

Guide 103. Conservation of Energy with Rotational Motion Related reading: Walker 106. The notation and discussion in the guide below is more specific in some respects than that in the textbook. Note the use of subscripts, the general conservation of energy equation, and the stipulation that rolling occurs without sliding, hence, v_{cm} = Rω. When you studied Chapter 8, you used conservation of energy to solve problems involving objects moving in curved paths. We need to make it clear, though, that rotation of the objects wasn't involved in those problems. The objects were treated as point particles and any movement along a curved surface was sliding movement rather than rolling. We make that point, because the present guide deals with conservation of energy for rigid objects that have extent in space and undergo rotation about an axis. The kinetic energy of rotation of a rigid object rotating about an axis is defined as K_{rot} = ½Iω^{2}, where I is the moment of inertia of the object about the axis and ω is the angular velocity. The object may also have translational motion in which case its translational kinetic energy is K_{trans} = ½Mv_{cm}^{2}, where M is the mass of the object and v the linear velocity of the center of mass. Note the use of subscripts to clearly distinguish the two types of kinetic energy. The total kinetic energy of an object that is both rotating and translating is K_{total} = K_{rot} + K_{trans}. In using the law of conservation of energy to solve problems, the total kinetic energy is calculated. Thus, the term ΔK in W_{ext} = ΔK + ΔU includes changes in both forms of kinetic energy, both rotational and translational, in the same way that the term ΔU includes changes in all forms of potential energy. Rotational kinetic energy and rolling motion Many situations that we consider involving rotation are of an object rolling on a surface. One assumes unless indicated otherwise that the object rolls without sliding. In that case, one can say that v_{cm }= Rω. Note that this also assumes the object has a circular cross section of radius R. Of course, this must be the case if the object rolls. With these assumptions, the rotational kinetic energy can be written as follows. K_{rot} = ½I(v_{cm}/R) ^{2} This leads to a simplification due to the fact that the rotational inertia of many symmetric objects with uniform mass distribution can be expressed as I = fMR^{2}, where f is a parameter that depends on the shape. For example, f = 2/5 for a solid sphere rotating about the center of mass. With the substitution I = fMR^{2}, we can write the kinetic energy of rotation as follows. K_{rot} = ½fMR^{2}(v_{cm}/R) ^{2} = ½fMv_{cm}^{2} Thus, the total kinetic energy is the following. K_{total} = ½fMv_{cm}^{2} + ½Mv_{cm}^{2} = (1 + f)(½Mv_{cm}^{2}) = (1 + f)K_{trans}. Note again that the above form is not general but rather requires that the object roll without slipping and that f = I/(MR^{2}). About friction and rolling In order for an object to roll, it must experience static friction. Therefore, one might think that static friction is an external force contributing to the W_{ext} term in the conservation of energy formula, W_{ext} = ΔK + ΔU. In fact, we typically take W_{ext} to be 0 in such situations. This is because the static friction is what causes the object to roll, thereby giving it rotational kinetic energy. Since we include ΔK_{rot} as part of the ΔK term, the energy change caused by static friction is accounted for. Note that this assumes that the object doesn't slide. If the object slides, then there is also kinetic friction, which would do work on the system.


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