CHAPTER 10

ROTATIONAL MOTION

Prof. Ilias Perakis

Introduction

So far in this course we studied the motion of point-like objects. However, objects have finite size. As we discussed in the previous Chapter, the translational overall motion of an object can be approximated by that of a pointlike particle located at the center of mass. However, an extended system can also rotate without any displacement of its center-of-mass.

In this Chapter we study the rotational motion of finite-sized systems consisting of many point-like particles. These can be either rigid objects (which rotate without distorting their shape) or many disconnected particles (e.g. the stars of a galaxy rotating around their common center). In the case of rotational motion, the point-like particles comprising the system move in circles whose centers lie on the rotation axis. Starting from Newton's second law and using what we learned about circular motion, we develop here the equations that describe rotational motion.

Kinematics of rotational motion

In order to describe the motion of a pointlike particle, we introduced the concepts of displacement, velocity, and acceleration. Similarly here we seek to describe the motion of a rotating rigid body, which consists of many pointlike particles moving together. Such particles move on circles on a plane perpendicular to the rotation axis. Therefore, it is more convenient to describe their positions in terms of polar coordinates (recall Chapter 1) $(r,\theta$), where r is the distance of a particle from the center of its circular path and $\theta$ is the angle by which the entire system has turned. During rotational motion, every single particle of a rigid body turns by the same angle.

We can fully characterize the rotational motion around a fixed axis by introducing the concepts of angular displacement $\theta(t)$ (i.e. the angle by which the object has turned), (instantaneous) angular velocity $\omega(t)= \frac{d \theta(t)}{dt}$(the rate at which the object is turning), and (instantaneous) angular acceleration $\alpha(t) = \frac{d \omega(t)}{dt}$.Such angular quantities can be related to the speed v(t), the centripetal (radial) acceleration a_r(t), and the tangential acceleration a_t(t) of a particle at a distance r from the rotation axis (recall Chapter 3 for the definitions of these quantities). A pointlike particle that rotates by an angle $\Delta \theta$ on a circle of radius r moves by a distance of $\Delta s = r \Delta \theta$ and therefore its speed (velocity magnitude) $v = \frac{ \Delta s}{\Delta t}$ can be expressed in terms of the angular velocity $\omega(t)$ as follows:

$$v(t) = r \ \frac{ \Delta \theta}{\Delta t} = r \ \omega(t).$$

By taking the time derivatives of both sides of the above equation we can express the tangential acceleration $a_{t} = \frac{ d v(t)}{dt}$in terms of the angular acceleration $\alpha(t)$:

$$a_{t}(t) = \ \alpha(t)\ r.$$

Using the above expression for the speed we obtain for the radial acceleration $a_{r}= \ \frac{v^{2}}{r}$

$$a_{r}(t) = \ \omega^{2}(t) \ r.$$

It is important to remember that all the constituent particles of a rigid body rotating around a fixed axis move by the same angle, angular speed, and angular acceleration, while their speed and radial and tangential accelerations depend on their distance r from the axis.

The simplest form of rotational motion corresponds to a constant angular acceleration $\alpha$.In this case, we obtain by integrating both sides of the equation that defines the angular acceleration, $d \omega = \alpha \ dt$,

$$\omega(t) = \omega_{0} + \alpha t,$$

where $\omega_{0}$ is the angular velocity at time t=0. By integrating both sides of the above equation we obtain for the angular displacement

$$\theta(t)= \theta_{0} + \omega_{0} t + \frac{1}{2} \alpha t^{2},$$

where $\theta_{0}$ is the angle at time t=0. Finally, by eliminating the time variable from the above two equations, we obtain similar to Chapter 2

$$\omega^{2} = \omega_{0}^{2} + 2 \alpha (\theta - \theta_{0} ).$$

Notice the similarity to the Chapter 2 kinematics equations for a pointlike particle ($x \rightarrow \theta, v \rightarrow \omega, a \rightarrow \alpha$).

Kinetic energy of a rotating rigid body

The kinetic energy K of an extended object is equal to the sum of the kinetic energies of all the pointlike particles that make up the object: $K = \sum_{i} \frac{1}{2} m_{i} v_{i}^{2},$where m_i and v_i denote the mass and the speed of the i-th particle. In the case of rotational motion, the latter moves on a circle of radius r_i (distance from the rotation axis) at an angular speed of $\omega$ (same for all the particles). Substituting $v_{i} = \omega \ r_{i}$into the above equation we obtain for the rotational kinetic energy

$$K = \omega^{2} \sum_{i} \frac{1}{2} m_{i} r_{i}^{2} = \frac{1}{2} I \omega^{2},$$

where

$$I =\sum_{i} m_{i} r_{i}^{2}$$

is the moment of inertia (more on this later).

