Combined rotational and translational kinetic energy

[e2m2a]

How would I determine the total kinetic energy of a long thin rod which is connected at one end to an axis and which rotates around this axis with a constant angular velocity, and the axis moves with a constant translational velocity? I am not sure how to go about this with this unsymmetric situation.

 

[Bob S]

Calculate the rotational energy about its center of mass, then use the Parallel Axis Theorem.

 

[astrokreng]

To determine the total kinetic energy of a long thin rod in this scenario, we can use the equation for combined rotational and translational kinetic energy:

KE = 1/2 * (m * v^2) + 1/2 * (I * ω^2)

Where:
- KE is the total kinetic energy
- m is the mass of the rod
- v is the translational velocity of the axis
- I is the moment of inertia of the rod
- ω is the angular velocity of the rod

To calculate the moment of inertia, we can use the formula for a thin rod rotating around one end:

I = 1/3 * m * L^2

Where L is the length of the rod.

Substituting this into the equation for combined kinetic energy, we get:

KE = 1/2 * (m * v^2) + 1/2 * (1/3 * m * L^2 * ω^2)

Since the axis is moving with a constant translational velocity, we can consider the translational kinetic energy to be constant. Therefore, the total kinetic energy of the rod will depend on the rotational kinetic energy, which is directly proportional to the square of the angular velocity.

In summary, to determine the total kinetic energy of a long thin rod connected to an axis that is rotating and translating at constant velocities, we can use the equation for combined rotational and translational kinetic energy, taking into account the moment of inertia for a thin rod rotating around one end.