11.1. Rotational variablesIn this chapter we will be dealing with the rotation of a rigid body about a fixed axis. Every point of the body moves in a circle, whose center lies on the axis of rotation, and every point experiences the same angular displacement during a particular time interval.
Figure 11.1. Relation between s and [theta].Suppose the z-axis of our coordinate system coincides with the axis of rotation of the rigid body. The x-axis and the y-axis are taken to be perpendicular to the z-axis. Each part of the rigid body moves in a circle around the z-axis. Suppose a given point A on the body covers a linear distance s during the rotation (see Figure 11.1). During one complete revolution point A covers a distance equal to 2[pi]r. In that case, the angle of rotation is equal to 2[pi] radians. For the situation shown in Figure 11.1, the angle of rotation can be easily calculated:
If the angle of rotation [theta] is time dependent, it makes sense to introduce the concept of angular velocity and angular acceleration. The angular velocity [omega] is defined as
In order to describe rotation around a point (rather than a fixed axis) the concept of an angular velocity vector is introduced. The magnitude of the angular velocity vector is equal to the absolute value of the angular velocity for rotation around a fixed axis (as defined above). The direction of the velocity vector is parallel to the rotation axis and the right-hand rule needs to be used to determine whether the vector points upwards or downwards.
A wheel rotates with an angular acceleration a given by
To solve this problem, we start with looking at the relation between the angular acceleration and the angular velocity
11.2. Constant angular accelerationIf the angular acceleration a is constant (time independent) the following equations can be used to calculate [omega] and [theta] at any time t:
A wheel starting from rest, rotates with a constant angular acceleration of 2.0 rad/s2. During a certain 3.0 s interval it turns through 90 rad. (a) How long had the wheel been turning before the start of the 3.0 s interval ? (b). What was the angular velocity of the wheel at the start of the 3.0 s interval ?
Time t = 0 s is defined as the moment at which the wheel is at rest. Therefore, [omega]0 = 0 rad/s. The rotation angle at any later time is measured with respect to the position of the body at time t = 0 s: [theta]0 = 0 rad. The equations of rotation are now given by
11.3. Relation between linear and angular variablesAn example of the relation between angular and linear variables has already been discussed. Figure 1 illustrates how the distance s, covered by point A, is related to the radius of the circle and the angle of rotation
Figure 11.2. Components of the acceleration of point A.We can conclude that when a rigid body is rotating around a fixed axis, every part of the body has the same angular velocity [omega] and the same angular acceleration a, but points that are located at different distances from the rotation axis have different linear velocities and different linear accelerations.
11.4. Kinetic energy of rotationThe total kinetic energy of a rotating object can be found by summing the kinetic energy of each individual particle:
11.5. Calculation of rotational inertiaTo calculate the moment of inertia of a rigid body we have to integrate over the whole body
Sample Problem 11-8
Determine the moment of inertia of a uniform rod of mass m and length L about an axis at right angle with the rod, though its center of mass (see Figure 11.3).
The mass per unit length of the rod is m/L. The mass dm of an element of the rod with length dx is
Figure 11.3. Sample Problem 11-8.
Figure 11.4. Sample Problem 11.8.Example: Moment of Inertia of Disk
Figure 11.5. Moment of inertia of a disk.A uniform disk has a radius R and a total mass M. The density of the disk is given by
11.6. TorqueSuppose a force F is applied to point A (see Figure 11.6). Point A is part of a rigid body with an axis of rotation going through the origin. Suppose the angle between the force F and the position vector r is [phi]. The force F can be decomposed into two components: one parallel to the position vector and one perpendicular to the position vector. It is obvious that the component parallel to the position vector can not cause a rotation of the rigid body.
The magnitude of the component of the force perpendicular to the position vector is given by
Figure 11.6. TorqueThe tangential component of the applied force F will produce a rotation of the object; the actual angular velocity will depend not only on the applied force but also on the distance between the axis of rotation and point A. To describe the effect of the force, the concept of torque is introduced.
Sample Problem 11-11
Figure 11.7 shows a uniform disk with mass M and radius R. The disk is mounted on a fixed axle. A block with mass m hangs from a light cord that is wrapped around the rim of the disk. Find the acceleration of the falling block, the angular acceleration of the disk, and the tension of the cord.
Figure 11.7. Sample Problem 11-10.The forces acting on the mass and the disk are shown in Figure 11.8. Since the mass m is moving downward, the gravitational force m . g must exceed the tension T in the cord. The linear acceleration of the mass m is defined to be positive if it points down:
Figure 11.8. Sample Problem 11-10.The moment of inertia of the disk is given by
Figure 11.9. Work done by a force.Suppose a particle with mass m is connected to the end of a rod (with negligible mass). Under the influence of a force F, the system rotates through an angle [Delta][theta] (see Figure 11.9). The work done by the force F is determined by the tangential component of F