If your solution to stated problem does not match the given solution, you should self-critique per instructions at

 

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

 

Your solution, attempt at solution. 

 

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it.  This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

 

032.  Moment of inertia

 

Question:  `q001.  Note that this assignment contains 4 questions.

 

The moment of inertia of a concentrated mass m lying at a distance r from  the axis of rotation is m r^2.  Moments of inertia are additive--that is, if an object with a moment of inertia about some axis is added to another object with  its moment of inertia about the same axis, the moment of inertia of the system about that axis is found by simply adding the  moments of inertia of the two objects.

 

Suppose that a uniform steel disk has moment of inertia .0713 kg m^2 about an axis through  its center and perpendicular to its plane.  If a magnet with mass 50 grams is attached to the disk at a point 30 cm from the  axis, what will be the moment of inertia of the new system?

 

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence rating:
 

Given Solution: 

A mass of m = .05 kg at distance r = .30 meters from the axis of rotation has moment of inertia I = m r^2 =  .05 kg * (.30 m)^2 = .0045 kg m^2.

 

The moment of inertia of the new system will therefore be the sum .0713 kg m^2 + .0045 kg  m^2 = .0758 kg m^2 of the moments of inertia of its components, the disk and the magnet.

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique rating:
 

Question:  `q002.  A uniform rod with mass 5 kg is 3 meters long.  Masses of .5 kg are added at the ends and at .5 meter intervals along the rod.  What is the moment  of inertia of the resulting system about the center of the rod?

 

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence rating:
 

Given Solution: 

The rod itself, being rotated about its center, has moment of inertia 1/12 M L^2 = 1/12 * 5 kg * (3 m)^2 =  3.75 kg m^2.

 

The added masses are at distances 1.5 meters (the two masses masses on the ends), 1.0 meters (the two masses .5 m  from the ends), .5 meters (the two masses 1 m from the ends) and 0 meters (the mass at the middle of the rod) from the center of  the rod, which is the axis of rotation.

 

At 1.5 m from the center a .5 kg mass will have moment of inertia m r^2 = .5 kg * (1.5 m)^2 =  1.125 kg m^2; there are two such masses and their total moment of inertia is 2.25 kg m^2.

 

The  two masses lying at 1 m from the center each have moment inertia m r^2 = .5 kg * (1 m)^2 = .5 kg m^2, so the total of the two masses is double is, or 1 kg  m^2.

The two masses lying at .5 m from the center each have moment of inertia m r^2 = .5 kg ( .5 m)^2 = .125 kg  m^2, so their total is double this, or .25 kg m^2.

 

The mass lying at the center has r = 0 so m r^2 = 0; it therefore makes no  contribution to the moment of inertia.

 

The total moment of inertia of the added masses is therefore 2.25  kg m^2 + 1 kg m^2 + .25 kg m^2 = 3.5 kg m^2.  Adding this to the he moment of inertia of the rod itself, total moment of inertia is 3.75 kg m^2 + 3.5 kg m^2 = 7.25 kg m^2.

 

We note that the added masses, even including the one  at the center which doesn't contribute to the moment of inertia, total only 3.5 kg, which is less than the mass of the rod; however  these masses contribute as  much to the moment of inertia of the system as the more massive uniform rod.

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique rating:
 

Question:  `q003.  A uniform disk of mass 8 kg and radius .4 meters rotates about an axis through its center and perpendicular to its plane.  A uniform rod with mass 10  kg, whose length is equal to the diameter of the disk, is attached to the disk with its center coinciding with the center of the disk.  The system is subjected to a torque of .8 m N.  What will be its acceleration  and how to long will it take the system to complete its first rotation, assuming it starts from rest?

 

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence rating:
 

Given Solution: 

The moment of inertia of the disk is 1/2 M R^2 = 1/2 * 8 kg * (.4 m)^2 = .64 kg m^2.  The rod will be  rotating about its center so its moment of inertia will be 1/12 M L^2 = 1/12 * 10 kg * (.8 m)^2 = .53 kg m^2 (approx).

 

 ( Note that  the rod, despite its greater mass and length equal to the diameter of the disk, has less moment of inertia.  This can happen because  the mass of the disk is concentrated more near the rim than near the center (there is more mass in the outermost cm of the disk than in the innermost cm), while the mass of the rod is concentrated the same from cm to cm. ).

 

The  total moment of inertia of the system is thus .64 kg m^2 + .53 kg m^2 = 1.17 kg m^2.  The acceleration of the system when subject to a .8 m N torque will  therefore be

 

 

To find the time required to complete one revolution from rest we note that the initial angular velocity is 0, the angular displacement is 1 revolution or 2 `pi  radians, and the angular acceleration is .7 rad/s^2.  By analogy with `ds = v0 `dt + 1/2 a `dt^2, which for v0=0 is `ds = 1/2 a `dt^2,  we write in terms of the angular quantities `d`theta = 1/2 `alpha `dt^2 so that

 

`dt = +- `sqrt( 2 `d`theta / `alpha )

= +-  `sqrt( 2 * 2 `pi rad / (.7 rad/s^2))

= +-`sqrt( 12.56 rad / (.7 rad/s^2) ) = +-4.2 sec.

 

We choose the positive value of `dt, obtaining `dt = +4.2 sec..

 

INSTRUCTOR'S FURTHER CLARIFICATION

 

You have the acceleration and you know how far the system travels from rest.

If the system was accelerating along a line these quantities would be a, `ds and v0. The third equation of uniformly accelerated motion (`ds = v0 `dt + 1/2 a `dt^2) would apply, with `dt as the unknown. Since v0 = 0 the equation becomes `ds = 1/2 a `dt^2.

That equation isn't appropriate here because this is a rotating system. So instead of using `ds we use `dTheta, which is the symbol for angular displacement, and instead of a we use alpha, the symbol for angular acceleration.

Our equation therefore translates to

`dTheta = 1/2 alpha `dt^2.

Now using alpha = .7 rad/s^2 and `dTheta = 2 pi radians, we solve for `dt.

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique rating:

 

 

 

If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily.  If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.

 


 

Question:  `q004.  To a uniform rod of length 30 cm and mass 80 grams, initially at rest and constrained to rotate about an axis through its center, we add a 20-gram domino at one end and a 50-gram magnet on the other side of the axis, at a distance of 6 cm from the axis.

 

Show that the rod will balance about its center.

 

If the rod is subject to a torque of .03 m N for 5 seconds, what will be its angular velocity?

 

What centripetal force will be required to keep the domino moving around its circular path?  Answer the same for the magnet.

 

What will be the kinetic energy of the domino?  Answer the same for the magnet.

 

Your Solution:

 

Confidence Rating:

 

Self-critique (if necessary):

 

Self-critique rating: