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phy131studiof15:lectures:chapter10 [2015/09/24 20:55] mdawber [Momentum] |
phy131studiof15:lectures:chapter10 [2015/09/28 09:10] (current) mdawber [A thin uniform plate] |
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+ | ===== 10.P.011 ===== | ||

===== Center of mass ===== | ===== Center of mass ===== | ||

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$\vec{P}=M\vec{v}_{CM}$ | $\vec{P}=M\vec{v}_{CM}$ | ||

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+ | ===== 10.P.015 ===== | ||

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$\vec{x}_{CM}=\frac{1}{M}\int x\,dm=\frac{1}{M}\int_{0}^{l}\int_{0}^{w}\rho_{A}x\,dy\,dx=\frac{1}{M}\frac{1}{2}\rho_{A}l^{2}w=\frac{l}{2}\hat{i}$ | $\vec{x}_{CM}=\frac{1}{M}\int x\,dm=\frac{1}{M}\int_{0}^{l}\int_{0}^{w}\rho_{A}x\,dy\,dx=\frac{1}{M}\frac{1}{2}\rho_{A}l^{2}w=\frac{l}{2}\hat{i}$ | ||

- | $\vec{y}_{CM}=\frac{1}{M}\int y\,dm=\frac{1}{M}\int_{0}^{l}\int_{0}^{w}\rho_{A}y\,dy\,dx=\frac{1}{M}\frac{w}{2}\rho_{A}l^{2}l=\frac{w}{2}\hat{j}$ | + | $\vec{y}_{CM}=\frac{1}{M}\int y\,dm=\frac{1}{M}\int_{0}^{l}\int_{0}^{w}\rho_{A}y\,dy\,dx=\frac{1}{M}\frac{1}{2}\rho_{A}w^{2}l=\frac{w}{2}\hat{j}$ |

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The key to rocket propulsion is that the momentum of the expelled gas is equal and opposite to the forward momentum of the rocket. This does not rely on any interaction with the external atmosphere. | The key to rocket propulsion is that the momentum of the expelled gas is equal and opposite to the forward momentum of the rocket. This does not rely on any interaction with the external atmosphere. | ||

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+ | ===== 10.P.049 ===== | ||

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+ | ===== 10.P.045 ===== | ||

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