Ćwiczenia 04.05.09

$$\overrightarrow{M_{m}} = \frac{\sum_{}^{}{m_{i}\overrightarrow{r_{i}}}}{\sum_{}^{}m_{i}}\ = \frac{\int_{m}^{}{r_{i}\text{dm}}}{\int_{m}^{}\text{dm}}$$

I = ∫_Mr² dm

I = I_M + I_m

Liczymy po wsp. kreskowych

$$x_{s_{m}}^{m} = \ \frac{\int_{- 1}^{0}{\frac{m}{l}\text{dx}}}{m} = \ \int_{- 1}^{0}{x \bullet \frac{1}{l}dx = \frac{1}{l}\int_{- 1}^{0}{x\ dx = \frac{1}{l}\frac{x^{2}}{2}|\begin{matrix} 0 \\ - 1 \\ \end{matrix} = \frac{1}{l}\frac{x^{2}}{2}}}|\begin{matrix} 0 \\ - 1 \\ \end{matrix} = \frac{1}{l}0 - \frac{1}{l}\frac{l^{2}}{2} = \ - \frac{l}{2}$$

y_sm^m = 0

x_sm^M = −l

$$y_{s_{m}}^{M} = - \frac{l}{2}$$

$$x_{s_{m}}^{} = \frac{m\left( - \frac{l}{2} \right) + M( - l)}{m + M} = \frac{- l(\frac{m}{2} + M)}{m + M} = \ - \frac{l}{2}$$

$$y_{s_{m}}^{} = \frac{m0 + M( - \frac{l}{2})}{m + M}$$

I =  I_o + b²

I = I₁ + I₂

$$I_{1} = \frac{{m_{1}l_{1}}^{2}}{12} + m_{1}\left( \frac{l_{1}}{2} \right)^{2} = \frac{m_{1}l_{1}^{2}}{3}$$

$$I_{2} = \frac{{m_{2}l_{2}}^{2}}{12} + m_{2}({l_{1}}^{2} + \left( \frac{l_{2}}{2} \right)^{2})$$

$$I = \frac{m_{1}l_{1}^{2}}{3} + \frac{m_{2}l_{2}^{2}}{12} + m_{2}\left( l_{1}^{2} + \left( \frac{l_{2}}{2} \right)^{2} \right) = \ \frac{m_{1}l_{1}^{2}}{3} + m_{2}l_{1}^{2} + \frac{m_{2}l_{2}^{2}}{3}$$

$$I = \sum_{}^{}{r_{i}}^{2}m_{i} = \int_{M}^{}{r^{2}\text{dm}}$$

Dla walca

I = ∫r²dm = ∫r²•2πrdrlρ=2πlρ∫₀^Rr³dr

dm = 2πr • dr • L • ρ

$$\rho = \frac{M}{\pi R^{2}l}$$