- #1
muzialis
- 166
- 1
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 length 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 length 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!