radius of the disc, converted from cm to m)

Calculating \( I \):
\[ I = \frac{1}{2} \times 49 \times (0.5)^2 = \frac{1}{2} \times 49 \times 0.25 = \frac{49}{8} \, \text{kg m}^2 \]

### Step 3: Calculate the kinetic energy (K.E.) of the rotating disc.
The kinetic energy of a rotating object is given by:
\[ K.E. = \frac{1}{2} I \omega^2 \]
Substituting the values of \( I \) and \( \omega \):
\[ K.E. = \frac{1}{2} \times \frac{49}{8} \times (4\pi)^2 \]
Calculating \( \omega^2 \):
\[ (4\pi)^2 = 16\pi^2 \]
Now substituting back into the K.E. formula:
\[ K.E. = \frac{1}{2} \times \frac{49}{8} \times 16\pi^2 \]
\[ K.E. = \frac{49 \times 16\pi^2}{16} = \frac{49\pi^2}{1} = 49\pi^2 \, \text{J} \]

### Step 4: Calculate the numerical value of K.E.
Using \( \pi \approx 3.14 \):
\[ K.E. \approx 49 \times (3.14)^2 \approx 49 \times 9.8596 \approx 483.1 \, \text{J} \]

### Final Answer:
The kinetic energy of the rotating disc is approximately **484 J**.

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