Rotational energy

If we push on an object in the forward direction while the object is moving forward, we do positive work on the object.  The object accelerates, because we are pushing on it.  F = ma.  It gains kinetic energy.  The translational kinetic energy of an object with mass m, whose center of mass is moving with speed v is K = ½mv2.

Translational kinetic energy = ½ mass * speed2

Kinetic energy increases quadratically with speed.  When the speed of a car doubles, its energy increases by a factor of four.

A rotating object has kinetic energy, even when the object as a whole has no translational motion.  If we consider the object made up of a collection of particles, then each particle i has kinetic energy Ki = ½mivi2.
The total kinetic energy of the rotating object is therefore given by
K = ∑Ki = ∑½mivi2 = ∑½mri2ω2 = ½ω2 ∑mri2.
We write  K = ½(∑mri22 = ½Iω2.
The quantity in parenthesis is called the moment of inertia  I = ∑miri2 of the object about the axis of rotation.
The moment of inertia of a system about an axis of rotation can be found by multiplying the mass mi of each particle in the system by the square of its perpendicular distance ri from the axis of rotation, and summing up all these products,  I = ∑miri2.
For a system with a continuous mass distribution the sum turns into an integral, I = ∫r2dm.
The units of the moment of inertia are units of mass times distance squared, for example kgm2.

When an object is rotating about an axis, its rotational kinetic energy is K = ½Iω2.

Rotational kinetic energy = ½ moment of inertia * (angular speed)2.

When the angular velocity of a spinning wheel doubles, its kinetic energy increases by a factor of four.

When an object has translational as well as rotational motion, we can look at the motion of the center of mass and the motion about the center of mass separately.

The total kinetic energy is the sum of the translational kinetic energy of the center of mass (CM) and the rotational kinetic energy about the CM.


The moment of inertia of an object depends on the mass of the object, and on how this mass is distributed with respect to the axis of rotation.  The moment of inertia is always defined with respect to an axis of rotation.

The farther the bulk of the mass is from the axis of rotation, the greater is the rotational inertia (moment of inertia) of the object.

Example:

imageImagine two wheels with the same mass.  One is a solid wheel with its mass evenly distributed throughout the structure, while the other has most of the mass concentrated near the rim. 

The wheel with the mass near the rim has the greater moment of inertia.

Example:

imageThe moment of inertia of a circular disk spinning about an axis through its center perpendicular to the plane of the disk differs from the moment of inertia of a disk spinning about an axis through its center in the plane of the disk.

The moments of inertia of many objects with symmetric mass distribution of about different symmetry axes can be looked up in tables.

Link:  List of moments of inertia

Problem:

imageThree particles are connected by rigid rods of negligible mass lying along the y-axis as shown.
If the system rotates about the x-axis with angular speed of 2 rad/s, find
(a)  the moment of inertia about the x-axis and the total rotational kinetic energy evaluated from ½Iω2, and
(b)  the linear speed of each particle and the total kinetic energy evaluated from Σ½mivi2.

Solution:
(a) The moment of inertia is I = ∑miri2.  Here ri is the perpendicular distance of particle i from the x-axis.
I = (4 kg)(9 m2) + (2 kg)(4 m2) + (3 kg)(16 m2) = 92 kgm2.
The rotational kinetic energy is K = ½Iω2 = 46*4/s2 = 184 J.
(b) The linear speed of particle i is vi = ωri.
The linear speed of the 4 kg mass is v = 6 m/s, and its kinetic energy is ½mv2 = 72 J.
The linear speed of the 2 kg mass is v = 4 m/s, and its kinetic energy is ½mv2 = 16 J.
The linear speed of the 3 kg mass is v = 8 m/s, and its kinetic energy is ½mv2 = 96 J.
The sum of the kinetic energies of the three particles is 184 J.

Problem:

imageThe four particles in the figure on the right are connected by rigid rods.  The origin is at the center of the rectangle.   Calculate the moment of inertia of the system about the z-axis.

Solution:
The moment of inertia is I = ∑miri2.  Here ri is the perpendicular distance of particle i from the z-axis.
Each particle is a distance r = (9 + 4)½ m = (13)½ m from the axis of rotation.
I = (3 kg + 2 kg + 4 kg + 2 kg)*13 m2 = 143 kgm2.

Problem:

imageFind the moment of inertia of a very thin hoop of mass m and radius r about its symmetry axis.

Solution:
The mass is continuously distributed, so the ∑m -->  ∫dm.  All mass elements dm are a perpendicular distance r from the axis of rotation.
I = ∫ r2dm = r2∫dm  = mr2.


Parallel axis theorem

imageConsider a composite object, such as the two joined disks in the figure on the right.
The center of mass of this object is at the origin.  To find the moment of inertia of the object about the CM we can use the parallel axis theorem.  This theorem states that the moment of inertia of an object about any axis is equal to the sum of two terms.  The first term is of the moment of inertia of the object about a parallel axes through its center of mass.  The second term is the product of the mass of the object M times the square of the distance R of its center of mass from the axis in question.

I = ICM + MR2.

We can treat the composite object as the sum of its parts, and for each part calculate the moment of inertia about the z-axis.
For disk 1 we have I1 = ICM1 + M1R12, and for disk 2 we have I2 = ICM22 + M2R22.
The moment of inertia of the composite object about the z-axis then is I = I1 + I2.
For a uniform disk of mass M the moment of inertia about an axis through its center and perpendicular to the plane of the disk is ½MR2.  For the object in the figure we therefore have for the moment of inertia about the z-axis

I = (3/2)MR2 + (3/2)MR2 = 3MR2.


Rolling

imageConsider a wheel of radius r and mass m rolling on a flat surface in the x-direction. 
The displacement Δx and the angular displacement Δθ are related through Δx = rΔθ.
The magnitudes of the linear velocity and the angular velocity are related through vCM = rω. 
The kinetic energy of the disk is the sum of the kinetic energy of the motion of the center of mass ½mvCM2 = ½mr2ω2, and the kinetic energy of the motion about the center of mass, ½Iω2.
Thee total kinetic energy is

KEtot = ½mr2ω2 + ½Iω2 = ½[mr2 + I]ω2 = ½[m + I/r2]v2.

Example:

Assume the wheel is a uniform disk.  The moment of inertia I of a uniform disk about an axis perpendicular to the plane of the disk through its CM is ½mr2
The kinetic energy of the disk therefore is KEtot = (3/4)mr2ω2.

Question:

Suppose you are designing a race bicycle and it comes time to work on the wheels.  You are told that the wheels need to be of a certain mass but you may design them either as wheels with spokes (like traditional bike wheels) or you may make them as having solid rims all the way through.  Which design would you pick given that the racing aspect of the machine is the most important?  Please explain!