A force applied to a body that causes it to rotate creates torque. Similar to force, torque acts to angularly accelerate a spinning object (in other words, make it spin faster). The equation for torque (expressed as G here, but other textbooks may use t) looks very much like your good ol’ F = ma:

G = Ia
Instead of mass, we have rotational inertia. Instead of linear acceleration, we have angular acceleration.

Now, if a force is applied linearly to make an object move, its torque is defined as:

G = F × r
r is defined as the distance from the axis of rotation at which the force was applied. The line from the axis of rotation to the place where the force is applied is called the moment arm. Notice how this is a cross product. So, if the force was acting in line with the moment arm (either 0° or 180°), there would be no torque, and you would be moving the object only translationally.


Rotational Energy

The work that a rotating body does is also similar to the definition of translational work. Instead of force, we use torque. Instead of displacement, we use angular displacement. Thus:

W = integral(Gamma d(theta))

And since power is the rate of work, we have:

P = Gw
Note also how similar this is to power for linear motion, P = Fv.

Now, we can also extend the analogy for kinetic energy as well. Remember we used kinetic energy to derive the concept of moment of inertia in the first place? Here is the equation again:

K[rotational] = 0.5I(omega^2)

A Hoop, a Disk, and a Sphere on a Ramp VRML ModelNow that we have this, let’s make an example to put all our knowledge together. We have a hoop, a disk, and a sphere that will roll down a ramp with an angle (q) of 17° as shown on the illustration at right (click on the illustration for a VRML model of the ramp).
Each of them have the same mass and radius and will be released simultneously from rest at the top of a ramp whose length (L) is 1.5 m. How fast are each of the objects moving at the bottom of the ramp?

We can do this using a conservation of energy method. At the top, potential energy is Mgh or MgL sin q for each of them, right? They are not moving or rotating at the top so they don’t have any translational or rotational kinetic energy. So, their total energy at the beginning is just MgL sin q.

Now, after they roll down, they will lose all their potential energy but should have both translational and rotational kinetic energy. Therefore, we set the sum of the two of them equal to the energy before they roll:

MgL sin (theta) = 0.5mv^2 + 0.5I(omega^2)


Since each of the objects’ rotational inertias are different, we have to solve the above equation three times.

For the hoop, the velocity is:

Click to view Equation 5-14-hoop


For the disk:

Click to view Equation 5-14-disk


And finally for the sphere:

Click to view Equation 5-14-sphere


And we see that the sphere leaves the ramp with the highest velocity.

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