A rod revlving 60 times in a minute about an axis passing through an end at right angles to the length, has kinetic energy of 400 J. find the moment of inertia of the rod.

Answer

To find the moment of inertia of the rod, we can follow these steps:

### Step 1: Convert the rotational speed from revolutions per minute to radians per second.
The rod is revolving at 60 revolutions per minute (rpm). To convert this to radians per second (rad/s), we use the following conversion:

\[ \omega = \text{revolutions per minute} \times \frac{2\pi \text{ radians}}{1 \text{ revolution}} \times \frac{1 \text{ minute}}{60 \text{ seconds}} \]

Substituting the values:

\[ \omega = 60 \times \frac{2\pi}{60} = 2\pi \text{ rad/s} \]

### Step 2: Use the formula for rotational kinetic energy.
The formula for rotational kinetic energy (KE) is given by:

\[ KE = \frac{1}{2} I \omega^2 \]

Where:
- \( KE \) is the kinetic energy (400 J)
- \( I \) is the moment of inertia (which we need to find)
- \( \omega \) is the angular velocity (which we calculated as \( 2\pi \text{ rad/s} \))

### Step 3: Substitute the known values into the kinetic energy formula.
We can rearrange the formula to solve for \( I \):

\[ 400 = \frac{1}{2} I (2\pi)^2 \]

### Step 4: Simplify the equation.
First, calculate \( (2\pi)^2 \):

\[ (2\pi)^2 = 4\pi^2 \]

Now substitute this back into the equation:

\[ 400 = \frac{1}{2} I (4\pi^2) \]

This simplifies to:

\[ 400 = 2\pi^2 I \]

### Step 5: Solve for the moment of inertia \( I \).
Rearranging gives:

\[ I = \frac{400}{2\pi^2} = \frac{200}{\pi^2} \]

### Step 6: Calculate the numerical value of \( I \).
Using \( \pi \approx 3.14 \):

\[ I \approx \frac{200}{(3.14)^2} \approx \frac{200}{9.8596} \approx 20.26 \text{ kg m}^2 \]

### Final Answer:
The moment of inertia of the rod is approximately \( 20.26 \, \text{kg m}^2 \).

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