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.  The object 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 K_i = ½m_iv_i².
The total kinetic energy of the rotating object is therefore given by
K = ∑K_i = ∑½m_iv_i² = ∑½mr_i²ω² = ½ω² ∑mr_i².
We write  K = ½(∑mr_i²)ω² = ½Iω².
The quantity in parenthesis is called the moment of inertia  I = ∑m_ir_i² 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 m_i of each particle in the system by the square of its perpendicular distance r_i from the axis of rotation, and summing up all these products,  I = ∑m_ir_i².
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:

Imagine 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:

The 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:

Three 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 Σ½m_iv_i².

Solution:

• Reasoning:
  The moment of inertia is I = ∑m_ir_i².  Here r_i is the perpendicular distance of particle i from the x-axis.
  The linear speed of particle i is v_i = ωr_i.
• Details of the calculation:
  (a) 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 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:

The 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:

• Reasoning:
  The moment of inertia is I = ∑m_ir_i².  Here r_i is the perpendicular distance of particle i from the z-axis.
• Details of the calculation:
  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:

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

Solution:

• Reasoning:
  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

Consider 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 = I_CM + 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 = I_CM1 + M₁R₁², and for disk 2 we have I₂ = I_CM2² + 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

The kinetic energy of an object with translational and rotational motion is the sum of its translational and its rotational kinetic energy.
Translational kinetic energy = ½mv_CM².
Rotational kinetic energy = ½Iω².
Total kinetic energy = ½mv_CM² + ½Iω².

Consider 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 v_CM = rω. 

The kinetic energy of the wheel is the sum of the kinetic energy of the motion of the center of mass ½mv_CM² = ½mr²ω², and the kinetic energy of the motion about the center of mass, ½Iω².
Thee total kinetic energy is

KE_tot = 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 KE_tot = (3/4)mr²ω².

The ratio of the translational to the rotational kinetic energy is  E_trans/E_rot = mr²/I.
If two rolling object have the same total kinetic energy, then the object with the smaller moment of inertia has the larger translational kinetic energy and the larger speed.

Problem:

Assume a disk and a ring with the same radius roll down an incline of height h and angle theta. If they both start from rest at t = 0, which one will reach the bottom first?

Solution:

• Reason it out yourself and then watch this video clip.  Was your answer correct?

Module 8: Question 2

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!

Discuss this with your fellow students in the discussion forum!

Module 8: Question 3

Describe the energy transformations involved when a yo-yo is thrown downward and then climbs back up its string to be caught in the user's hand.

Discuss this with your fellow students in the discussion forum!