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phy131studiof15:lectures:chapter12 [2015/07/22 10:24] mdawber created |
phy131studiof15:lectures:chapter12 [2015/10/12 09:19] (current) mdawber [Equations of motion for rotational motion] |
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The units of angular acceleration can be expressed as either $\mathrm{rad\,s^{-2}}$ or $\mathrm{s^{-2}}$ | The units of angular acceleration can be expressed as either $\mathrm{rad\,s^{-2}}$ or $\mathrm{s^{-2}}$ | ||

+ | |||

+ | ===== 12.P.008 ===== | ||

+ | |||

+ | ===== Equations of motion for rotational motion ===== | ||

+ | |||

+ | At the beginning of the course we derived, using calculus, a set of equations for motion under constant acceleration. | ||

+ | |||

+ | $v= v_{0}+at$ | ||

+ | |||

+ | $x= x_{0}+v_{0}t+\frac{1}{2}at^2$ | ||

+ | |||

+ | $v^{2}=v_{0}^2+2a(x-x_{0})$ | ||

+ | |||

+ | We can equally derive similar equations for our rotational quantities. Indeed as we can see that the relationships between the new rotational quantities we have now are exactly the same as those between the translational quantities we can simply rewrite the translational motion equations in terms of rotational variables. | ||

+ | |||

+ | $\omega= \omega_{0}+\alpha t$ | ||

+ | |||

+ | $\theta= \theta_{0}+\omega_{0}t+\frac{1}{2}\alpha t^2$ | ||

+ | |||

+ | $\omega^{2}=\omega_{0}^2+2\alpha(\theta-\theta_{0})$ | ||

+ | |||

+ | ===== 12.P.005 ===== | ||

+ | ===== 12.P.026 ===== | ||

===== From angular to tangential quantities ===== | ===== From angular to tangential quantities ===== | ||

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$a=\alpha r$ | $a=\alpha r$ | ||

+ | |||

+ | ===== 12.P.038 ===== | ||

===== Useful relationships concerning the angular velocity ===== | ===== Useful relationships concerning the angular velocity ===== | ||

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$\arctan\frac{\alpha}{\omega^{2}}$ away from the radial direction. | $\arctan\frac{\alpha}{\omega^{2}}$ away from the radial direction. | ||

+ | |||

+ | ===== 12.P.031 ===== | ||

+ | |||

===== Pseudovector representation of angular velocity and acceleration ===== | ===== Pseudovector representation of angular velocity and acceleration ===== | ||

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- | ===== Equations of motion for rotational motion ===== | ||

- | At the beginning of the course we derived, using calculus, a set of equations for motion under constant acceleration. | ||

- | $v= v_{0}+at$ | ||

- | $x= x_{0}+v_{0}t+\frac{1}{2}at^2$ | ||

- | $v^{2}=v_{0}^2+2a(x-x_{0})$ | ||

- | |||

- | We can equally derive similar equations for our rotational quantities. Indeed as we can see that the relationships between the new rotational quantities we have now are exactly the same as those between the translational quantities we can simply rewrite the translational motion equations in terms of rotational variables. | ||

- | |||

- | $\omega= \omega_{0}+\alpha t$ | ||

- | |||

- | $\theta= \theta_{0}+\omega_{0}t+\frac{1}{2}\alpha t^2$ | ||

- | |||

- | $\omega^{2}=\omega_{0}^2+2\alpha(\theta-\theta_{0})$ | ||

===== Combining translation motion with rotational motion - rolling ===== | ===== Combining translation motion with rotational motion - rolling ===== | ||

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The units of torque are $\mathrm{Nm}$ | The units of torque are $\mathrm{Nm}$ | ||

+ | |||

+ | ===== 12.P.041 ===== | ||

+ | |||

===== Production of Torque in an engine ===== | ===== Production of Torque in an engine ===== | ||

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$\frac{d}{dt}(\vec{A}\times\vec{B})=\frac{d\vec{A}}{dt}\times\vec{B}+\vec{A}\times\frac{d\vec{B}}{dt}$ | $\frac{d}{dt}(\vec{A}\times\vec{B})=\frac{d\vec{A}}{dt}\times\vec{B}+\vec{A}\times\frac{d\vec{B}}{dt}$ | ||

+ | |||

+ | |||

+ | ===== 12.P.047 ===== | ||

+ | |||

+ | ===== 12.P.048 ===== | ||

===== Acceleration due to torque ===== | ===== Acceleration due to torque ===== | ||

