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Moment of Inertia |
 When we studied linear motion we started with the description of linear motion which we called kinematics. From there we went on to dynamics which provided the reason for the motion, i.e. the cause or influence. In a similar fashion we introduced rotational kinematics relating t, q, w, and a. Continuing the correspondence between linear and rotational motion, we now introduce rotational dynamics. Our most important equation in linear dynamics was Newton's Second Law, F = ma. This equation stated that a force applied to a mass at its center of mass will result in the mass undergoing an acceleration. The important point is that the net force is acting on the center of mass of the object, so its effect can't cause any rotation of the object. What will happen if the force is applied at a location which is not the center of mass? Start with F = ma, multiply by r, rF = rma substitute a = ar, rF = rmar, let rF = t t = mr2a t = Ia, where I = mr2. The quantity, I, is called the moment of inertia, and it expresses the resistance to rotational acceleration similar to the way that inertia, as measured by mass, was related to the resistance to acceleration. The unique thing about the moment of inertia is that its value depends upon the axis of rotation, whereas the the usual mass is constant.
The general expression for the moment of inertia is I = Smiri2, where the summation extends over all masses comprising the object of interest. For example, consider two masses, M=2 kg and m=1 kg, connected by a 1 meter light rod. Attach a perpendicular axis at a point 0.25 m from the small mass. The moment of inertia calculated relative to the axis of rotation is, I = 2kg (0.75m)2 + 1kg(0.25m)2 = 1.1875 kgm2 If we move the axis of rotation to the center of the 1 m rod, the moment of inertia becomes, I = 2kg (0.5m)2 + 1kg(0.5m)2 = 0.75 kgm2 A new result expressing a smaller resistance to rotational acceleration.
For objects that can be broken into discrete components calculations of this type will enable the determination of the moments of inertia. For objects that are continuous we would need the tools of calculus to determine the moments of inertia. For our purposes we will only need the moments of inertia of specific objects, such as a sphere, a rod, a disk and a hoop. Torque is the rotational influence of a force applied away from the center of mass of an object. The application of an unbalanced torque will create rotational acceleration. The definition of torque, t , is t = rFsinq, where the angle, q, is between the direction of r and F. Notice that the torque has a maximum value for an angle of 90 degrees. Torques create clockwise or counterclockwise rotational influences. Clockwise torques are positive and counterclockwise torques are negative. The net torque on an object is calculated from the sum of the counterclockwise torques minus the sum of the clockwise torques. The result of this net torque acting upon an object of moment of inertia, I, is expressed by, t = Ia . In analyzing mechanical problems we now have the necessary tools to calculate linear motion using Newton's Second Law and rotational motion using the above equation. Equilibrium Before we do any problems or lab exercises we need to define the concept of equilibrium. We have enough mechanics information to do this. As you sit in your chair, ask yourself whether you are in equilibrium. Of course your answer is yes, I am in equilibrium. Why, what defines equilibrium? Well, you are not moving nor are you accelerating either linearly or rotationally. So, if you are not accelerating in any sense, there must be a balance of forces and torques acting upon you. In Newtonian terms that means,
SFi = 0 and Sti = 0 . When these conditions are met complete static equilibrium exists.
Even though rotating objects are not moving horizontally they still possess kinetic energy. You know that if you've ever tried to grab a rotating tire and stop it. It takes effort, i.e. work, on your part to stop its rotation. Anything that requires work in order to influence it must possess energy. Rotational kinetic energy is defined as, K = 1/2 Iw2
The angular momentum is defined as, L = Iw Applications of Mechanical Principles
The principle of conservation of energy still applies
when dealing with rotational motion. The only added consideration is
that, now, we must account for the rotational kinetic energy contribution
to the object's motion. For example, a sphere rolling along the
surface of a table has two kinetic energy parts, i.e. translational and
rotational: K = 1/2 Iw2+ 1/2mv2 with I = 2/5MR2. Example: A thin ring of radius, r = 10 cm, and mass, m = 1 kg, is placed on an incline 50 cm above the level surface of a table. The ring is released and allowed to roll down the incline to the level surface. What is the ring's speed as it reaches the table surface? Solution: This is a conservation of energy problem. What is the total energy when the ring is sitting on the incline? Eo = K + U = 0 + mgy = (1 kg)(9.8m/s2)(.5 m) Eo = 4.9 Joules What is the total energy at the bottom of the incline? E = K + U = 1/2 Iw2+ 1/2mv2 + 0 E = 1/2mR2 (v/R)2+ 1/2mv2 = mv2 E = (1 kg) v2 Conservation of energy: Eo = 4.9 Joules = E = (1 kg) v2 so, v = 2.2 m /s How would this change is the sphere slid down the plane rather than rolled? All the energy at the bottom would be translational kinetic energy! 4.9 J = 1/2mv2 9.8 = v2 v = 3.13 m/s As you recall when we discussed conservation of momentum for linear motion, we started with Newton's Second Law, F = ma. So, let's start with the rotational equivalent of Newton's Second Law for rotational motion, t = Ia . t = IDw/Dt . t = D(Iw)/Dt . t = DL/Dt . Once again, if we consider a system which is isolated so that no external torques are applied, then we find DL = 0 or Lo = Lf . The classic example of this is a spinning ice skater the pulls her arms in as she spins.
As you know, her rotational speed increases dramatically as she pulls her arms in. Why? Because her angular momentum before bringing her arms in is, Lo = Iowo and after her arms are pulled in, Lf = Ifwf . By bringing her arms in or out she is reducing or increasing her moment of inertia. Since her angular momentum must stay constant, then when she reduces her moment of inertia by pulling her arms in her angular speed must increase in order for the angular momentum to remain the same. So her final angular speed will be,
Ifwf = Iowo wf = (Io/If)wo Summary of Rotational Motion Equations: I = Smiri2, t = rFsinq, St = Ia . SFi = 0 and Sti = 0 K = 1/2 Iw2 L = Iw DL = 0 or Lo = Lf . q = wavet wave = (wo + w)/2 w = wo + at q = wot + ½ at2 w2 = wo2 + 2aq
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