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Summary: This module explains the physics of rolling objects in a format that is accessible to blind students.
This module is part of a collection (see http://cnx.org/content/col11294/latest/ ) of modules designed to make physics concepts accessible to blind students. The collection is intended to supplement but not to replace the textbook in an introductory course in high school or college physics.
This module explains the physics of rolling objects in a format that is accessible to blind students.
In addition to an Internet connection and a browser, you will need the following tools (as a minimum) to work through the exercises in these modules:
The minimum prerequisites for understanding the material in these modules include:
I recommend that you open another copy of this document in a separate browser window and use the following links to easily find and view the figures while you are reading about them.
I recommend that you also study the other lessons in my extensive collection of online programming tutorials. You will find a consolidated index at www.DickBaldwin.com .
What happens when objects with different moments of inertia (rotational inertia) roll down a hill? Does the moment of inertia effect how the objects roll? Those are the kinds of questions that we will explore in this module.
A symmetrical object rolling down an incline
Imagine an object that is symmetrical about its center of mass (such as a sphere or a cylinder) rolling down an incline. The center of mass experiences translational motion as it rolls down the incline. In addition, the object is rotating about an axis that passes through the center of mass.
The translational velocity
If the object is rolling without slipping, the translational velocity of the center of mass is given by
Vcm = W*R
where
A well-defined relationship
This means that there is a well-defined relationship between the rolling object's translational and rotational kinetic energies. The total kinetic energy of a rolling object is the sum of its translational and rotational kinetic energies.
A clue
This may give you a clue as to where we are heading with this. As a preview, when an object rolls down an incline it exchanges potential energy for kinetic energy. Some of that potential energy is transformed into translational kinetic energy and some is transformed into rotational kinetic energy. The manner in which that potential energy is distributed between the two forms of kinetic energy has an impact on how the object rolls.
Cylinders, disks, and spheres
Imagine a cylinder-like, disk-like, or sphere-like object that has a mass M and a radius R rolling down an incline. The moment of inertia for each of four different geometrical examples is shown in Figure 1 .
| Examples of moment of inertia. | |
|---|---|
|
A very important constant
As you can see from Figure 1 , for these four shapes at least, the moment of inertia is equal to a constant multiplied by the product of the mass and the square of the radius. (As far as I know, this constant hasn't been given a widely-recognized name.)
In these four cases, we can write an expression for the moment of inertia for rotation about the center of mass as
Icm = Q*M*(R^2)
where
The purpose of the constant
The purpose of the constant, Q, is to specify how the mass is distributed relative to the axis of rotation. Larger values of Q generally indicate that the mass is distributed further from the axis of rotation. Similarly, larger values of Q also indicate larger moments of inertia.
Given the relationships
Icm = Q*M*R^2, and
Vcm = W*R, and
Krot = (1/2)*Icm*W^2
where
We can substitute and write
Krot = (1/2)*(Q*M*R^2)*(Vcm/R)^2, or
Krot = (1/2)*(Q*M*R^2)*(Vcm^2/R^2), or
Krot = (1/2)*(Q*M)*(Vcm^2), or
Krot = Q*(1/2)*(M)*(Vcm^2)
Given that
Ktr = (1/2)*(M)*(Vcm^2)
where
A relationship between rotational and translational kinetic energy can be derived. That relationship is given by the expression shown in Figure 2 .
| Relationship between rotational and translational kinetic energy. | |
|---|---|
|
A very interesting equation
The equation shown in Figure 2 is very interesting. It shows that the relationship that determines the distribution of kinetic energy between translational kinetic energy and rotational kinetic depends solely on the constant Q, and is independent of the mass, the radius, etc.
The constant Q depends on how the mass is geometrically distributed in the rolling object. Generally speaking, the more the mass is distributed toward the outer edge of the object, the greater will be the value of Q.
