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phy131studiof15:lectures:m2review1 [2015/10/22 09:47] mdawber [Vector cross product] |
phy131studiof15:lectures:m2review1 [2015/10/23 09:17] (current) mdawber [Ballistic Pendulum] |
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$\frac{1}{2}m_{A}^{2}v_{A}+\frac{1}{2}m_{B}v^{2}_{B}=\frac{1}{2}m_{A}v'^{2}_{A}+\frac{1}{2}m_{B}v'^{2}_{B}$ | $\frac{1}{2}m_{A}^{2}v_{A}+\frac{1}{2}m_{B}v^{2}_{B}=\frac{1}{2}m_{A}v'^{2}_{A}+\frac{1}{2}m_{B}v'^{2}_{B}$ | ||

+ | |||

+ | ===== 1D elastic collisions with one body initially at rest===== | ||

+ | |||

+ | $v'_{B}=\frac{2m_{A}}{m_{A}+m_{B}}v_{A}$<html>        </html> $v'_{A}=\frac{m_{A}-m_{B}}{m_{A}+m_{B}}v_{A}$ | ||

+ | |||

+ | |||

+ | $m_{A}=m_{B}$ | ||

+ | | ||

+ | {{elastic1.gif}} | ||

+ | |||

+ | $m_{A}>m_{B}$ | ||

+ | |||

+ | {{elastic2.gif}} | ||

+ | |||

+ | $m_{B}>m_{A}$ | ||

+ | |||

+ | {{elastic_3.gif}} | ||

+ | |||

+ | ===== 2D elastic collisions with one body initially at rest ===== | ||

+ | |||

+ | | Conservation of momentum<html>     </html> \\ $v_{A}=v'_{Ax}+v'_{Bx}$ \\ $0=v'_{Ay}+v'_{By}$ |Conservation of energy \\ $v_{A}^2=v'^{2}_{A}+v'^{2}_{B}$ | | ||

+ | |||

+ | {{snookerballs.png}} | ||

+ | |||

+ | |$v'_{Bx}=v'_{B}\cos\theta$ <html>     </html>\\ $v'_{By}=v'_{B}\sin\theta$ | $v'_{Ax}=v'_{A}\sin\theta$ \\ $v'_{Ay}=v'_{A}\cos\theta$ | | ||

+ | |||

+ | |||

+ | |||

+ | Square the conservation of momentum equations | ||

+ | |||

+ | $v'^{2}_{Ay}=v'^{2}_{By}$ | ||

+ | |||

+ | $v_{A}^2-2v_{A}v'_{Ax}+v'^{2}_{Ax}=v'^{2}_{Bx}$ | ||

+ | |||

+ | Add these together | ||

+ | |||

+ | $v_{A}^2+v'^{2}_{A}-2v_{A}v'_{Ax}=v'^{2}_{B}$ | ||

+ | |||

+ | Use kinetic energy equation to eliminate $v'_{B}$ | ||

+ | |||

+ | $v_{A}^2+v'^{2}_{A}-2v_{A}v'_{Ax}=v_{A}^{2}-v'^{2}_{A}$ → $v'^{2}_{A}=v_{A}v'_{Ax}$ | ||

+ | |||

+ | $v'^{2}_{A}=v_{A}v'_{A}\sin\theta$ → $v'_{A}=v_{A}\sin\theta$ | ||

+ | |||

+ | and conservation of energy equation gives us $v'_{B}=v_{A}\cos\theta$ | ||

+ | ===== Ballistic Pendulum (1D inelastic collision) ===== | ||

+ | |||

+ | |||

+ | We can use conservation of momentum for the collision and conservation of energy for the **subsequent motion** in a [[wp>Ballistic_pendulum|ballistic pendulum]] to [[https://www.youtube.com/watch?v=MdwVrrnRaCE|find the velocity of a bullet]] fired in to a block from the height that the bullet and block rises to. | ||

+ | |||

+ | {{slide10.jpg}} | ||

+ | |||

+ | $mv=(M+m)v'$ | ||

+ | |||

+ | $\frac{1}{2}mv^2\neq\frac{1}{2}(M+m)v'^{2}$ | ||

+ | |||

+ | but | ||

+ | |||

+ | $\frac{1}{2}(M+m)v'^{2}=(M+m)gh$ → $\frac{1}{2}v'^{2}=gh$ | ||

+ | |||

+ | so | ||

+ | |||

+ | $v=\frac{M+m}{m}\sqrt{2gh}$ | ||

+ | |||

+ | ===== Perfectly Inelastic collisions in 2 dimensions ===== | ||

+ | |||

+ | {{inelasticcollision.png}} | ||

+ | |||

+ | In a perfectly inelastic collision the two object stick together after the collision so conservation of momentum now is written | ||

