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15. Parallel Axis Theorem and Torque
A) Overview
In this unit we will continue our study of rotational motion. In particular, we will first prove a very useful theorem that relates moments of inertia about parallel axes. We will then move on to develop the equation that determines the dynamics for rotational motion. In so doing, we will introduce a new quantity, the torque, that plays the role for rotational motion that force does for translational motion. We will find, once again, that the rotational analog of mass will be the moment of inertia.
B) Parallel Axis Theorem
We have shown earlier that the total kinetic energy of a system of particles in any reference frame is equal to the kinetic energy of the center of mass of the system, defined to be one half times the total mass times the square of the center of mass velocity, plus the kinetic energy of the motion of all of the parts relative to the center of mass.
K system , lab = KREL + K CM
For a solid object the only possible motion relative to the center of mass is rotation. Therefore, the kinetic energy relative to the center of mass is just equal to one half times the moment of inertia about a rotation axis through the center of mass times the square of the angular velocity about the center of mass. 2 2
(^21) 2
1
K system , lab = ICM ω CM + MvCM
We can use this result to calculate the moment of inertia about a chosen axis if the moment of inertia about a parallel axis that passes through the center of mass is known.
For example, Figure 15.1 shows the thin rod from the last unit rotating about an axis perpendicular to the page at a distance D from the center of mass. We can describe this motion, as the center of mass rotating about the axis with angular velocity ω, plus the rod rotating about its center of mass with that same angular velocity ω. Thus, for every rotation the center of mass makes about the axis, the rod makes one revolution about its center of mass.
We start from our previous result that the total kinetic energy about the chosen axis can be written in terms of the moment of inertia about that axis. 2 2
(^1) ω
K total = Itotal
Now we know this kinetic energy must be equal to the sum of the kinetic energy of the center of mass and the kinetic energy relative to the center of mass. 2 2
(^21) 2
(^1) ω
K total = MvCM + ICM
Figure 15.
A thin rod rotates with angular velocity ω
about an axis perpendicular to the page located a distance D from the center of mass of the rod.
Now the velocity of the center of mass in the lab frame is just equal to the product of D
and ω , and the kinetic energy relative to the center of mass is just equal to ½ the product
of the moment of inertia about this parallel axis and the square of the angular velocity. 2 2
(^21) 2
K total = M D + ICM
We can cancel a factor of one half omega squared from each term in the kinetic energy equation to determine that the total moment of inertia about the chosen axis is just the moment of inertia about a parallel axis passing through the center of mass plus the moment of inertia of the center of mass, treated as a point particle, about the chosen axis.
I total = MD + I CM
2
This result is known as the parallel axis theorem and it will prove to be very useful in the next few units. You can verify that this parallel axis theorem predicts the relationship we obtained last time for the two moments of inertia for a thin rod.
C) Example: Moment of Inertia of a Dumbbell
We can now use this parallel axis theorem to calculate the moment of inertia of a dumbbell made up of two solid spheres connected by a solid rod about an axis that is perpendicular to the rod and passes through its center as shown in Figure 15.2.
We’ll start by using our result from the last unit that the moment of inertia of the dumbbell about the given axis is equal to the sum of the moments of inertia of its components, the rod and the two spheres, about that same axis. To find the moments of inertia of the spheres about the axis through the center of the rod, we will apply the parallel axis theorem we developed in the last section.
Namely, we know that the moment of inertia of a solid sphere about an axis passing through its center is equal to 2/5 the product of its mass and the square of its radius.
2 ,
I sphere CM = MR
Applying the parallel axis theorem, we see that the moment of inertia of each sphere about the given axis is just given by:. 2 2 ,
L
I sphereaxis MR M
To find the total moment of inertia of the dumbbell, we just add the contributions from the two spheres to the moment of inertia of the thin rod about its center to get the result:
= + +
2 2 2 5 2
2 2 12
1 L I (^) dumbbell mrodd MR M
Figure 15. A dumbbell composed of two spherical masses of radius R and mass M separated by a distance d are connected by a rod with mass mrod.
door. To close this door, you need to push on it. Here we see a force F being applied a distance r from the hinge. The door is heavy and barely moves as you push on it. What would you change about the way you are pushing in order to close it quicker? Your intuition tells you that you would either push harder, or you would push on the door at a point further from the hinge. You would certainly not try pushing closer to the hinge, and you would also not change the direction you were pushing to be more parallel to the door. In other words, the biggest effect you can have on the door is to push on it as hard as possible, in a direction perpendicular to the door, at a point far from the hinge.
The equations we derived in the previous section tell us exactly the same thing! We expect the angular acceleration to be biggest when the torque is biggest – in other words the door will close fastest when the torque is large. The torque is largest when the distance r between the axis and the perpendicular force is biggest, and when the perpendicular force itself is biggest. If we push parallel to the door, there is no perpendicular force component and we don’t expect the door to start moving at all!
In the next section we will generalize this description of rotations by defining torque, angular velocity, and angular acceleration as vector quantities and by defining the cross product of two vectors..
F) Torque and the Cross Product
Figure 15.5 shows a spinning top. So far, we have described this motion in terms of one dimensional variables, the angular displacement, the angular velocity and the
Figure 15. An overhead view of a closing door. The force F applies a torque about the hinge which causes an angular acceleration.
Figure 15. A top spins on its axis. A counterclockwise rotation is defined to be positive.
angular acceleration, all defined relative to the axis of rotation. We will need to generalize this description to three dimensions when we allow the direction of the rotation axis itself to change.
If a rotating object is viewed in a reference frame in which the rotation axis is perpendicular to the page, as shown in Figure 15.5, it is conventional to define a counter- clockwise rotation as positive and a clockwise rotation as negative. We adopt this convention in order to match the usual measurement of the angle theta relative to the x- axis in a right-handed Cartesian coordinate system.
There is a simple rule involving the right hand that can be used to define the directions of the angular velocity vector. Namely, if you curl the fingers of your right hand in the direction of rotation of the object, your thumb will point in the direction of the angular velocity vector of the object. In other words, if some object is spinning in the counter-clockwise direction in the x-y plane, curling the fingers of your right hand in this direction results in your thumb pointing in the + z direction which we define to be the direction of the angular velocity vector. Similarly, if the object is rotating in the clockwise direction, the same exercise results in your thumb pointing in the – z direction.
We use the same right hand rule to define the direction of the angular acceleration. For example, if the magnitude of the angular velocity increases in time, then the angular acceleration vector has the same direction as that of the angular velocity. If the magnitude of the angular velocity decreases in time, then the angular acceleration vector has the opposite direction as that of the angular velocity.
One nice feature of using the right hand rule to define the direction of vectors in systems involving rotation is that these vectors then do not rely on the choice of a coordinate system at all; the rotation vector becomes a property of the object itself.
Now that we have defined the direction of the angular velocity and angular acceleration vectors, we will now define the torque as a vector quantity and obtain the vector equation that determines rotational dynamics.
We have shown that the magnitude of the torque is given by the product of the length of the line connecting the axis to the point where the force acts, r , and the component of the force F that is perpendicular to this line. If we define θ as the angle
Figure 15. The right hand rule for determining the direction of the angular veloicty
ω: Curl your fingers in the direction
of rotation and your thumb will
point along the direction of ω.
