In problems involving linear acceleration, we found that the resultant force is directly proportional to the linear acceleration with the constant of proportionality being the mass of the object; i.e. F = ma. It can be demonstrated experimentally that the resultant torque on an object is proportional to its angular acceleration. However, the constant of proportionality is not a simple property of the object. It will depend on both the mass of the object, where the axis is chosen, and the distribution of the mass. A simple example will serve to clarify this.

Suppose that we have a massless disk on a frictionless axis through its center. Suppose that there is a small object of mass m, located at a point on the rim of the disk at a distance R from the axis of rotation. Now let us apply force on the mass tangent to the disk. At the moment of application of the force, the object will accelerate in the direction of the force with 

As soon as the object starts to move, the disk will rotate, and we must change the direction of the force if it is to remain tangent to the disk and keep accelerating the object with the same acceleration. The easiest way to do this is to wrap a string around the rim of the disk and pull the end of hte string in a straight line.

Now let us change our notation to angular notation. If we multiply Equation 10 by R on both sides we obtain:

But torque, τ, is force times radius, and linear acceleration is angular acceleration, a, multiplied by the radius. Therefore, we have:

If we call the quantity mR², the moment of inertia of the object, we can rewrite equation 12 as:

Clearly this is a very special object with all its mass concentrated at a single point. In general an object will have its mass distributed over a larger area. In this case it can be shown that the moment of inertia can be determined by mentally breaking the object into a large number of small point masses and adding the quantity mR² for each to find the moment of inertia for the object. Normally this is a problem for a calculus student so we will not persue this topic any further here. Formulae for the moment of inertia for objects with simple shapes are readily available and some are shown in Table 6.3.1.

^{Table 6.3.1}

Once we have the concept of the moment of inertia, our study of rotational motion can be extended to include topics such as rotational kinetic energy and rotational (or angular) momentum. It is possible to define almost all rotational quantities by analogy to the corresponding linear quantities. For example, linear work is force multiplied by distance moved in the direction of the force. By analogy, rotational work will be equal to torque multiplied by the angle moved in the direction of the torque. Linear power is either linear work divided by time, force multiplied by distance divided by time, or force multiplied by velocity. Rotational power will be either rotational work divided by time, torque multiplied by angle moved divided bytime, or torque multiplied by angular velocity. Since linear kinetic energy is mv², rotational kinetic energy must be Iω², and must it be a scalar. Finally rotational or angular momentum will be Iω, and must be a vector which points in the direction of the angular velocity. Table 6.3.2 summarizes these relationships.

^{Table 6.3.2}

Sample Problem #1

A hoop of radius 3 m weighs 1600 N.  Its center of mass has a speed of 0.50 m/s as it rolls across a horizontal surface.  How much work must be done to stop it? 

Sample Solution #1

The mass of the object is found using:

The moment of inertia, I, can be found from the Table 2 above:

The angular velocity, ω can be found as follows:

The work done to stop the hoop is equal to the translational kinetic energy plus the rotational kinetic energy of the hoop:

1.

An automobile engine develops 100 HP when rotating at a speed of 1800 rev/min. What torque does it develop?

2.

If the earth were a sphere of uniform density, what would its rotational kinetic energy be? (Note: Earth's radius is 6.4x10³ km and its mass is 6.0 x 10²⁴ kg.)

3.

Suppose earth's rotational kinetic energy could be could be harnessed for man’s use. For how many years could the earth supply 1 Kw of power to each of the 4 X 10⁹ people on the earth?

4.

A box 2 m high by 1 m wide by 1 m deep contains a refrigerator. The box is sitting vertically in the back of a truck. The weight of the refrigerator and box is 1500 N. The box is tipped over by the truck's acceleration. What is the minimum value that this acceleration must have had? (Assume the box's weight is uniformly distributed)

5.

The center of mass of a car is 2.5 ft above the road. The width between the wheels is 4.5 ft. If the car races around an unbanked curve with 100 ft radius, and does not skid, what is the largest speed possible before the car overturns?

6.

A meter stick is held vertically with one end on the floor. The stick is released and allowed to fall. Find the speed of the other end when it hits the floor. (Assume that the end on the floor does not slip.)

7.

By what amount would the diameter of the earth need to shrink in order to speed its rotation by one second per day?

8.

A man is on a frictionless rotating platform which is rotating with a speed of 1 rev/sec. His hands are outstretched and he holds a weight in each hand. With his hands in this position the total rotational inertia of the man and platforms is 6 kg-m². If by drawing in the weights the man decreases the rotational inertia to 2 kg-m². What is the resulting angular speed of the platform?

9.

In the previous problem, by how much is the kinetic energy increased?

10.

Where does the extra kinetic energy in the previous problem come from?

11.

A wheel with a moment of inertia of 5.5 kg·m² is rotating with an angular speed of 500 rev/min on a shaft whose rotational inertia is negligible. A second wheel with a moment of inertia of 2.25 kg·m², initially at rest, is suddenly coupled to the same shaft. What is the angular speed of the resulting combination?