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We should note that the distance $R$ is the distance of each mass with respect to the axis of rotation, which need not be the center of mass, and is not an intrinsic property of an object! | We should note that the distance $R$ is the distance of each mass with respect to the axis of rotation, which need not be the center of mass, and is not an intrinsic property of an object! | ||

+ | |||

+ | ===== 12.P.053 ===== | ||

+ | |||

===== Two weights on a thin bar ===== | ===== Two weights on a thin bar ===== | ||

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The axis of rotation coincides with the center of mass only if $m_{1}R_{1}=m_{2}R_{2}$ | The axis of rotation coincides with the center of mass only if $m_{1}R_{1}=m_{2}R_{2}$ | ||

- | |||

- | ===== Moment of inertia for extended objects ===== | ||

- | |||

- | So far our definition of moment of inertia is really only practical for systems comprised of one or more point like objects at a distance from an axis of rotation. | ||

- | |||

- | For extended objects we are much better served by considering an object as being made up of infinitesimally small mass elements, each a distance $R$ from the axis of rotation, and integrating over these mass elements to find the moment of inertia. | ||

- | |||

- | $I= \Sigma_{i} m_{i}R_{i}^{2}$ โ $I=\int R^2\,dm$ | ||

- | |||

- | ===== Hoop, thin walled cylinder and solid cylinder ===== | ||

- | |||

- | {{hoopcyl.png}} | ||

- | |||

- | Considering rotation axis through the center of the circle | ||

- | |||

- | $I=\int R^2\,dm$ | ||

- | |||

- | Hoop | ||

- | |||

- | $I=\frac{MR^2}{2\pi}\int_{0}^{2\pi}\,d\theta=MR^2$ | ||

- | |||

- | Thin walled cylinder | ||

- | |||

- | $I=\frac{MR^2}{2\pi h}\int_{0}^{h}\int_{0}^{2\pi}\,d\theta\,dz=MR^2$ | ||

- | |||

- | Solid Cylinder | ||

- | |||

- | $I=\frac{M}{\pi h R_{C}^2}\int_{0}^{R_{C}}\int_{0}^{h}\int_{0}^{2\pi}R^3\,d\theta\,dz\,dR=\frac{M}{\pi h R_{C}^2}\frac{ 2\pi h R_{C}^4}{4}=\frac{1}{2}MR_{C}^{2}$ | ||

- | |||

- | |||

- | |||

- | |||

- | ===== Parallel axis theorem ===== | ||

- | |||

- | Usually a rotation axis that passes through the center of mass of an object will be one of the easiest to find the moment of inertia for, because as we saw in the last lecture the center of mass usually reflects the symmetry of the object. More moments of inertia on [[http://en.wikipedia.org/wiki/List_of_moments_of_inertia|wikipedia]]. | ||

- | |||

- | If we know the moment of inertia of an object around an axis that passes through it's center of mass there is a theorem that can help us find the moment of inertia around a different axis parallel to the axis through the COM. | ||

- | |||

- | If the axis of rotation is a distance $h$ from the axis through the COM then | ||

- | |||

- | $I=I_{COM}+Mh^{2}$ | ||

- | |||

- | where M is the total mass of the object. | ||

- | |||

- | | {{spherearoundpole.png}} | As an example, a solid sphere stuck to a turning pole will have moment of inertia\\ $I=\frac{2}{5}MR^{2}+MR^2=\frac{7}{5}MR^2$ | | ||

- | |||

- | ===== When can we approximate a mass as a point? ===== | ||

- | |||

- | {{pointmassmI.png}} | ||

- | |||

- | Frequently we will want to approximate a mass at some distance from it's center of rotation as a point mass. i.e we would like to simply write | ||

- | |||

- | $I=mR_{1}^{2}$ | ||

- | |||

- | The parallel axis theorem tells us that in fact (if the mass is spherical and solid) | ||

- | |||

- | $I=mR_{1}^{2}+\frac{2}{5}mR_{2}^{2}$ | ||

- | |||

- | The fractional error introduced by the above approximation is | ||

- | |||

- | $\frac{\frac{2}{5}mR_{2}^{2}}{mR_{1}^{2}+\frac{2}{5}mR_{2}^{2}}=\frac{\frac{2}{5}}{\frac{R_{1}^{2}}{R_{2}^2}+\frac{2}{5}}$ | ||

