On A Roll! The Theoretical Background A. D. Andrew February 2005 Introduction In the first project this semester, you calculated the time required for a bead to slide down a wire connecting two points, and then tried to find a fast path between the points. The calculation of descent time began with the statement of conservation of energy —the potential energy lost as the bead slides is converted to kinetic energy. In this project you'll consider bodies rolling down an inclined plane. The calculations involve the use of triple integration to find the moment of inertia of a solid body. The project is comprised of three documents. The theoretical background is explained in this document. The Maple illustrations and exercises are in the Maple worksheet, On A Roll! Examples and Exercises, which is available from my web pages for this course. The Cover Sheet, which you should attach to your project when you submit it is also posted on my web pages. Part I. Kinetic Energy of a Rotating Body In these problems, the conservation of energy statement is slightly more complicated than in the case of the sliding beads -- part of the kinetic energy comes from 1 the translation of the body ( m v 2 , just as in the case of the sliding bead), and part of it 2 comes from the rotation of the body. Now this is especially interesting, since we'll see that the fraction of kinetic energy due to rotation depends on the geometry of the body. To begin, assume there are several objects at the top of a ramp, with the ramp making an angle with horizontal. The objects are a solid sphere, a hollow sphere (or a spherical shell), a solid cylinder, and a hollow cylinder. If they are released at the same time, in what order do they reach the bottom of the ramp? Here are two diagrams of this situation. 1 Figure 1 Objects on a Ramp To calculate the time required to roll down the ramp, we again equate potential energy (mgH) at the top of the ramp with kinetic energy at the bottom of the ramp. What we need to determine is the kinetic energy of a body rotating with angular velocity about an axis L. Suppose that a solid S, having mass density (x,y,z) is rotating about an axis L with angular velocity radians per second. Such a solid is pictured in Figure 2. To calculate its kinetic energy, note that we may approximate the body by a bunch of little 1 m v 2 , and we can volumes as shown. Each of these little volumes has kinetic energy 2 express this in terms of . The volume element dV shown below has mass dV, and it's moving in a circle of radius d , so it has velocity v = d. Thus the kinetic energy of the volume element dV is dE 1 ( d) 2 dV , 2 and the total kinetic energy of the rotating body is E 1 2 d 2 2 dV S 1 I 2 , 2 where I is the moment of inertia I d dV 2 . S Notice that d is the distance to the axis of rotation L. Thus if the axis of rotation is the 2 2 2 2 2 2 z-axis, then d x y and if the axis of rotation is the x -axis, d y z . The exact form of d will depend on the coordinate system you are using and the location of the axis. 2 Figure 2 A rotating body Note the similarities between the expressions 1 1 m v 2 and I 2. 2 2 Part II. Calculation of the Moment of Inertia Here is an example of the calculation of a moment of inertia. We consider a cylinder of uniform density with base radius a and height h rotating about its axis, as shown below. 3 We'll calculate the moment of inertia using cylindrical coordinates ( r, ,z) so the distance d to the axis is r and dV = r dr d dz. Thus 2 I h a r dr dz d 3 0 z 0 r 0 m (m is the total mass of the cylinder and V is its volume). This V is worked out easily by hand, and we illustrate the calculation with Maple on the worksheet for this project. where is the density 4 Part III. Calculation of the Descent Time Suppose the height of the ramp is H, and that the angle between the ramp and the horizontal is . Suppose further that a body with radius a, after starting at rest from the top of the ramp, has rolled part way down the ramp, achieving a velocity v after having descended a vertical distance y. Figure 3 It has lost potential energy mgy , and it has total kinetic energy 1 1 mv 2 I 2. 2 2 If the object is rolling without slipping, then v and are related by v a . v We may then replace by , and equate the potential energy lost with the kinetic a energy. That is, 2 1 1 v m g y m v 2 I 2 2 a I 2 2 g y v 1 . m a 2 I is called the normalized moment of inertia, and it's the parameter m a2 I to keep track of. Let's denote it by J = , and see how it determines the descent time. m a2 We now see that The ratio v2 2gy , 1 J and since the vertical component of velocity is v sin(), we obtain 5 dy v sin( ) dt 2g y sin( ) . 1 J Separating variables yields y 12 2g sin( ) dt . 1 J dy Finally, denoting the descent time by T, we integrate this expression to obtain H y 12 T dy 0 0 2g sin( )dt . 1 J Thus 2 H T 2g sin( )T , and solving for T: 1 J 2H (1 J) . 2 gsin ( ) Let's meditate on this formula for descent time. First, if the ramp height H is increased, T is increased. Second, if the angle of elevation is small, the descent time is large. Third, if the acceleration of gravity g is increased (say by rolling objects on Jupiter) the descent time T is decreased. All of these features of the descent time agree with our intuition. Finally, note the parameter J is the only parameter controlled by the geometry of the rolling body. The smaller J is, the shorter the descent time. Part IV. Some figures You may find these figures useful when you do the exercises. A hollow sphere having mass m, inner radius ai , and outer radius ao 6 A hollow cylinder of mass m, length l, inner radius ai , and outer radius ao . References Andrew, Cain, Crum, Morley, Calculus Projects Using Mathematica, McGraw-Hill, 1996. Drucker, A Mathematical Roller Derby, The College Mathematics Journal, Vol 23, No. 5, 1992, 396-401 Salas, Hille, Etgen, Calculus: one and several variables, ninth edition, John Wiley and Sons, 2003. Strang, Calculus, Wellesley-Cambridge Press, Wellesley, MA, 1991. 7