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 phy131studiof15:lectures:chapter10 [2015/07/22 09:33]mdawber created phy131studiof15:lectures:chapter10 [2015/09/28 09:10] (current)mdawber [A thin uniform plate] 2015/09/28 09:10 mdawber [A thin uniform plate] 2015/09/24 21:10 mdawber [Rockets] 2015/09/24 21:00 mdawber [Center of mass] 2015/09/24 20:57 mdawber [10.P.005] 2015/09/24 20:55 mdawber [Momentum] 2015/07/22 09:37 mdawber [Momentum] 2015/07/22 09:33 mdawber created Next revision Previous revision 2015/09/28 09:10 mdawber [A thin uniform plate] 2015/09/24 21:10 mdawber [Rockets] 2015/09/24 21:00 mdawber [Center of mass] 2015/09/24 20:57 mdawber [10.P.005] 2015/09/24 20:55 mdawber [Momentum] 2015/07/22 09:37 mdawber [Momentum] 2015/07/22 09:33 mdawber created Line 20: Line 20: $F_{ave}=\frac{\Delta p}{\Delta t}=\frac{mv_{2}-mv_{1}}{\Delta t}$ $F_{ave}=\frac{\Delta p}{\Delta t}=\frac{mv_{2}-mv_{1}}{\Delta t}$ + + ===== 10.P.005 ===== + + + ===== 10.P.011 ===== + + ===== Center of mass ===== + + We can  consider the motion of any system of particles or extended mass to be composed of different kinds of motion with respect to the center of mass. + + In general the displacement vector for the center of mass of a system of particles $m_{i}$ can be written as + + $\vec{r}_{CM}=\frac{\Sigma_{i}m_{i}\vec{r_{i}}}{\Sigma_{i}m_{i}}=\frac{\Sigma_{i}m_{i}\vec{r_{i}}}{M}$ + + Differentiating with respect to time gives + + $M\frac{d\vec{r}_{CM}}{dt}=\Sigma_{i}m_{i}\frac{d\vec{r}}{dt}$ or  $M\vec{v_{CM}}=\Sigma_{i}m_{i}\vec{v}_{i}$ + + and doing it once more + + $M\vec{a_{CM}}=\Sigma_{i}m_{i}\vec{a}_{i}$ + + $\Sigma_{i}m_{i}\vec{a}_{i}=\Sigma_{i}F_{i}=\Sigma \vec{F}_{ext}$ + + so we have obtained a new form of Newton'​s Second Law that works for a system of particles. + + $M\vec{a_{CM}}=\Sigma \vec{F}_{ext}$ + + the total momentum of a system can also be written in these terms + + $\vec{P}=M\vec{v}_{CM}$ + + ===== 10.P.015 ===== + + + + ===== Center of mass for extended objects ===== + + For point like objects at given displacement the center of mass vector is + + $\vec{r}_{CM}=\frac{\Sigma_{i}m_{i}\vec{r_{i}}}{M}$ + + For solid objects it is more useful to consider the center of mass in integral form. + + $\vec{r}_{CM}=\frac{1}{M}\int \vec{r}\,​dm$ ​ + + which we should recall is the same as having a set of equations for each of the components, ie. + + $x_{CM}=\frac{1}{M}\int x\,dm$ , $y_{CM}=\frac{1}{M}\int y\,dm$ , $z_{CM}=\frac{1}{M}\int z\,​dm$ ​ + + ===== Mass Density ===== + + + + + For object'​s of uniform density we can express the mass element over which we integrate as a spatial element which in 3 dimensions is $dm=\rho\, dV$, in 2 dimensions is $dm=\rho_{A}\,​ dA$ in one dimension is $dm=\lambda\,​dx$,​ where $\rho$, $\rho_{A}$ and $\lambda$ are the density, areal density or linear density of the object we consider. + + In cases where we know the total mass and total size (either volume, area or length) of an object the density can be found by dividing the total mass by the total size. + + ===== A uniform rod ===== + + {{COMrod.png}} + + We can consider a uniform rod to be a one dimensional object. + + If we want to find the COM we could place the origin of our coordinate system at the center of the rod and then + + $\vec{r}_{CM}=\frac{1}{M}\int \vec{r}\,​dm=\frac{1}{M}\int_{-l/​2}^{l/​2}\lambda x\,​dx=\frac{1}{M}\frac{\lambda}{2}((\frac{l}{2})^2-(-\frac{l}{2})^2)=0$ + + + {{COMrodleft.png}} + + On the other hand if we were to place the origin at the left end of the rod then