The total kinetic energy is given by
K = Ktr + Krot, or
K = (1/2)*M*(Vcm)^2 + (1/2)*Icm*W^2
where
Through substitution we can write
K = Ktr + Krot, or
K = Ktr + Q*Ktr, or
K = (1 + Q)*Ktr, or
K = (1 + Q)*(1/2)*M*(Vcm)^2
Thus, an object with a given mass and a given translational velocity has a total kinetic energy that is proportional to (1+Q). The larger the value of Q, the greater will be the kinetic energy possessed by the object.
From Figure 1 , a rolling hollow cylinder with a Q value of 1 would have a greater total kinetic energy than a solid cylinder with the same mass and a Q value of 0.5 rolling with the same translational velocity. This is because the cylinder would have more rotational kinetic energy.
If there were no friction, an object would not roll down an incline. Instead it would simply slide down the incline. In the absence of torque, the rotational inertia of the object would prevent it from experiencing angular acceleration.
The frictional force parallel to the incline and pointing up the incline produces a torque that causes the object to experience angular acceleration.
In this section, I will explain some example scenarios involving various aspects of rolling solid and hollow cylinders.
A cylindrical shell and a solid cylinder, each of unknown mass and unknown radii roll down an incline of unknown height and unknown angle.
Which object will have the greater translational velocity when they reach the bottom of the incline?
Solution:
From Figure 1 , the rotational inertia for a thin hollow cylinder is given by
I = M*R^2
The rotational inertia for a solid cylinder is given by
I = (1/2)*MR^2
Let the cylindrical shell be identified as M1 and the solid cylinder be identified as M2.
The total kinetic for each object at the bottom is equal to potential energy for that object at the top. Therefore,
M1*g*h = K1 = (1+Q1)*M1*(V1cm)^2, and
M2*g*h = K2 = (1+Q2)*M2*(V2cm)^2
Eliminating the mass from both sides of both equations yields
g*h = (1+Q1)*(V1cm)^2, and
g*h = (1+Q2)*(V2cm)^2
Solving each equation for the velocity yields
V1cm = ((g*h)/(1+Q1))^(1/2), and
V2cm = ((g*h)/(1+Q2))^(1/2)
Taking the ratios of the velocities yields
V1cm/V2cm = ((1+Q2)/(1+Q1))^(1/2)
Solving for V1cm in terms of V2cm yields
V1cm = V2cm * ((1+Q2)/(1+Q1))^(1/2)
From Figure 1 ,
For the cylindrical shell, Q1 = 1 and
For the solid cylinder, Q2 = (1/2)
Substituting values yields
V1cm = V2cm * ((1+(1/2))/(1+1))^(1/2)
Solving with the Google calculator yields
V1cm = V2cm * 0.866
Therefore, the translational velocity of the cylindrical shell is less than the translational velocity the solid cylinder by a factor of 0.866. This is because more of the potential energy is transformed into rotational kinetic energy in the cylindrical shell, resulting in less translational kinetic energy and less translational velocity.
Note that this result is independent of the mass of the objects, the radii of the objects, the height of the incline, and the angle of the incline.
What percentage of the total kinetic energy is translational kinetic energy for each of the cylinders in Part 1 ?
Solution:
From above ,
K = (1 + Q)*(1/2)*M*(Vcm)^2
We know from earlier modules that
Ktr = (1/2)*M*(Vcm)^2
Taking the ratio of Ktr to K gives us
Ktr/K = ((1/2)*M*(Vcm)^2)/((1 + Q)*(1/2)*M*(Vcm)^2)
Simplification yields
Ktr/K = 1/(1+Q), or
Ktr = K * (1/(1+Q))
For the cylindrical shell
Ktr = K * (1/(1+1)) = K * (1/2)
Half or 50 percent of the total kinetic energy is translational kinetic energy for the cylindrical shell.
For the solid cylinder
Ktr = K * (1/(1+Q)) = K * (1/(1+(1/2))) = K * 0.67
Sixty-seven percent of the total kinetic energy is translational kinetic energy for the solid cylinder.
