Physics 1110 Fall 2004L39 - 1Lecture 39 24 November 2004Announcements• Reading Assignment for Mon, Nov. 29: Knight 14.1-14.5Rotational Kinetic EnergyNow that we have learned how to calculate angular accelerationusing the net torque, it is time to advance to our energy andmomentum techniques. In fact, the energy technique is quiteimportant, especially as applied to rolling motion.Let’s consider some object rotating about an axis. If we look at themotion of all the little pieces of mass that make up the object, theyare all following circular paths with some tangential speed vt (differentfor each piece of mass). Thus, to find the kinetic energy from thismotion, we just add up the regular kinetic energy from each piece I:Krot=12imivt,i2.But here we can use the fact that all of the pieces of mass have thesame angular velocity , so we can replace vt,I with ri . This givesthe following expression:Krot=12imiri22.Since the factors 1/2 and 2 are common to all terms, we can factorthem out to getKrot=122miri2i.At this point, we can see that the summation is in fact the definition ofthe moment of inertia I and we actually have a quite simple formula:Krot=12I2.We could have guessed this formula just by using the analogy withregular translational motion; this is 1/2 times “angular mass” timeangular velocity squared.In general now, when we calculate any kinetic energy, we need toinclude the translational kinetic energy plus the rotational. However,Physics 1110 Fall 2004L39 - 2when you actually apply this to problems, you have to be a littlecareful not to double count the motion of each piece of mass. Thetype of problems fall into two categories: fixed axis and free motion.1. Fixed axis – In these problems, the axle or pivot is fixed tosomething so that is cannot move. Even though the center ofmass of the whole object may well be able to move, it will stillbe a type of rotation, so we only use the rotational kineticenergy, not both rotational and translational. If we tried toinclude both we would essentially double count the contributionfrom each little piece of mass.2. Free motion – In this more general case, the object can bothtravel through space (translational motion) and rotate aboutsome axis, so we need to include both terms in the total kineticenergy formula: K =12mvcm2+12I2.In the case of free motion, there is no particular relation between thecenter of mass velocity and the angular velocity (for example, think ofa football thrown with some spin across a field). However there is aspecial case of free motion where there is a relation: rolling motion.If we have a circular object of radius r rolling along a surface withoutslipping then we know that the object (and hence its center of mass)moved by 2r every time it rotates one revolution. This is just ourusual looking relation (though it is actually quite different) in radians,s = r, and this means that we also have a relation between thevelocity of the center of mass and the angular velocity if we take thetime derivative of both sides of the equation:vcm= r .Angular Work and Angular Potential EnergyWe can also have energy transfer by work from forces; you canderive them by considering the circular motion, but here we’ll juststate the result, which you could guess by analogy: W = . The dotproduct, which is necessary for regular motion, is already taken careof in the definition of torque. You now have the tools to perform arelatively sophisticated energy analysis. For example, in our rollercoaster car problem, you could include the effect of the rotationalPhysics 1110 Fall 2004L39 - 3kinetic energy of the wheels, as well as the frictional energy loss atthe axle/wheel bearings.Angular potential energy is not very common, but there does exist aspecial class of springs, called torsional springs, which provide atwisting force (you may have seen a common type of clock, usuallybrass and inside a glass jar which has a bottom fixture with 4 ballsthat rotates back and forth). The torque from such a spring looks justlike Hooke’s law (though the constant is not a regular springconstant): = ˜ k (e). The potential energy from such a spring isalso very similar to a regular spring: U =12˜ k (e)2.Angular Momentum, “The Vector Returns”A few weeks ago we briefly looked at angular momentum L for aparticle of mass m on a circular path of radius r, where it can besimply calculated as rmvt. However, we now need to consider theangular momentum of more complicated systems and objects withmore complicated motion, so we need to calculate the full angularmomentum, which is actually a vector, not a scalar. (Recall regularmomentum is also a vector.)For an example just to think about, consider my favorite rotatingobject, a bicycle wheel, shown below.We will calculate a quantity called the angular momentum. If youremember regular momentum, it gave a measure of a quantity thatrequired a net force to change (think of a car rolling with somevelocity). In this case, we’re talking about a property of some objectthat is rotating and requires some torque to change. In the case ofregular momentum, the direction is just the direction of motion, whichseems so obvious that you never questioned it. But what direction willwork for a rotating object? Every piece of mass of the object has aslightly different direction of motion at some instant, and even forPhysics 1110 Fall 2004L39 - 4unchanging rotational motion, the direction of any piece of masswould change in the next instant.The only directions that are common to all the pieces of mass areactually parallel the axis of rotation! However, we still have to make achoice of which direction along the axis of rotation to take as thedirection of angular momentum. The choice is arbitrary but has someconsequences. By consensus everyone (sorry we forgot to ask you)has agreed to choose the direction of the angular momentumaccording to a rule, called the right hand rule. The right hand rule canbe stated several different ways. For rotation, the rule says to curvethe fingers of your right hand along the direction of rotation. Thethumb of your right hand now points in the direction of the angularmomentum. If we use the right hand rule to define the direction of theangular momentum, we must also use a right-handed coordinatesystem. To make a right-handed coordinate system we use a slightlydifferent version of the right hand rule:
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