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Moment of Inertia Consider two objects: a hoop and a solid disc. Let's say the hoop and the disc have the same mass and radius as shown below.  If we allowed them to start from the same position on an inclined ramp, which one would make it to the bottom of the ramp first? According to everything learned to this point, they should tie at the bottome because mass does not matter right?  Not exactly. They would both tie if they slid down the ramp, but we are now going to consider the rolling aspect. It turns out that the disc would beat the hoop to the bottom of the ramp! Before you throw your hands in the air in frustration because this goes against everything discussed before now, listen to the reason. This is the first time we have considered the aspect of rolling and its consequences on the linear speed of objects. Remember inertia is the object's resistance to changing states of motion. This is linear motion. Objects also have a resistance to changing states of rotation, which is called the Moment of Inertia. The moment of inertia, just like inertia, depends on the mass of the object. However, mass is not enough to account for the object's resistance to rotate, or stop rotating. Consider the hoop and disc. The disc had less resistance than the hoop to rotate, which is why it made it to the bottom of the ramp first. Remember, it was stated above that they had the same mass and radius. So, what is different between these two objects that would make the hoop resist more than the disc? Obviously the hoop has a big hole through the center. As a result, all of the mass is distributed farther away from the center of rotation as compared to the disc where some of the mass is far and some is very close.  Given this information, the moment of inertia seems to depend on not only the mass, but how that mass is distributed with respect to the center of rotation. In fact, this is exactly what it depends on. For the sake of making a point, I will repeat exactly what determines the moment of inertia: it depends on how much mass an object has, and how that mass is distributed with respect to the center of rotation! This means when the mass of a rotating object is closer to the center of rotation, it will be easier to either make it rotate or stop rotating. On the other hand, objects with the mass farther away from the center, such as the hoop, will be more difficult to make rotate or stop rotating. This concept has so many applications; you may have used moment of inertia and not even known. Have you ever walked across a narrow path such as a balance beam or fence? What did you do to help keep balance? Most people stick their arms out. Why? By sticking your arms out from your sides, you increased your moment of inertia thereby increasing your resistance to rotate, which helped you stay upright. This is why tight rope walkers use a long pole. If they begin to fall, they push down on the pole. Since the pole is long, it has a greater resistance to start rotating, which gives the person enough resistance to regain his or her balance. Consider the two scenarios below. Which pole would be easier to balance: the pole with the extra weight attached to the top or bottom?  A lot of people would say the pole with the weight on the bottom. Let's analyze this logic. If the extra weight is on the bottom, it is closer to the center of rotation, which in this case is the point where the finger is touching the pole. That is the center of rotation because that is the center of the circle the pole would make when it falls.  Therefore, the weight that is at the bottom actually decreases the moment of inertia of the pole, resulting in it being more difficult to balance. Having the weight at the top of the pole increases the moment of inertia because it is farther from the center of rotation. Because of the greater resistance to rotate, the pole does not move from side to side as quickly and readily as the pole with the weight on the bottom. This means the person has more time to react to the pole's movements, and gives him/her a better chance to regain balance.  Due to the diverse nature of objects, there isn't just one way to express the moment of inertia, but each way does include two things: mass and how far that mass is from the center of rotation. This may seem difficult to express because how do you account for the fact that some of the mass of a disc for example is found close and far. How do you know how much mass is located at a certain distance from the center? Fortunately, calculus makes this task very easy. We will not go through the steps here, but just be aware that their are mathematical methods to develop these equations. It turns out that the moment of inertia, represented by I, for a hoop is,  This means the moment of inertia for a hoop is equal to the hoop's mass times the square of the radius of the hoop. (This applies to a hoop that has a very thin ring so that the two radii, one for the inner part and one for the outer part, are almost equal). For a disc, the moment of inertia is,  Because of the 1/2, it is twice as hard to make a hoop start or stop rotating as compared to a disc with the same mass and radius. We knew the disc had a smaller I, but now we know exactly how much less. Below is a list of I's for various objects.  It is interesting to note that the moment of inertia for all of those objects have two things in common: the mass and the square of the radius. The only thing that separates them is the fraction found in front. The fraction dictates the overall size of the resistance. For example, think about which scenario would be easier: grabbing a yard stick at the center and twisting your wrist back and forth, or grabbing the yard stick at the very end and repeating that motion?  From experience, it is easier to rotate the stick from the center than the end. Notice the fraction in front of the thin rod with axis at the center compared to the thin rod with axis at the end. The fraction for the center example is 1/12 where as the for the end example is 1/3. 