+ | |||

+ | $m_{A}\vec{v_{A}}+m_{B}\vec{v_{B}}=(m_{A}+m_{B})\vec{v}'$ | ||

===== Extended objects and sets of objects ===== | ===== Extended objects and sets of objects ===== | ||

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+ | ===== Atwood machine with rotation ===== | ||

+ | | {{atwoodwithr.png}} | We can apply our new knowledge about moment of inertia to our old friend the Atwood machine.If we take in to account the mass $M$ and radius $R$ of the pulley the tensions in the ropes on either side of the pulley need not be the same. | | ||

+ | |||

+ | The sum of the torques on the pulley will be given by | ||

+ | |||

+ | $\Sigma \tau =(T_{2}-T_{1})R$ | ||

+ | |||

+ | and as we saw that the moment of inertia for solid cylinder is $I=\frac{1}{2}MR^2$ we can find the angular acceleration of the pulley | ||

+ | |||

+ | $\large \alpha=\frac{\Sigma \tau}{I}=\frac{2(T_{2}-T_{1})}{MR}$ | ||

+ | |||

+ | This can be related to the tangential acceleration of a point on the edge of the pulley by multiplying by R as $a=\alpha R$ | ||

+ | |||

+ | $\large a=\frac{\Sigma \tau}{I}=\frac{2(T_{2}-T_{1})}{M}$ | ||

+ | |||

+ | As we did before when we neglected rotation we should write Newton's Second Law for the two weights | ||

+ | |||

+ | $m_{2}a=m_{2}g-T_{2}$ → $T_{2}=m_{2}g-m_{2}a$\\ | ||

+ | $m_{1}a=T_{1}-m_{1}g$ → $ T_{1}=m_{1}g+m_{1}a$\\ \\ | ||

+ | |||

+ | $T_{2}-T_{1}=m_{2}g-m_{2}a-m_{1}g-m_{1}a$ | ||

+ | |||

+ | |||

+ | $\frac{1}{2}Ma=m_{2}g-m_{2}a-m_{1}g-m_{1}a$ | ||

+ | |||

+ | $\large a=g\frac{m_{2}-m_{1}}{\frac{1}{2}M+m_{1}+m_{2}}$ | ||

===== Angular Momentum ===== | ===== Angular Momentum ===== | ||

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For any problem I will ask you you can use $L=I\omega$ | For any problem I will ask you you can use $L=I\omega$ | ||

+ | |||

+ | ===== Conservation of angular momementum ===== | ||

+ | |||

+ | In the absence of a net external torque | ||

+ | |||

+ | $\Sigma \tau=\frac{dL}{dt}=0$ | ||

+ | |||

+ | and angular momentum is conserved. | ||

+ | |||

+ | $L=I\omega=\mathrm{constant}$ | ||

+ | |||

+ | ===== Changing the moment of inertia of a spinning object ===== | ||

+ | |||

+ | Suppose I with two weights in my hands can be approximated by an 80 kg cylinder of radius 15 cm. My moment of inertia if I am spinning around an axis going down my center will be | ||

+ | |||

+ | $I=\frac{1}{2}MR^{2}=0.9\,\mathrm{kgm^2s^{-1}}$ | ||

+ | |||

+ | With my arms (which we approximate as 3.5 kg and a length of 0.75 m from my shoulder) extended holding 2.3 kg weights my moment of inertia will be considerably higher. When I am holding my arms and weights out I should remove their mass from the cylinder | ||

+ | |||

+ | $I=\frac{1}{2}68.4\times0.15^2=0.77\,\mathrm{kgm^2s^{-1}}$ | ||

+ | |||

+ | The moment of inertia of the two weights around the axis when my arms are extended is | ||

+ | |||

+ | $I=2\times2.3\times(0.9)^{2}=3.726\,\mathrm{kgm^2s^{-1}}$ | ||

+ | |||

+ | The [[wp>List_of_moments_of_inertia|moment of inertia]] of an arm if it is rotated around it's center of mass is | ||

+ | |||

+ | $I=\frac{1}{12}\times3.5\times(0.75)^2=0.16\,\mathrm{kgm^2s^{-1}}$ | ||

+ | |||

+ | But if it rotates around a point $0.375+0.15\,\mathrm{m}$ from it's center of mass then its moment of inertia is | ||

+ | |||

+ | $I=\frac{1}{2}\times3.5\times(0.75/2)^2+3.5\times(0.375+0.15)^2$\\ | ||

+ | $=0.16+0.96=1.12\,\mathrm{kgm^2s^{-1}}$ | ||

+ | |||

+ | So the total moment of inertia due to the extended arms, weights and the cylinder is | ||

