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## Main Question or Discussion Point

Hi All,

I have a very simple question on a calculation, I would not doubt it correctness if it were not for the fact I am told by a colleague it is wrong, and I cannot see why (good job college times are gone).

We have a cylinder of lenght L, density $$\rho$$ constant along the cylinder, one end is clamped, one end is moved according to a function $$\delta(t)$$ describing the imposed displacement over time.

Each section is supposed to move as described by the function $$u = \frac{\delta}{L} x$$, u being the displacement.

Then the total kinetic energy equals

$$E = \frac{1}{2} \rho \int_{0}^{L} \dot{u(x)}^{2} \mathrm{d}x$$

yielding $$E = \frac{1}{6}\rho L \dot{\delta}^2$$, so 3 times less than the kinetic energy of a translation at speed $$\delta$$.

It seems all reasonable to me...maybe some rest will help..

Thanks!

I have a very simple question on a calculation, I would not doubt it correctness if it were not for the fact I am told by a colleague it is wrong, and I cannot see why (good job college times are gone).

We have a cylinder of lenght L, density $$\rho$$ constant along the cylinder, one end is clamped, one end is moved according to a function $$\delta(t)$$ describing the imposed displacement over time.

Each section is supposed to move as described by the function $$u = \frac{\delta}{L} x$$, u being the displacement.

Then the total kinetic energy equals

$$E = \frac{1}{2} \rho \int_{0}^{L} \dot{u(x)}^{2} \mathrm{d}x$$

yielding $$E = \frac{1}{6}\rho L \dot{\delta}^2$$, so 3 times less than the kinetic energy of a translation at speed $$\delta$$.

It seems all reasonable to me...maybe some rest will help..

Thanks!