Moment of Inertia

Class Notes Physics I, 11/16/98

Moment of Inertia

The quiz problem concerned the moment of inertia of three concentric hoops, as depicted below, with radii .4 m, .8 m and 1.2 m and respective masses 2 kg, 4 kg and 6 kg. We were also to determine the moment of inertia and the acceleration resulting from the application of a force of 8 Newtons in the most advantageous direction upon the first loop.

The most advantageous direction is at a point where the force is perpendicular to a radial line, as indicated in the figure.
This force will also be tangent to the hoop.

The moment of inertia is as always the sum of all the m r^2 contributions of all the point masses.

Since every point of the 2 kg hoop lies a distance of .4 meters from the common center of the hoops, the total m r^2 contributions from this hoop is 2 kg (.4 m)^2.

By similar reasoning, the other three hoops contribute 4 kg (.8 m)^2 and 6 kg (1.2 m)^2 to the moment of inertia.
Calculating these quantities, we find that the contributions of the three hoops are respectively .32 kg meters ^ 2, 2.56 kg meters ^ 2, and 8.64 kg meters ^ 2.
The moment of inertia is therefore the total, which is 11.5 kg meters ^ 2.

To determine the acceleration we use Newton's Second Law for angular quantities.

The torque resulting from the application of the 8 Newton force, since the force is exerted at a perpendicular to the radial line from the axis of rotation, is .4 m * 8 N = 3.2 m N.

(Note that we write the units as m N rather than N m to distinguish torque from work/energy; this is not absolutely necessary but is a good idea then we will follow it here).

Newton's Second Law for angular quantities tells us that angular acceleration `alpha is equal to the torque divided by the moment of inertia (`alpha = `tau / I), so that here the angular acceleration is `alpha = `tau / I = 3.2 m N /(11.5 kg m^2) = .28 rad/s^2.

http://youtu.be/AhhCIdDbWEA

We extend our analysis to the situation for the force is exerted at an angle of 60 degrees to the most advantageous direction.

In the figure below we see how this force is equivalent to its two components 8 N cos(30 deg) = 7 N (approximately) and 8 N sin(30 deg) = 4 N in the direction of the radial line and perpendicular to this direction, respectively.
Only the component perpendicular to the radial line has a rotational effect.

As a result we find that the torque of this force is .4 m (4 Newtons) = 1.6 m Newtons.

The resulting acceleration is therefore .14 radians/second/second.

We next find the kinetic energy of the system under the influence of the original force after the time `dt has elapsed, with the system initially at rest.

We recall that the acceleration in this case was .28 radians/second/second.
From this we conclude that the angular velocity attained was `omega = .28 rad / s / s * `dt.
The kinetic energy at this angular velocity is KE = sum(m v^2).
The velocity of any point on the innermost hoop is v = r `omega = .4 m (`omega); the velocity for a point on the second hoop is similarly found to be v = .8 m (`omega), while the same quantity for a point on the third hoop is v = 1.2 m (`omega).

The kinetic energy is the total of all these 1/2 m v^2 contributions, as shown below.

We see that the kinetic energy of the first hoop is .16 kg m^2 * `omega^2, which as indicated in the fifth line of the figure below is the product of half of the moment of inertia of this hoop and the square of the angular velocity (we designate the angular velocity of the first hoop by I1).
Similarly we see that the kinetic energies of the other two hoops are respectively 1.28 kg m^2 `omega^2 = 1/2 I2 `omega^2 and 4.31 kg m^2 * `omega^2 = 1/2 I3 `omega^2.
The total kinetic energy is 5.8 kg meters ^ 2 * `omega^2, which is equal to 1/2 I `omega^2, where I is the total moment of inertia of the system.

http://youtu.be/EwuY0d3GyXA

We next construct an approximate model of the moment of inertia of a uniform circular disk of radius R and mass M.

Our model will divide the disk into a central disk and two rings defined by the circles of radius 1/3 R, 2/3 R and R.
We first note that the mass of a the disk of radius 1/3 R must be 1/9 of the mass of the entire disk, since a circle of radius 1/3 R must have area 1/9 that of the entire circle (since area is proportional to the square of radius).

Similarly, a disk of radius 2/3 R must have (2/3)^2 = 4/9 the area of the entire disk, and therefore 4/9 the mass of the entire disk.

Thus the masses of the inner disk and the disk of radius 2/3 R will be 1/9 M and 4/9 M.