+ | |||

+ | $I=0.77+3.726+2.24=6.736\,\mathrm{kgm^2s^{-1}}$ | ||

+ | |||

+ | |||

+ | ===== Consequence of conservation of angular momentum ===== | ||

+ | |||

+ | From the previous calculation we have | ||

+ | |||

+ | With arms in $I=0.9\,\mathrm{kgm^2s^{-1}}$ | ||

+ | |||

+ | With arms out $I=6.736\,\mathrm{kgm^2s^{-1}}$ | ||

+ | |||

+ | If I am rotating with angular velocity $\omega$ with my arms out and I bring them in then my angular velocity afterward is $\omega'$ and from conservation of momentum | ||

+ | |||

+ | $L=I\omega=I'\omega'$ | ||

+ | |||

+ | $\frac{\omega'}{\omega}=\frac{I}{I'}=\frac{6.916}{0.9}=7.5$ | ||

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Where the torques can be calculated around **any** axis. Some axes are more sensible than others, a good idea is to choose an axis at place where a force is acting (as the torque due to that force around that axis will be zero.) | Where the torques can be calculated around **any** axis. Some axes are more sensible than others, a good idea is to choose an axis at place where a force is acting (as the torque due to that force around that axis will be zero.) | ||

+ | ===== Hanging sign ===== | ||

+ | |||

+ | {{f11finalq3fig.png}} | ||

+ | |||

+ | $\theta=\tan^{-1}\frac{30}{60}=26.57^{o}$ | ||

+ | |||

+ | Vertical forces, up is positive $T\sin\theta+Fh_{y}=m_{sign}g+m_{rod}g$ | ||

+ | |||

+ | Horizontal forces, right is positive $-T\cos\theta+Fh_{x}=0$ | ||

+ | |||

+ | Torques $0.6T\sin\theta=0.8m_{sign}g+0.4m_{rod}g$ | ||

+ | |||

+ | $T=\frac{0.8\times3\times9.8+0.4\times1\times9.8}{0.6\times\sin26.57^{o}}=102.24\mathrm{N}$ | ||

+ | |||

+ | $Fh_{y}=4.981-102.24\sin26.57^{o}=-6.53\mathrm{N}$ | ||

+ | |||

+ | So force points down. | ||

+ | |||

+ | $Fh_{x}=102.24\cos26.57^{o}=91.4\mathrm{N}$ | ||

+ | |||

+ | Force points to the right | ||

+ | ===== Will the ladder slip? ===== | ||

+ | |||

+ | {{ladderslip.png}} | ||

+ | |||

+ | The forces that act on the ladder of length $l$ are the weight of the ladder that acts down at it's center of mass, $m\vec{g}$, a normal force exerted by the wall, $\vec{f}_{NW}$, an normal force exerted by the ground $\vec{f}_{NG}$, and a force due to the friction of the ground, $\vec{F}_{Fr}$. The friction between the ladder and the wall is typically neglected. | ||

+ | |||

+ | For equilibrium the sum of the forces in both the horizontal and vertical directions must be zero and also the sum of the torques around an appropriate axis must be equal to zero. | ||

+ | |||

+ | Balance of the forces in the horizontal direction implies that | ||

+ | |||

+ | $F_{NW}=F_{Fr}$ | ||

+ | |||

+ | and in the vertical direction | ||

+ | |||

+ | $F_{NG}=m_{l}g$ | ||

+ | |||

+ | If we calculate the torques around the point at which the ladder touches the ground then they will be | ||

+ | |||

+ | $F_{Nw}l\sin\theta-m_{l}g\frac{l}{2} \sin(90^{o}-\theta)=0$ | ||

+ | |||

+ | $F_{Nw}l\sin\theta-m_{l}g\frac{l}{2}\cos\theta=0$ | ||

+ | |||

+ | This can be written in terms of the frictional force by using the equation we got from the horizontal forces. | ||

+ | |||

+ | $F_{fr}l\sin\theta-m_{l}g\frac{l}{2}\cos\theta=0$ | ||

+ | |||

+ | The maximum force that can be provided by friction is $\mu F_{NG}$ and so the maximum safe angle is when this frictional force is provided | ||

+ | |||

+ | $\mu F_{NG}l\sin\theta_{max}-m_{l}g\frac{l}{2}\cos\theta_{max}=0$ | ||

+ | |||

+ | $\mu mgl\sin\theta_{max}-m_{l}g\frac{l}{2}\cos\theta_{max}=0$ | ||

+ | |||

+ | $\tan\theta_{max}=\frac{1}{2\mu}$ | ||

+ | |||

+ | Of course if a person is standing on the ladder, their weight would need to be taken in to account both in the balance of the vertical forces and the torques. |