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 phy131studiof16:lectures:chapter12 [2016/07/21 12:07]127.0.0.1 external edit phy131studiof16:lectures:chapter12 [2016/10/25 10:20] (current)mdawber [Conditions for stable equilibrium] 2016/10/25 10:20 mdawber [Conditions for stable equilibrium] 2016/10/25 10:20 mdawber [Circus Trick] 2016/07/29 12:21 mdawber [14.P.042] 2016/07/29 12:20 mdawber [14.P.051] 2016/07/29 12:20 mdawber [14.P.030] 2016/07/29 12:20 mdawber [14.P.025] 2016/07/29 12:20 mdawber 2016/07/21 12:07 external edit Next revision Previous revision 2016/10/25 10:20 mdawber [Conditions for stable equilibrium] 2016/10/25 10:20 mdawber [Circus Trick] 2016/07/29 12:21 mdawber [14.P.042] 2016/07/29 12:20 mdawber [14.P.051] 2016/07/29 12:20 mdawber [14.P.030] 2016/07/29 12:20 mdawber [14.P.025] 2016/07/29 12:20 mdawber 2016/07/21 12:07 external edit Line 1: Line 1: ~~SLIDESHOW~~ ~~SLIDESHOW~~ - ====== Chapter 12 - Rotation I: Kinematics ​and Dynamics ​====== + ====== Chapter 12 - Static equilibrium,​ Elasticity ​and Fracture ​====== - ===== Polar coordinates for rotational motion ===== + Equilibrium describes a state where both the linear acceleration and angular acceleration of an object or system are zero. This means the object is either at rest or moving with constant velocity. - {{angulardisplacement.png}} + Here we will focus on situations where the system is a rest, so called static equilibrium,​ as opposed to dynamic equilibrium,​ where the system has constant velocity. - If we consider two points on a turning disc at distance $r_{1}$ and $r_{2}$ from the center of the disc we can make a number of observations. + ===== Conditions for Equilibrium ===== - First the distance traveled by the points is quite different, points that are far from the center go through a greater distance. + For an object to have no translational acceleration - However, the angular displacement is the same for both points, suggesting that angular displacement is a good variable to describe the motion of the whole rotating object. + $\Sigma \vec{F}=0$ - We would like to be able to have a straight forward relationship between ​the actual distance traveled and the angular displacement,​ which we can have if we express our angular displacement in [[wp>​Radian|radians]]. + and for this to be true the sum of **each component** of the forces must be zero. - As the relationship between the arc length ​on a circle and the angle which it subtends ​in radians is + An object may have no net force on it, but be able to turn, which means for an object to be in equilibrium - $l=r\theta$ + $\Sigma \tau = 0$ - a change in the angular displacement of $\Delta\theta$ results in a change in the displacement ​ + 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.) - $\Delta l=r\Delta\theta$ + ===== 2 Balls on a rod ===== - ===== Angular velocity===== + {{2ballsonarod.png}} - For change in angular displacement $\Delta \theta$ in a time interval ​$\Delta t$ we can define an average angular velocity + In the above example we calculate the torques around the pivot point. This is a sensible choice, but we could have also chosen, for example, an axis that goes through the second mass $m_{2}$. - $\bar{\omega}=\frac{\Delta \theta}{\Delta t}$ + In this case - As we are now accustomed, we can also define an instantaneous velocity + $\Sigma \tau=m_{1}g4L-({m_{1}+m_{2}})g3L=0$ - $\omega=\lim_{\Delta t\to 0}\frac{\Delta \theta}{\Delta t}=\frac{d\theta}{dt}$ + $m_{1}gL=m_{2}g3L$ - The units of angular velocity can be expressed as either ​$\mathrm{rad\,s^{-1}}$ or $\mathrm{s^{-1}}$ + $\large m_{2}=\frac{m_{1}}{3}$ - ===== Angular acceleration ===== - Previously we only considered circular motion in which the angular velocity $\omega$ remained constant ​with time, but of course we can also define an angular acceleration + ===== Crane boom with arm at 90 degrees ===== - $\bar{\alpha}=\frac{\Delta \omega}{\Delta t}$ - $\alpha=\lim_{\Delta t\to 0}\frac{\Delta \omega}{\Delta t}=\frac{d\omega}{dt}$ + There are four forces which act on the arm. Forces do not need to be directed along the arm as it is a rigid object. On the other hand, the force due to tension ​ **must** be directed along the rope (otherwise it would change it's shape). To take the weight of the arm in to account we can treat the weight of the rod as if it acts through the center of mass of the object, which we will approximate as being halfway along the arm. - The units of angular acceleration can be expressed as either $\mathrm{rad\,s^{-2}}$ or $\mathrm{s^{-2}}$ + {{boomarmwitharmweigt.png}} - ===== 12.P.008 ===== - ===== Equations ​of motion for rotational motion ===== + Sum of forces on arm - At the beginning of the course we derived, using calculus, a set of equations for