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Question Number 142401 by mohammad17 last updated on 31/May/21

Commented by mohammad17 last updated on 31/May/21

$$helpmesirplease$$

Answered by mr W last updated on 31/May/21

$$Q1$$$$pressureatr:p\left(r\right)$$$$r=250mm=0.25m$$$$r_0=r-t=250-120=130mm=0.13m$$$$dp=\rho r{\omega }^{2}dr$$$${\int }_0^{p}dp={\int }_{r_0}^r\rho r{\omega }^{2}dr$$$$\Rightarrow p=\frac{\rho {\omega }^2(r^2-r_0^2)}{2}$$$$=\frac{1}{2}\times 2200\times {160}^2\times (0.{25}^2-0.{13}^2)$$$$=1284096Pa\approx 1.284MPa$$

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