1Experiment 6 Moment of Inertia Preparation Prepare for this week's experiment by studying angular velocity, acceleration, rotational momentum and kinetic energy, and moment of inertia. Principles An object traveling in a straight line has kinetic energy,21K= mv2, and linear momentum, mpv= . An applied force will change the linear momentum, and that change will be in the direction of the force, dmdpFat==∑. An object that is rotating about a fixed axis will have rotational kinetic energy 2R1K= Iω2 and angular momentum L = Iω, where ω is the angular velocity. To change angular momentum one must apply a torque. The change in angular momentum will be in the same direction as the torque, dIdtLτα==∑. The quantity I is called the moment of inertia of the object. It depends on the magnitude and the distribution of the mass of an object. You can think of mass as the amount of resistance an object offers linear acceleration and the moment of inertia as the resistance an object offers to angular acceleration. For a point mass, like a very small rock tied to a string, the moment of inertia is 2I=mr , where m is the mass of the object and r is the distance from the mass to the center of rotation. If several masses rotate about an axis, like a cheerleader's baton, the net moment of inertia is simply the sum of the individual moments such that n2ii1I=mr∑. Moments of inertia can be added and subtracted as long as they are about the same axis of rotation. For an extended body, the summation becomes an integral and the moment of inertia becomes 2I=rdm∫,2where r is the distance from each mass element, dm, to the axis of rotation. For an extended object, r is variable and dm must be expressed as a function of r so that the expression can be integrated. In this experiment we are going to calculate the theoretical moment of inertia of a ring and a disk. Actually, the disk is simply a special type of ring, so one integral will work for both problems. Start by assuming that the ring has an inner radius Ri, an outer radius Ro, and a constant height h. The object has uniform density, ρ, which is the mass per unit volume, M/V. Since M=ρV , if ρ is constant, dm =ρdV . This does not seem like progress until you rewrite the volume element, dV, in polar coordinates. Each element of volume is the product of three infinitesimal length elements. The element along the direction of the ring’s height is dz, the element along the radial direction is dr, and the element along the curve of the ring is rdθ . Therefore dm =ρdV = ρ rdz dθ dr . Now we can find the total mass of the object by integrating dm over the limits of the ring. Along the direction of the ring’s height, z ranges from 0 to h. The radial variable, r, extends from the inner to the outer radius. The angular extent of the ring, θ, goes from 0 to 2π . The total mass then is the product of three integrals oih2π R00 RM= dm=ρdV = ρ dV = ρ dz dθ rdr∫∫ ∫ ∫∫∫. Evaluating the integrals gives oiR222oiRrM=ρ(h)(2π)=ρ h π (R - R )2. A disk is just a ring whose inner radius equals zero. Now that dV is written in polar coordinates, the moment of inertia can be calculated in a similar fashion to that used to calculate the mass, 222io1I=rdm= M(R+R)2∫. In this experiment you will also find the moments of inertia of a ring, a disk, and a ring and a disk together experimentally and compare our results to the calculated theoretical moments. In this experiment you will first find the moment of inertia of a rotating base (sometimes called a spider). Then put the ring on top of the spider and determine the moment of the system, the base and ring together. Since moments of inertia can be added and subtracted as long as they are3about a common axis of rotation, we can find the experimental moment for the ring by subtracting ring system baseI=I -I . There are two ways to derive the experimental moment of inertia. Force and Acceleration Study the diagram. A torque will be applied to the base by putting weight on a string that winds around the axle. The tension in the string will provide the force that produces the torque and will also overcome the sliding friction in the bearings and pulleys. The magnitude of the frictional force, f, will be determined experimentally by finding the mass, fm, which makes the base rotate at constant velocity, so that ff=m g. An additional mass, m, will then be added to make the base accelerate. Applying Newton’s Second Law to the masses allows us to show that the total tension in the string is related to the acceleration of the masses by fT=(m+m )(g-a). The magnitude of the net force which rotates the spider is this tension force minus the opposing frictional force, f, such that netF=T-f. The torque, τ, which accelerates the spider will equal netF×r. Since netF and r are perpendicular in this case, the magnitude of the torque is simply equal to netFr. The magnitude of the torque also equals Iα. Remember that the acceleration of the mass and the angular acceleration of the spider are not independent, there is a relationship between a and α.4 The magnitude of the acceleration, a, can be determined be finding the time it takes the to masses fall from rest through a known distance, s, such that 21s= at2. Combining these equations for force, torque, acceleration, and angular acceleration, the moment of inertia of each system can be experimentally found using the formula 22fgtI=r m -1 -m2s⎡⎤⎛⎞⎢⎥⎜⎟⎝⎠⎣⎦. Work and Energy This equation can also be derived using the concepts of work and energy. The initial potential energy is mostly converted to the linear kinetic energy of the two masses and the rotational kinetic energy of the spider. However some of the initial potential energy is removed from the system by the (negative) work of friction, a non-conservative force. Therefore ffifffiiW=∆E=E -E ,W=U+K-U-K. Its useful to note that the final kinetic energy is comprised of two terms, linear and rotational, which can be written as 22fff f11K= (m+m)v+ Iω22. Remember that the final velocity of the masses and the final angular velocity of the spider are not independent: v=rω. The time it takes for the masses to fall a known distance can be related to their final velocity. The average velocity is given by fiv+vsv= =2t. Since the falling masses start from rest, vi equals zero, so we can solve for vf. These