A solid cylinder is rolling down an incline without slipping. What is the translational acceleration of the cylinder? How does that acceleration compare with the acceleration of the same object sliding down the incline?
Assume that
Solution:
The only forces acting on the cylinder are the force of gravity and an unknown frictional force parallel to the incline and pointing up the incline. The force of gravity does not produce a torque on the object acting about the center of mass. Therefore, the only torque is produced by the unknown force of friction.
Angular acceleration is commonly represented by the Greek letter alpha. However, your Braille display probably won't display the Greek letter alpha, so we will let angular acceleration be represented by "An" and will let translational acceleration be represented by "Atr".
We know that the relationship between torque, angular acceleration, and moment of inertia is
Sum of torques = I*An
We only have one torque. Therefore,
T = I*An
We also know that
T = R*f
where
Therefore,
T = I*An, or
An = T/I, or
An = R*f/I, or
f = An*I/R
where
As we discussed earlier, when an object is rolling with a given angular velocity, the translational velocity of the axis of rotation is proportional to the radius of the object. The greater the radius, the greater will be the translational velocity. Thus, the translational velocity is proportional to the angular velocity with the radius being the proportionality constant. We can write
Vcm = W*R
If the object is not slipping, the rate of change of the translational velocity must also be proportional to the rate of change of the angular velocity through the same proportionality constant, which is the radius.
Therefore, we can write
Atr = R*An, or
An = Atr/R
The translational acceleration is also given by the net force divided by the mass. The net force is the component of the weight pointing down the incline minus the force of friction pointing up the incline. Therefore, we can write
M*g*sin(U) - f = M*Atr
Substitution from above yields
M*g*sin(U) - (An*I/R) = M*Atr
Further substitution from above yields
M*g*sin(U) - ((Atr/R)*I/R) = M*Atr, or
M*g*sin(U) - (Atr*I/(R^2)) = M*Atr
Solving for Atr yields
M*Atr + (Atr*I/(R^2)) = M*g*sin(U), or
Atr * (M + I/(R^2)) = M*g*sin(U), or
Atr = (M*g*sin(U))/(M + I/(R^2)), or
Atr = (g*sin(U))/(1 + I/(M*R^2))
For a solid cylinder,
I = (1/2)*M*R^2
By substitution
Atr = (g*sin(U))/(1 + ((1/2)*M*R^2)/(M*R^2)), or
Atr = (g*sin(U))/(1 + (1/2)), or
Atr = (g*sin(U))/(3/2), or
Therefore, the translational acceleration of the rolling solid cylinder is given by
Atr = (2/3)*(g*sin(U))
If the cylinder were sliding in the total absence of friction, from what you learned in earlier modules, the acceleration would simply be
Atr = g*sin(U)
Therefore, if there is sufficient friction to cause the cylinder to roll without slipping, the translational acceleration will be only (2/3) of the sliding acceleration. Once again, this is the result of a portion of the potential energy being transformed into rotational kinetic energy, resulting in less translational velocity, and less translational acceleration.
I encourage you to work through the computations that I have presented in this lesson to confirm that you get the same results. Experiment with the scenarios, making changes, and observing the results of your changes. Make certain that you can explain why your changes behave as they do.
I will publish a module containing consolidated links to resources on my Connexions web page and will update and add to the list as additional modules in this collection are published.
This section contains a variety of miscellaneous information.
Financial : Although the Connexions site makes it possible for you to download a PDF file for this module at no charge, and also makes it possible for you to purchase a pre-printed version of the PDF file, you should be aware that some of the HTML elements in this module may not translate well into PDF.
I also want you to know that I receive no financial compensation from the Connexions website even if you purchase the PDF version of the module.
Affiliation : I am a professor of Computer Information Technology at Austin Community College in Austin, TX.
-end-
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This work is licensed by Richard Baldwin under a Creative Commons Attribution License (CC-BY 3.0), and is an Open Educational Resource.
Last edited by Richard Baldwin on Oct 8, 2012 1:46 pm -0500.
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