- | |||

- | If we want to be accurate to say 1% we only need $\frac{R_{1}}{R_{2}}\approx\sqrt{40}\approx 6$ | ||

- | |||

- | ===== 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 | ||

- | |||

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

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

- | |||

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

- | |||

- | |||

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

- | |||

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

- | |||

- | ===== Rotational Kinetic Energy ===== | ||

- | |||

- | Each piece of mass in a rotation problem that has velocity $v$ should have kinetic energy | ||

- | |||

- | $K=\frac{1}{2}mv^2$ | ||

- | |||

- | In terms of the angular velocity this is | ||

- | |||

- | $K=\frac{1}{2}m\omega^2R^2$ | ||

- | |||

- | and if we sum over all the masses | ||

- | |||

- | $K=\frac{1}{2}(\Sigma m_{i}R_{i}^2)\omega^{2}=\frac{1}{2}I\omega^2$ | ||

- | |||

- | |||

- | |||

- | ===== Conservation of energy with rotation ===== | ||

- | |||

- | For a rolling object $v=\omega r$ | ||

- | |||

- | The kinetic energy of a rolling object is therefore | ||

- | |||

- | $\large K=\frac{1}{2}mv^{2}+\frac{1}{2}I\omega^{2}=\frac{1}{2}mv^{2}+\frac{1}{2}I\frac{v^{2}}{r^2}$ | ||

- | |||

- | The kinetic energy thus depends on the moment of inertia of the object. | ||

- | |||

- | Suppose we release a hoop and disk from the top of a slope. They begin with the same potential energy, which one gets to the bottom of the slope first? | ||

- | |||

- | {{hoopdisk.png}} | ||

- | |||

- | ===== Hoop and disk solution ===== | ||

- | |||

- | Hoop | ||

- | |||

- | $\large K=\frac{1}{2}mv^{2}+\frac{1}{2}I\omega^{2}=\frac{1}{2}mv^{2}+\frac{1}{2}mr^{2}\frac{v^{2}}{r^2}=mv^{2}$ | ||

- | |||

- | $mgh=mv^{2}$ | ||

- | |||

- | $v=\sqrt{gh}$ | ||

- | |||

- | Disk | ||

- | |||

- | $\large K=\frac{1}{2}mv^{2}+\frac{1}{2}I\omega^{2}=\frac{1}{2}mv^{2}+\frac{1}{2}\frac{1}{2}mr^{2}\frac{v^{2}}{r^2}=\frac{3}{4}mv^{2}$ | ||

- | |||

- | $mgh=\frac{3}{4}mv^{2}$ | ||

- | |||

- | $v=\sqrt{\frac{4}{3}gh}$ | ||

- | |||

- | Recall that for a sliding object (without friction) $v=\sqrt{2gh}$ | ||

- | |||

- | Solid sphere $I=\frac{2}{5}mr^{2}$\\ | ||

- | Hollow sphere $I=\frac{2}{3}mr^{2}$ | ||

- | |||

- | ===== Work energy theorem for rotation ===== | ||

- | |||

- | $W=\int \vec{F}\cdot\,d\vec{l}=\int F_{\perp}R\,d\theta=\int_{\theta_{1}}^{\theta_{2}}\tau\,d\theta$ | ||

- | |||

- | $\tau=I\alpha=I\frac{d\omega}{dt}=I\frac{d\omega}{d\theta}\frac{d\theta}{dt}=I\omega\frac{d\omega}{d\theta}$ | ||

- | |||

- | $W=\int_{\omega_{1}}^{\omega_{2}}I\omega\,d\omega=\frac{1}{2}I\omega_{2}^2-\frac{1}{2}I\omega_{1}^2$ | ||

- | |||

- | Therefore the work done in rotating an object through an angle $\theta_{2}-\theta_{1}$ is equal to the change in the rotational kinetic energy of the object. | ||

- | |||

- | ===== Power and Torque ===== | ||

- | |||

- | |||

- | $W=\int_{\theta_{1}}^{\theta_{2}}\tau\,d\theta$ | ||

- | |||

- | $P=\frac{dW}{dt}=\tau\frac{d\theta}{dt}=\tau\omega$ | ||

- | |||

- | This equation can help us understand the two "figures of merit" often given for a car engine, horsepower and torque. | ||