we could show that + + + $\vec{r}_{CM}=\frac{1}{M}\int \vec{r}\,​dm=\frac{1}{M}\int_{0}^{l}\lambda x\,​dx=\frac{1}{M}\frac{\lambda}{2}l^2$ + + and as $M=\lambda l$ + + $\vec{r}_{CM}=\frac{l}{2}\hat{i}$ + ===== A thin uniform plate ===== + + {{COMplate.png}} + + We now look at a 2 dimensional object, in this case a thin rectangular plate. + + The mass interval $dm=\rho_{A}\,​dA=\rho_{A}\,​dx\,​dy$ + + The COM of the plate in the x direction is + + $\vec{x}_{CM}=\frac{1}{M}\int x\,​dm=\frac{1}{M}\int_{-l/​2}^{l/​2}\int_{-w/​2}^{w/​2}\rho_{A}x\,​dy\,​dx=0$ + + $\vec{y}_{CM}=\frac{1}{M}\int y\,​dm=\frac{1}{M}\int_{-l/​2}^{l/​2}\int_{-w/​2}^{w/​2}\rho_{A}y\,​dy\,​dx=0$ + + We can see that symmetry often enables us to identify the center of mass of an object, equally distributed mass on either side of the origin cancels out. + + + {{COMplatebottomleft.png}} + + If we instead put the bottom left hand corner of the plate at the origin then + + + $\vec{x}_{CM}=\frac{1}{M}\int x\,​dm=\frac{1}{M}\int_{0}^{l}\int_{0}^{w}\rho_{A}x\,​dy\,​dx=\frac{1}{M}\frac{1}{2}\rho_{A}l^{2}w=\frac{l}{2}\hat{i}$ + + $\vec{y}_{CM}=\frac{1}{M}\int y\,​dm=\frac{1}{M}\int_{0}^{l}\int_{0}^{w}\rho_{A}y\,​dy\,​dx=\frac{1}{M}\frac{1}{2}\rho_{A}w^{2}l=\frac{w}{2}\hat{j}$ + + + ===== Center of mass of the human body ===== + + How symmetric do you think the human body is? Here's an [[http://​hypertextbook.com/​facts/​2006/​centerofmass.shtml|experiment to determine the center of mass of people]]. + + + + ===== Polar coordinates ===== + + {{polarcoords.png}} + + For circular objects and rotational motion we will find [[wp>​Polar_coordinate_system|polar coordinates]] to be advantageous. + + To transform from polar coordinates in to Cartesian coordinates + + $x=r\cos\theta$ + + $y=r\sin\theta$ + + and from Cartesian to polar + + $r=\sqrt{x^{2}+y^{2}}$ + + $\large\tan\theta=\frac{y}{x}$ + + ===== COM of a thin uniform disk ===== + + {{COMdisk.png}} + + $\vec{r}_{CM}=\frac{1}{M}\int \vec{r}\,​dm=\frac{1}{M}\int\vec{r}\rho_{A}\,​dA$ + + $dm=\rho_{A}\,​dA=\rho_{A}r\,​dr\,​d\theta$ + + $x=r\cos\theta$ + + $y=r\sin\theta$ + + $x_{CM}=\frac{1}{M}\int x\rho_{A} \, dA=\frac{\rho_{A}}{M}\int_0^{2\pi}\int_{0}^{R}r^2\cos\theta\,​dr\,​d\theta=\frac{\rho_{A} R^3}{3M}\int_0^{2\pi}\cos\theta\,​d\theta$ + $=\frac{\rho_{A} R^3}{3M}[\sin(2\pi)-\sin(0)]=0$ + + $y_{CM}=\frac{1}{M}\int y\rho_{A} \, dA=\frac{\rho_{A}}{M}\int_0^{2\pi}\int_{0}^{R} r^2\sin\theta\,​dr\,​d\theta=\frac{\rho_{A} R^3}{3M}\int_0^{2\pi}\sin\theta\,​d\theta$ + $=\frac{\rho_{A} R^3}{3M}[-\cos(2\pi)+\cos(0)]=0$ + + ===== COM of a thin uniform half-disk ===== + + {{COMhalfdisk.png}} + + $\vec{r}_{CM}=\frac{1}{M}\int \vec{r}\,​dm=\frac{1}{M}\int\vec{r}\rho_{A}\,​dA$ + + $x_{CM}=\frac{1}{M}\int x\rho_{A} \, dA=\frac{\rho_{A}}{M}\int_0^{\pi}\int_{0}^{R}r^2\cos\theta\,​dr\,​d\theta=\frac{\rho_{A} R^3}{3M}\int_0^{\pi}\cos\theta\,​d\theta$ + $=\frac{\rho_{A} R^3}{3M}[\sin(\pi)-\sin(0)]=0$ + + $y_{CM}=\frac{1}{M}\int y\rho_{A} \, dA=\frac{\rho_{A}}{M}\int_0^{\pi}\int_{0}^{R}r^2\sin\theta\,​dr\,​d\theta=\frac{\rho_{A} R^3}{3M}\int_0^{\pi}\sin\theta\,​d\theta$ + $=\frac{\rho_{A}R^3}{3M}[-\cos(\pi)+\cos(0)]=\frac{2\rho_{A}R^3}{3M}$ + + + We can convert this in to something more useful by considering that $M=\rho_{A}\frac{1}{2}\pi R^{2}$ + + which tells us that + + $y_{CM}=\frac{4R}{3\pi}$ ===== Conservation of Momentum ===== ===== Conservation of Momentum ===== Line 68: Line 235: 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. + + ===== 10.P.049 ===== + + ===== 10.P.045 ===== + + 