Since the mass density of this object is uniform, we can write. We chose to orient the rod along the x -axis for convenience—this is where that choice becomes very helpful. Note that a piece of the rod dl lies completely along the x -axis and has a length dx ; in fact,. The distance of each piece of mass dm from the axis is given by the variable x , as shown in the figure. Putting this all together, we obtain.
The last step is to be careful about our limits of integration. The rod extends from. Next, we calculate the moment of inertia for the same uniform thin rod but with a different axis choice so we can compare the results.
We would expect the moment of inertia to be smaller about an axis through the center of mass than the endpoint axis, just as it was for the barbell example at the start of this section. This happens because more mass is distributed farther from the axis of rotation. Now consider the same uniform thin rod of mass M and length L , but this time we move the axis of rotation to the end of the rod.
The quantity dm is again defined to be a small element of mass making up the rod. Just as before, we obtain. However, this time we have different limits of integration. Note the rotational inertia of the rod about its endpoint is larger than the rotational inertia about its center consistent with the barbell example by a factor of four.
The similarity between the process of finding the moment of inertia of a rod about an axis through its middle and about an axis through its end is striking, and suggests that there might be a simpler method for determining the moment of inertia for a rod about any axis parallel to the axis through the center of mass.
Such an axis is called a parallel axis. There is a theorem for this, called the parallel-axis theorem , which we state here but do not derive in this text. Then we have. This result agrees with our more lengthy calculation from above. This is a useful equation that we apply in some of the examples and problems. What is the moment of inertia of a cylinder of radius R and mass m about an axis through a point on the surface, as shown below?
Integrating to find the moment of inertia of a two-dimensional object is a little bit trickier, but one shape is commonly done at this level of study—a uniform thin disk about an axis through its center Figure. Since the disk is thin, we can take the mass as distributed entirely in the xy -plane. We again start with the relationship for the surface mass density , which is the mass per unit surface area. Since it is uniform, the surface mass density. Now we use a simplification for the area.
The area can be thought of as made up of a series of thin rings, where each ring is a mass increment dm of radius r equidistanct from the axis, as shown in part b of the figure. The infinitesimal area of each ring dA is therefore given by the length of each ring. The full area of the disk is then made up from adding all the thin rings with a radius range from 0 to R.
This radius range then becomes our limits of integration for dr , that is, we integrate from. Note that this agrees with the value given in Figure. Now consider a compound object such as that in Figure , which depicts a thin disk at the end of a thin rod. This cannot be easily integrated to find the moment of inertia because it is not a uniformly shaped object.
It is important to note that the moments of inertia of the objects in Figure are about a common axis. In the case of this object, that would be a rod of length L rotating about its end, and a thin disk of radius R rotating about an axis shifted off of the center by a distance. The moment of inertia of the disk about its center is.
Adding the moment of inertia of the rod plus the moment of inertia of the disk with a shifted axis of rotation, we find the moment of inertia for the compound object to be. The merry-go-round can be approximated as a uniform solid disk with a mass of kg and a radius of 2. Find the moment of inertia of this system. This problem involves the calculation of a moment of inertia.
We are given the mass and distance to the axis of rotation of the child as well as the mass and radius of the merry-go-round. Since the mass and size of the child are much smaller than the merry-go-round, we can approximate the child as a point mass. The notation we use is. The value should be close to the moment of inertia of the merry-go-round by itself because it has much more mass distributed away from the axis than the child does.
Find the moment of inertia of the rod and solid sphere combination about the two axes as shown below. The rod has length 0. Masses that are further away form the axis of rotation have the greatest moment of inertia. Angular momentum of an object, rotating about an axis, is a measure of the amount of rotation of that object when no external torques are acting upon it, with torque being defined as the moment of force and is a measure of how much force is needed to cause the rotation of an object.
Angular momentum is a conserved quantity, which means that it stays constant providing no external torques act upon it, and is the product of moment of inertia multiplied by angular velocity.
When the body has an increased radius, i. While in the pike position the body decreases in radius as each segment moves closer to the axis of rotation, resulting in angular velocity increasing and a decrease of moment of inertia. Thus, during a dive, angular momentum is constant meaning that moment of inertia is inversely proportional to angular velocity. Moments of Inertia were found for a backward and a forward pike dive by calculating the sum of the inertias for each segment of the body.
Both dives were performed by the same diver yet the moments of inertia are different due to the distance of each segment from the axis of rotation i. In order to do this they have to increase their angular velocity, consequently reducing their moment of inertia. This is done by changing their body configuration so as to decrease the distance between the centre of mass of each body segment and the axis of rotation, thus a tighter pike position gives the diver a smaller moment of inertia and greater angular velocity.
Once the diver leaves the board, there is no torque acting on the body. This means that angular momentum is conserved when no external torque acts on it, thus when the moment of inertia decreases angular velocity increases and vice versa. The dive is divided into 3 phases. The first phase is from when the diver leaves the board until entering the full pike position. Phase 2 is the execution of the somersaults in the pike position and the final phase is the release from the pike position and preparation for entry into the water.
In phase 1, once the diver leaves the board there is no external torque acting on him, thus angular momentum is conserved and remains so throughout the dive. When the diver first leaves the board, moment of inertia is high due the limbs i. Towards the end of phase 1, the diver is assuming the pike position meaning that all body segments are pulled as close as possible to the axis of rotation thus decreasing the moment of inertia and increasing the angular velocity.
Moment of inertia in the second phase varies in accordance with the angular velocity so as to conserve angular momentum. By the third phase the moment of inertia increases as the diver prepares to enter the water and releases from the pike position. This is because the arms are stretched over the head and are thus further away from the axis of rotation, similar to the initial position.
In experiments using a pair of inclined planes facing each other, Galileo observed that a ball would roll down one plane and up the opposite plane to approximately the same height. If smoother planes were used, the ball would roll up the opposite plane even closer to the original height. Galileo reasoned that any difference between initial and final heights was due to the presence of friction.
Galileo postulated that if friction could be entirely eliminated, then the ball would reach exactly the same height. Galileo further observed that regardless of the angle at which the planes were oriented, the final height was almost always equal to the initial height. If the slope of the opposite incline were reduced, then the ball would roll a further distance in order to reach that original height. Galileo's reasoning continued - if the opposite incline were elevated at nearly a 0-degree angle, then the ball would roll almost forever in an effort to reach the original height.
And if the opposing incline was not even inclined at all that is, if it were oriented along the horizontal , then Isaac Newton built on Galileo's thoughts about motion. Newton's first law of motion declares that a force is not needed to keep an object in motion. Slide a book across a table and watch it slide to a rest position. The book in motion on the table top does not come to a rest position because of the absence of a force; rather it is the presence of a force - that force being the force of friction - that brings the book to a rest position.
In the absence of a force of friction, the book would continue in motion with the same speed and direction - forever! Or at least to the end of the table top. A force is not required to keep a moving book in motion. In actuality, it is a force that brings the book to rest. All objects resist changes in their state of motion. All objects have this tendency - they have inertia. But do some objects have more of a tendency to resist changes than others? Absolutely yes! The tendency of an object to resist changes in its state of motion varies with mass.
Mass is that quantity that is solely dependent upon the inertia of an object. The more inertia that an object has, the more mass that it has. A more massive object has a greater tendency to resist changes in its state of motion. Suppose that there are two seemingly identical bricks at rest on the physics lecture table. Yet one brick consists of mortar and the other brick consists of Styrofoam.
Without lifting the bricks, how could you tell which brick was the Styrofoam brick? You could give the bricks an identical push in an effort to change their state of motion. The brick that offers the least resistance is the brick with the least inertia - and therefore the brick with the least mass i. A common physics demonstration relies on this principle that the more massive the object, the more that object resist changes in its state of motion.
The demonstration goes as follows: several massive books are placed upon a teacher's head. A wooden board is placed on top of the books and a hammer is used to drive a nail into the board. Due to the large mass of the books, the force of the hammer is sufficiently resisted inertia. This is demonstrated by the fact that the teacher does not feel the hammer blow.
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