1/12 is certainly smaller than 1/3. In fact, 1/3 is exactly four times bigger. This means that it is four times more difficult to rotate a stick from the end as opposed to the center. Moment of inertia has a direct relationship to the mass of the object. This means if the mass is increased, the moment increase by the same and vice versa. But, it is a direct square relationship with the radius. So, if you increase the radius, the moment will increase by the square of that amount, just like the relationsip with work and velocity. If the radius is doubled, the moment is quadrupled. If the radius is reduced by one-half, the moment reduces by one-fourth. This is helpful information when considering the sizes of amusement park rides. If a ride needs to rotate, a motor must be used. The overall radius of the ride cannot be extended too much, because as the resistance to rotate increases, the motor has to work that much harder. In addition, you have to consider the extra weight of the passengers, as well as where they sit with respect to the center of rotation. All of these factors must be considered to determine how powerfull the motor must be, because they all contribute to increasing the ride's moment of inertia. Rotational Kinetic Energy The energy associated with an object's motion is called kinetic energy. Now that we will be discussing rotational energy, we need to use the term Translational Energy to represent the energy that makes the object move a linear distance. This will help to avoid confusion about which energy we are dealing with. We know that translational energy depends on the mass of the object and the speed, and is represented by the equation,  Rotational energy is the energy possessed by the object by virtue of its angular motion. This is energy that does not require an object to actually move a linear distance like translational energy does. A wheel can be rotating in the air and not go anywhere. In this case, it has rotational energy but no translational energy. The equation for rotational energy can easily be developed if we just replace the translational quantities with their angular counterparts. For example, mass is the measure of inertia. In the angular sense, this is the moment of inertia. Velocity is the rate of linear motion. The counterpart is the angular velocity, which is the rate of rotation.  Substituting these quantities will give rotational energy.  Now if an object is rolling across the floor, not only does it have translational energy, but rotational as well. Therefore, the total kinetic energy would be,  This brings up an interesting question: would an object have a faster linear speed at the bottom of a hill if it slid down, assuming no friction, or rolled down? Let's consider a disc for this example.  We first need to establish the fact that both hills are the same height. There is a way to express the linear speed of the disc in terms of just the height and g. Let's start with the slidding disc first. We are going to use the conservation of energy to solve this problem. First, when the disc is at the top before it moves, all of its energy is in the form of potential given by mgh. At the bottom of the hill, all of that energy is converted to translational kinetic energy. Since it is not rolling, the rotational energy is zero.  Therefore, since the initial energy equals the final energy, we have,  Solving for v in this equation, we can cancel the masses, multiply both sides by 2, and take the square root to give,  It should not be a surprise that v equals the square root of 2gh. That is the same speed an object will have when it is just dropped from a certain height as discussed in chapter four. Finding the linear speed for the rolling disc is a little more complicated. It still has the same initial energy, mgh, but the final energy is divided between the translational and rotational energy.  As a result, the energy conservation equation would become,  In order to solve for the linear speed v, we need to make a substitution for I and omega. We can do this because we know the moment of inertia for a disc, 1/2mr^2, and we also know that v = (r)(omega). Solving for omega and squaring we get,  Plugging that result in for omega and 1/2mr^2 for I we have,  Now every term has a mass, and the r^2 from the I cancels with the r^2 from the omega.  Therefore, it becomes,  We can now combine the v^2 terms and get,  Now we have the linear speed for both situations.  The two terms are exactly the same except for the 2 and the 4/3. Since 2 is larger than 4/3, skidding down the hill is faster than rolling. (Again, we have to assume no friction with the skidding. For example, a car on an icy road approximates no friction although there still is some). Why is it faster to skid rather than roll? The answer has to do with rotational energy. When the object skids, all of the potential energy is converted to translational energy, which is the energy that contributes to the object's linear speed. However, when the object rolls, some of the potential has to go into the rotational energy, which takes away from the translational. This results in a smaller energy, which contributes to a smaller linear speed. Let's see just how much of the potential energy went to the rotational in this example. The total energy at the bottom for the rolling disc would be the addition of the translational and rotational energies.  We can cancel the r^2's still, and then combine the two terms.  To determine what fraction of the potential went into the rotational energy, we have to take the rotational energy and divide by the total energy.  This means that one-third, or 33%, of the potential went into making the disk roll. Two-thirds, or 66%, was therefore left over to make the object move. Contrast this to the case of skidding where 100% of the potential went into making the object move. That is why skidding is faster than rolling. One more thing of interest to note with this section. Notice in the equations used above that both the radius and mass canceled out when solving for the linear speed at the bottom of the hill. This still means that mass does not matter when considering objects under the influence of gravity. The only thing that dictated the speed of the object was the fraction in the moment of inertia. Consider the equation below. When we dealt with this equation above, we canceled the masses and the r squared terms. The only thing left that affected the final linear speed was the 1/2 in the moment of inertia term. If that fraction was different, the linear speed at the end would be different. For example, if we considered a hoop for this equation, the fraction would be one, and the equation would be,  Solving for the linear speed would yield,  The mass or the radius of that hoop would not change the linear speed that we just found. So, you could take two hoops with different masses and radii, start them from the same height on a ramp, and they would tie at the bottom with the same linear speed. The only time two object's wouldn't tie would be when they have different moments of inertia. This is why Galileo was able to take different sizes of balls and show that they take the same amount of time to travel down varying inclines. It is a good thing he used the same type of object. If he would have used a ball and a disc, his results would have been different. |