motion under constant acceleration. + $F_{hx}=F_{Tx}$ - $v= v_{0}+at$ + $mg+m_{a}g=F_{Ty}+F_{hy}$ - $x= x_{0}+v_{0}t+\frac{1}{2}at^2$ + Sum of torques on arm around hinge - $v^{2}=v_{0}^2+2a(x-x_{0})$ + $F_{Ty}l=F_{T}l\sin\theta=mgl+m_{a}g\frac{l}{2}$ → $F_{T}\sin\theta=mg+\frac{1}{2}m_{a}g$ - 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. + $F_{T}=\frac{mg+\frac{1}{2}m_{a}g}{\sin\theta}$ - $\omega= \omega_{0}+\alpha t$ + Compressive force on arm is $F_{hx}=F_{Tx}=(mg+\frac{1}{2}m_{a}g)\frac{{\cos\theta}}{\sin\theta}=\frac{mg+\frac{1}{2}m_{a}g}{\tan\theta}$ - $\theta= \theta_{0}+\omega_{0}t+\frac{1}{2}\alpha t^2$ + ===== Crane boom with rope at 90 degrees ===== - $\omega^{2}=\omega_{0}^2+2\alpha(\theta-\theta_{0})$ + {{boomarmrope90.png}} - ===== 12.P.005 ===== + Forces on arm - ===== 12.P.026 ===== + - ===== From angular to tangential quantities ===== + $F_{hy}=mg+m_{a}g$ + $F_{hx}=F_{Tx}$ - A point at distance $r$ from the center of rotation will have at any time a tangential velocity of magnitude ​ + Torques on arm - $v=\omega r$ + $F_{T}l\sin\phi=(mgl+m_{a}g\frac{l}{2})\sin(90^{o}-\phi)=(mg+\frac{1}{2}m_{a}{g})l\cos\phi$ - and a tangential acceleration (not to be confused with the centripetal acceleration) of + $F_{T}=F_{Tx}=F_{hx}=\frac{mg+\frac{1}{2}m_{a}{g}}{\tan\phi}$ - $a=\alpha r$ + Compressive force on spring - ===== 12.P.038 ===== + $\frac{F_{hx}}{\cos\phi}=\frac{mg+\frac{1}{2}m_{a}{g}}{\sin\phi}$ - ===== Useful relationships concerning the angular velocity ​===== + ===== Crane boom with arm and rope at angle ===== - Rotational motion, when not accelerated,​ can be considered to be a form of periodic motion, and so relationships between the angular velocity, frequency and period are useful. + {{boomarmgeneral.png}} - $T=\frac{2\pi}{\omega}$ + Forces on arm - $f=\frac{\omega}{2\pi}$ + $F_{hy}+F_{Ty}=F_{hy}+F_{T}\sin\theta=mg+m_{a}g$ - $\omega=2\pi f$ + $F_{hx}=F_{Tx}=F_{T}\cos\theta$ - We'd also like to be able to express the centripetal acceleration in terms of $\omega$ + Torques on arm - $a_{R}=\frac{v^{2}}{r}=\frac{(\omega r)^{2}}{r}=\omega^{2}r$ + $F_{T}l\sin(\phi+\theta)=(mgl+m_{a}g\frac{l}{2})\sin(90^{o}-\phi)=(mg+\frac{1}{2}m_{a}{g})l\cos\phi$ - ===== About that acceleration.. ===== + $F_{T}=\frac{(mg+\frac{1}{2}m_{a}{g})\cos\phi}{\sin(\phi+\theta)}$ - {{totalrotacc.png}} + $F_{Tx}=\frac{(mg+\frac{1}{2}m_{a}{g})\cos\phi\cos\theta}{\sin(\phi+\theta)}=F_{hx}$ - The total acceleration of an object in accelerated rotational motion will be the vector sum of two perpendicular vectors, the tangential acceleration $\vec{a}_{tang}$,​ and the radial acceleration $\vec{a}_{R}$. + Compressive force on spring - The magnitude of the total acceleration is + $\frac{F_{hx}}{\cos\phi}=\frac{(mg+\frac{1}{2}m_{a}{g})\cos\theta}{\sin(\phi+\theta)}$ - $\vec{a}=\sqrt{a_{tang}^2+a_{R}^2}=\sqrt{\alpha^2r^2+\omega^{4}r^2}=r\sqrt{\alpha^2+\omega^4}$ - and it is directed an angle - $\arctan\frac{\alpha}{\omega^{2}}$ away from the radial direction. + ===== Hanging sign problem ===== - ===== 12.P.031 ===== + {{f11finalq3fig.png}} + After numerous unfortunate incidents involving polar bears and tourists who have seen too many Coke<​sup>​TM​ ads, Santa has decided to install a new sign at the North Pole. To do this he uses a uniform metal rod of length 0.8 m and a cable which he attaches 20 cm from the end of the metal rod. The new sign has mass 3 kg and hangs from the end of the rod, while the rod itself has mass 1 kg. The cable is attached the top of the pole and the rod is held by a hinge 30 cm down from the top of the pole. - ===== Pseudovector representation of angular velocity and acceleration ===== + What angle θ does the cable make with the rod? - As we know, translational velocity and acceleration are vector quantities. While we have defined angular velocity and acceleration we can see that they can represent points that are moving ​in directions that change with time, which would make them difficult to represent with vectors defined in cartesian coordinates. + What is the tension ​in the cable? - We can however consider an axial vector, ie. a vector that points in the direction about which rotation occurs as a good way of representing ​the direction of these rotational quantities. + What are the horizontal and vertical components ​of the force exerted by the hinge on the rod? Take the positive directions to be up and to the right. - We give the direction of rotation around the axes according to a [[wp>​Right-hand_rule|right-hand rule]]. + ===== Hanging sign solution ===== - {{rhgriprule.png}} + {{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. - ===== Combining translation motion with rotational motion - rolling ===== + $Fh_{x}=102.24\cos26.57^{o}=91.4\mathrm{N}$ - /​*<​html>​ + -