The mass of the middle ring is the difference between the mass of the inner disk and the mass of the disk of radius 2/3 R, or 4/9 R - 1.9 R = 3/9 R = 1/3 R.

The mass of the outer ring is easily found by similar reasoning to be 5/9 M.

We will model the central disk and the two rings by averaging the inner and outer radii of each and assuming that the total mass of each is concentrated at this radius.

This assumption is not completely accurate, since among other things there is more mass on the far side of this radius than on the near.
However, as we will see, we obtain a pretty good approximation of the actual moment of inertia.

The resulting '3-hoop' model therefore has masses M/9 at distance R/6 from the center of rotation, 1/3 M at distance R/2, and 5/9 M at distance 5R / 6.

The resulting moment of inertia of each hoop is easily calculated, as indicated below, by multiplying its mass by the square of its distance from the center of rotation (i.e., by finding its mr^2).

Adding up these masses (using a common denominator of 324) we obtain total moment of inertia 158 / 324 * m R^2, or about .485 M R^2.

As we will see, this compares very favorably with the precise moment of inertia .5 M R^2, found by dividing the circle into more and more rings and taking the limiting value of the resulting moments of inertia.

http://youtu.be/pQhIsGkuKBk

We now determine the moment of inertia of a ring in this disk whose average radius is r and whose width is `dr.

To determine the moment of inertia of this ring, we need only find its mass then determine m r^2.
If this ring was cut out then 'unrolled', its top surface would form a shape which is very nearly a rectangle of length 2 `pi r equal to its circumference and width `dr.
The area of this rectangle would therefore be (2 `pi r `dr) / (`pi R^2) of the area of the entire disk.
Its mass would therefore be in the same proportion to the total mass of the disk, so we would have `dm = (2 `pi r `dr) / (`pi R^2) M.

This much we can understand without calculus. We now make a brief digression to the calculus to obtain the total moment of inertia.

We must sum the `dm contributions for all rings, for radius 0 to radius R.
As we allow the width `dr to approach zero, the sum of all the `dm contributions approaches the integral of `dm * r^2, as indicated below.
This integral is easily evaluated and the result is I = M R^2 / 2, or I = .5 M R^2.
Compare this precise result with the I = .485 M R^2 approximation obtained for only three rings (with `dr = R / 3).

http://youtu.be/y66JYSgg_gE

Without going into detail (the details are in the Physics 241 text), we give in the figure below the moments of inertia for

a sphere rotating about an axis through its center I = 2/5 M R^2, somewhat less than the 1/2 M R^2 for a disk because the mass of a sphere is more concentrated toward its center, where r is less;
a uniform rod rotating about its center (I = 1/12 M L^2); and
a uniform rod rotating about one of its ends (I = 1/3 M L^2

note that this is 4 times the moment of inertia of a rod rotating about its center, since when the rod is rotated about one of its ends each point mass is an average of twice as far from the point of rotation as when the rod rotates about its center, so that the average r^2 will be 4 times as great.

http://youtu.be/wD58uLBgwAE

We can now understand the mysterious 5/7 that was invoked when we considered the kinetic energy of a ball rolling without slipping down an incline.

Recall that we said that the kinetic energy of the motion of the ball through space was, even in the absence of friction and other dissipative forces, only 5/7 of the PE loss of the ball rather than being equal to the entire PE loss.
At that time we said that the other 2/7 of the potential energy loss went into rotational kinetic energy.

In the figure below we see that if they sphere roles over a surface without slipping, then when its center of mass moves a distance `dx, the point of contact on the rim of the sphere must move an equal distance.

It follows that the velocity of a point on the outer rim of the sphere must be moving at the same velocity v as the center of mass of the sphere.

Since the angular velocity of a rotating object is equal to v / r, where v is a velocity at distance r from the center, we see that this sphere has angular velocity `omega = v / R.

Thus the sphere has translational kinetic energy 1/2 M v^2 and rotational kinetic energy 1/2 * `omega^2 = 1/2 (2/5 M R^2) (v/R)^2 = 1/5 M v^2.

The total kinetic energy of the sphere is therefore seen to be 7/10 M v^2.

Its translational kinetic energy is 1/2 M v^2, which is 1/2 / (7/10) = 5/7 of its total kinetic energy.

By conservation of energy, in the absence of external forces the total kinetic energy is equal to the change in potential energy.

We therefore see that under these conditions, the translational kinetic energy is therefore 5/7 of the potential energy loss.

http://youtu.be/pbxJsGYFlqA