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A Chart of Common Moments of Inertia

For linear, or translational, motion an object's resistance to a change in its state of motion is called its inertia and is measured in terms of its mass, in kg. When a rigid body is rotated, its resistance to a change in its state or rate of rotation is called its rotational inertia, which is measured in terms of its moment of inertia, in kg m². This resistance has a two-fold property:

the amount of mass present in the object, and
the distribution of that mass about the chosen axis of rotation.

In general, the formula for a single object's moment of inertia is I_cm = kmr² where k is a constant whose value varies from 0 to 1. Different positions of the axis result in different moments of inertia for the same object; the further the mass is distributed from the axis of rotation, the greater the value of its moment of inertia.

That is, the smaller the coefficient of mr², the easier it is to accelerate the object. That is, spheres accelerate easier than cylinders, which accelerate easier than thin rings or hoops. Since an object's moment of inertia increases as its mass is moved further from its axis of rotation, hoops and rings would represent the greater inertia since all of their mass is concentrated at a constant distance, r, from the center of rotation.

Below is a series of diagrams illustrating how the moment of inertia for the same object can change with the placement of the axis of rotation. This is not an all inclusive list, but it is a "most used" list.

solid sphere I = 2/5 MR²	thin-walled sphere I = 2/3 MR²


thin rod (perpendicular at end) I = 1/3 ML²	thin rod (perpendicular at center) I = 1/12 ML²


solid cylinder (about central axis) I = 1/2 MR²	thin-walled cylinder/hoop/ring (about central axis) I = MR²	thick-walled cylinder (about central axis) I = 1/2 M(R₁² + R₂²)


solid cylinder (perpendicular to central axis) I = 1/4 MR² + 1/12 ML²	thin-walled cylinder (perpendicular to central axis) I = 1/2 MR² + 1/12 ML²

Rotational Dynamics: Rolling Spheres/Cylinders

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Rolling Spheres and Cylinders

Consider the following three diagrams. The first one shows the velocity vectors for an object experiencing only pure translation. The center diagram is for an object that is experiencing only rotation. The final diagram is a combination of the two - both translation and rotation.

Notice how the vectors add together. At the bottom of the object that is both rotating and translating, the contact point is instantaneously at rest. This is why static friction is used to calculate the torque that produces rotational motion. Also notice that the very top point on the wheel is moving with speed 2v_CM - faster than any other point on the wheel.

Rotational Dynamics

Now consider an object rolling down an incline plane. The first thing that we are going to do is draw a freebody diagram of the forces acting on this mass and then resolve those forces into their components which act parallel and perpendicular to the plane.

Let's consider the axis of rotation passing through the disk's center of mass, cm. Notice in this case that only the instantaneous static friction force will supply a torque since the lines of action of the other two forces (normal and weight) act through the center of mass and cannot produce a torque. As long as the mass rolls without slipping, we can use the relationships: v = r? and a = r?.

Rotationally,

net ? = I_CM?
? = f_sr

f_s = I_CM(?/r)

Remember linearly,

net F = ma
mg sin? - f_s = mr? where a = r?

Simultaneously eliminating f_s and solving for ? yields:

mg sin? - I_CM(?/r) = mr?
? = g sin? /(r + I_CM/mr)

The moment of inertia for a disk, or solid cylinder (see chart provided below), equals ½mr². Substituting in this value and simplifying gives us

? = 2g sin?/(3r)

Since this angular acceleration is uniform, you would be free to use any of the rotational kinematics equations to solve for final angular velocity, time to travel down the incline, or the number of rotations it completes as it rolls along the incline's surface.

Moments of Inertia

Below you will find a chart of the three most popular "rolling objects." Notice that their rotational inertia increases from left to right as the mass distribution gets farther from the axis of rotation that passes through their center of mass.

solid sphere I = 2/5 MR²	solid cylinder (about central axis) I = 1/2 MR²	thin-walled cylinder/hoop/ring (about central axis) I = MR²

	Based on this chart, which object in the following picture would you predict would reach the bottom of the incline first: the solid cylinder or the thin ring? To decide, consider the rotational inertia of each object and how inertia affects motion. Both objects have the same mass and equal diameters. picture courtesy of Arbor Scientific

Using Energy Methods

As a solid sphere rolls without slipping down an incline, its initial gravitational potential energy is being converted into two types of kinetic energy: translational KE and rotational KE.

