To find the linear velocity of the rod's ends, we can use the relationship between rotational kinetic energy and linear velocity.

Step 1: Set up the Known Information

• Mass of the rod, $m = 101.56 \, \text{kg}$
• Rotational kinetic energy, $KE = 1.62 \, \text{kJ} = 1620 \, \text{J}$
• The rod rotates about its center.

Step 2: Rotational Kinetic Energy and Moment of Inertia

The formula for rotational kinetic energy is: $KE = \frac{1}{2} I \omega^2$ where:

• $I$ is the moment of inertia of the rod about its center.
• $\omega$ is the angular velocity.

For a thin rod rotating about its center, the moment of inertia $I$ is: $I = \frac{1}{12} m L^2$ where $L$ is the length of the rod.

Step 3: Express $\omega$ in Terms of Linear Velocity

The linear velocity $v$ at either end of the rod is related to $\omega$ by: $v = \frac{L}{2} \cdot \omega$ which rearranges to give: $\omega = \frac{2v}{L}$

Step 4: Substitute $\omega$ into the Kinetic Energy Equation

Substituting $\omega = \frac{2v}{L}$ into the kinetic energy equation: $KE = \frac{1}{2} \cdot \frac{1}{12} m L^2 \cdot \left(\frac{2v}{L}\right)^2$

This simplifies to: $KE = \frac{1}{2} \cdot \frac{1}{12} m L^2 \cdot \frac{4v^2}{L^2}$ $KE = \frac{1}{6} m v^2$

Now, solve for $v$: $v^2 = \frac{6 \cdot KE}{m}$ $v = \sqrt{\frac{6 \cdot KE}{m}}$

Step 5: Substitute Values and Calculate

$v = \sqrt{\frac{6 \cdot 1620}{101.56}}$ $v = \sqrt{\frac{9720}{101.56}}$ $v = \sqrt{95.73}$ $v \approx 9.78 \, \text{m/s}$

Final Answer

The linear velocity at either end of the spinning rod is approximately: $\boxed{9.78 \, \text{m/s}}$

Would you like a deeper explanation on any part?

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1. What is the angular velocity of the rod?
2. How would changing the mass of the rod affect the velocity?
3. How is kinetic energy distributed in rotational systems?
4. Why is the moment of inertia important in rotational motion?
5. How would this problem change if the axis was at one end?

Tip: Rotational kinetic energy depends on both the moment of inertia and angular speed, showing the influence of distribution of mass and rotational rate.