

Guide 113. Dynamic Application: Combining Net Force and Net Torque Methods For problems involving both translation and rotation with acceleration, a combination of net force and net torque methods is used. In addition to the usual practices in carrying out these methods, see the practices below.
Here's an example to illustrate the method. The diagram to the right shows a solid, cylindrical disc of uniform composition with a thin, flat ribbon wound around it. One end of the ribbon is attached to a support at the top; the other end is attached to the disc. The ribbon is wound around the disc initially, and the disc is released from rest. As the disc falls, the ribbon unwinds. The mass of the disc is M, and its radius is R. Determine the tension force and the acceleration of the CM of the disc in terms of M and g only. Assume that the ribbon doesn't stretch, that it has neglibible mass compared to the disc, and that the ribbon unwinds without slipping. The diagram shows the directions of rotation and of translational velocity of the CM of the disc. The angular velocity will increase in the counterclockwise sense; hence, the angular acceleration will be positive. The translational displacement of the center of mass is down. For consistency, therefore, we will select the direction of positive translational displacement to be down. We will also pick the CM of the disc as the axis of rotation for calculating torques. Now consider the diagram to the left. The directions of positive displacement are shown. The two forces acting on the disc are shown in red. These are the weight, Mg, acting from the CM, and the tension, T, of the ribbon, acting on the outside of the disc. We use the usual practice in problems involving torque to draw the forces extending from their respective points of application. Our strategy will be to write the net force and net torque equations. Each will contain the two unknowns of tension force and acceleration. We'll eliminate T by solving the two equations simultaneously. We'll solve the result for the acceleration and then substitute that back into the net force equation to obtain an expression for the tension. With the direction of positive translational displacement being down, the net force equation is: F_{net} = Mg  T. Applying Newton's 2nd Law yields Ma _{}= Mg  T. Therefore, T _{}= Mg  Ma. The net torque equation is τ_{net}= TR. Note that the weight doesn't create torque, since the weight vector passes through the axis of rotation. The torque produced by the tension force is positive, because it produces counterclockwise rotation. The moment arm of the torque is just the radius of the disc. Using the relationship between net torque and angular acceleration, we have: Iα = TR. Solving for T yields T = Iα/R. Next we equate the two results for T: Mg  Ma = Iα/R. We'll solve this equation for the acceleration. These substitutions will be necessary: i) α = a/R, because the ribbon unwinds without slipping), and ii) I = MR^{2}/2, the moment of inertia of a disc rotating through an axis about its center. Mg  Ma = Iα/R Mg  Ma = (MR^{2}/2)(a/R)/R Mg  Ma = Ma/2 Solving for a yields a = 2g/3. This is the downward acceleration of the CM of the disc. We can substitute this result back into T = Mg  Ma to find that the tension force is T = Mg/3. Alternate solution We used the method above to solve the problem in order to illustrate the combined net torque/net force method. However, it's also possible to solve the problem using conservation of energy. We'll carry out this method as a check on the dynamic method. Take the system to be the disc, ribbon, and Earth. With this choice, there are no external forces. There are three energy changes to deal with. Both the translational and rotational kinetic energies of the disc increase, while the gravitational potential energy decreases We write the conservation of energy equation, letting h represent the height through which the disc falls. W_{ext}_{} = ΔE_{sys} 0 = ΔK_{rot} + ΔK_{trans} + ΔU_{g} = Iω^{2}/2 + Mv^{2}/2  Mgh Substituting for the moment of inertia and employing ω = v/R, we have the following. 0 = (MR^{2}/2)(v/R)^{2}/2 + Mv^{2}/2  Mgh Solving for v yields v = (4gh/3)^{0.5}. Since the acceleration is uniform, we can solve for that using a dvat: v^{2} = v_{0}^{2} + 2aΔy. The simplification is shown below. v^{2} = v_{0}^{2} + 2aΔy 4gh/3 = 0 + 2ah a = 2g/3, as before. 

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