If the sphere were to both roll and slip, then conservation of energy could not be used to determine its velocity at the base of the incline. The slipping would result in kinetic friction doing work on the sphere and dissipating energy in the form of heat.

Remember that when an object rolls down an incline, it's linear (also called translational) kinetic energy is always less than what it would have been had it was only slid down a frictionless incline.

An Application of the Parallel-Axis Theorem

The principle that can unite a rolling object's rotational and translational kinetic energy into one expression of total KE is called the Parallel Axis Theorem.

I _P = I_CM + Mh²

where

I _P represents the object's moment of inertia from any location, P
I _CM represents the object's moment of inertia about its center of mass
h represents the perpendicular distance from P to the center of mass

For our purposes, let P represent the point of contact where the rolling thin ring, cylinder, or sphere touches the incline's surface.

Total KE = ½I_P?²
Total KE = ½(I_CM + mh²)?²
Total KE = ½I_CM?² + ½mh²?²
Total KE = ½I_CM?² + ½m(r²?²)
Total KE = ½I_CM?² + ½mv²
Total KE = KE_rotational + KE_{translational}

This procedure can apply to any rolling object - just substitute in its correct moment of inertia.

Random Variables and Probability Distributions

A random variable assigns a value to the outcomes in a random situation. Random variables can be continuous, meaning that they can take on any value or they can be discrete. Discrete random variables can only take on values from a countable list.

Consider attending a baseball game at Shea Stadium, which is located next to LaGuardia Airport in New York. At a ballgame there are many random variables that may effect you while you watch the game. The temperature is considered a continuous random variable because it can take on any value. Although we usually round to the nearest degree, it is still a continuous variable. On the other hand, the number of planes that fly over during the game is a discrete random variable. There can only be 0, 1, 2, 3, 4, etc planes that fly over. There is a countable list of numbers that the number of planes comes from. In other words, there cannot be 3.7 planes that fly over. The number of people who attend the baseball game is another example of a discrete random variable. The length of time you wait in the security line is another example of a continuous random variable.

Once we have identified random variables and the type of variable it is, there are usually probabilities assigned for each possible value.

Let's Practice:

Let?s look at a discrete case in which you toss a coin two times. What are the possible outcomes and probabilities?

First, the possible outcomes.

HH, HT, TH, and TT

In other words, there can be 0, 1, or 2 tails:

the probability of 0 tails is 1/4,
the probability of one tail is 2/4 or 1/2 and
the probability of 2 tails is 1/4.

Most of the time this information is organized in table form and called a discrete probability distribution. It is describing a discrete random variable and it shows how the probabilities are distributed to all the outcomes. Hence the name - discrete probability distribution. For our example, it would look like

Number of Tails 0 1 2

Probability 1/4 1/2 1/4

Note that all the probabilities on the ?probability? line of the distribution must all add up to 1.

There are a variety of problems that you can be asked to solve based on a discrete probability distribution. You could be asked to do something as simple as complete a table or perform more complicated tasks such as finding the mean and standard deviation of a distribution. Also there are special kinds of discrete probability distributions, one of which is a binomial distribution. Examining these types of problems can take up an entire chapter in a probability/statistics book and are too numerous to explore in a single lesson. For additional information, ask your teacher for help or to direct you to an appropriate statistics textbook.

As discrete random variables can have distribution functions, there is a similar situation for continuous random variables. However, since each and every possibility cannot be listed for continuous variables as is the case with discrete random variables, we have a slightly different look and terminology.

When dealing with a continuous random variable and the assigned probabilities, we refer to a probability density function and deal with the area under a curve.

Suppose the shuttle service from the airport parking lot to the terminal arrives at the parking lot every 10 minutes. If you arrive at a random time, how long should you expect to wait? That is dependent on the density function, in this case a uniform density function. Another very common type of probability density function is the normal distribution. As with discrete probability distribution function, there are entire chapters in books devoted to this topic and too numerous to discuss in one, or even ten, lessons.

author-nitin

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