Torque and vector product

We now move to the dynamics of rotations. We know that a net force ${\bf F}$ acting on a pointlike object leads to an acceleration ${\bf a}= \frac{1}{m} {\bf F}$.In this section we obtain the relation between the applied forces causing rotational motion and the angular acceleration.

Consider a rigid body that can rotate around a fixed axis. All its pointlike particles move on circles centered on the rotation axis on a plane perpendicular to the axis. From experience we know that the rotation is only affected by the component of any applied force that is parallel to the plane of rotation. Furthermore, we know that the rotation depends not only on the force's magnitude and direction but also on where the force is applied. Maximum angular acceleration can be achieved by applying the force as far away from the rotation axis as possible. Indeed, to remove a nut with a wrench, you grasp the wrench as far out as possible and try to push or pull perpendicular to the wrench. You do this because you know intuitively that the rotation is determined by a quantity called the torque, denoted by $\tau$.

Consider a plane that includes the point where a force is applied, is perpendicular to the rotation axis, and intersects the latter at point O. We denote by ${\bf F}$ the component of the force parallel to that plane (this is the only one which causes rotation) and by r the displacement vector of the point where the force is applied from O. Then

$$\tau = F d = F r \sin \phi,$$

where $d= r \sin \phi$ ( called the moment (lever) arm of the force) is the distance between O and the point of application of the force in a direction perpendicular to that of the force, and $\phi$ is the angle between r and F.

On the right-hand-side of the above equation we see the product of the magnitude of two vectors ${\bf F}$ and ${\bf r}$ times the sine of their angle. Such a combination appears in different problems of physics and mathematics, so the mathematicians decided to define the vector or cross product ${\bf C} = {\bf A} \times {\bf B}$of two vectors A and B with relative angle $\phi$ (between 0^o and 180^o). The latter is a vector of magnitude

$$C = A B \sin \theta$$

and direction perpendicular to the plane formed by A and B and determined by the right-hand rule (see Serway). Torque is a vector defined by

$${\bf \tau } = {\bf r} \times {\bf F}$$

so that its direction coincides with the rotation axis. The above equation can be used to define the torque of any force F about a given point O, r being the displacement from O of the point where the force is applied.

Useful properties of the vector product include

$${\bf A} \times {\bf B} = - {\bf B} \times {\bf A} \ , \ {\bf A} \times {\bf A} =0.$$

An alternative way of calculating a cross product is to express the vectors ${\bf A}$ and ${\bf B}$ in terms of the unit vectors ${\bf i}$ and ${\bf j}$ and their cartesian components (see Chapter 1), multiply out the different terms, and use the above properties and

$${\bf i} \times {\bf j} = {\bf k} \ , \ {\bf j} \times {\bf k} = {\bf i} \ , \ {\bf k} \times {\bf i} = {\bf j}.$$

Torque and Angular Acceleration

Torque causes angular acceleration just like force causes linear acceleration. Torque may be thought of as describing a twist just like force describes a push or pull.

Consider the rotational motion of an object around a fixed axis resulting from the application of different forces, each of which leads to a torque. As shown in Serway, the net torque, given by the vector sum of all the individual torques, is related to the angular acceleration $\alpha$ as follows:

$$\tau_{\rm{net}} = \sum_{i} {\bf \tau_{i}} = I \ \alpha$$

where I is the moment of inertia introduced above. Be careful when summing torques: they could be parallel (positive) or antiparallel (negative) to the rotation axis. Note the similarity between this equation and Newton's second law. The moment of inertia is to rotational motion what mass is to motion along a straight line. Just like the mass resists accelerations, the moment of inertia resists rotational acceleration.

From the above we can conclude that, if an object is in equilibrium, the net torque due to all the forces acting on it must be zero so that its angular acceleration is zero (and thus rotates at a constant angular velocity). Also, the net force must be zero, otherwise the object's center of mass would be accelerating.

Angular momentum

We now add to our list of conserved quantities the angular momentum, the rotational analog of the linear momentum. As you recall from Chapter 8, the momentum vector p is related to the net force ${\bf F}$ via ${\bf F} = \frac{d {\bf p}}{dt}.$In an analogous way, the angular momentum vector ${\bf L}$ is related to the net torque ${\bf \tau}$ via

$${\bf \tau}=\frac{d {\bf L}}{dt}.$$

To see this, consider a pointlike particle of mass m, velocity v, and displacement r from a given point O. A force F acting on this particle leads to a torque

$${\bf \tau} = {\bf r} \times {\bf F} = {\bf r} \times \frac... ...bf p}}{dt} = \frac{d}{dt} \left( {\bf r} \times {\bf p}\right).$$

The angular momentum of a pointlike particle is thus defined as

$${\bf L} = {\bf r} \times {\bf p}.$$

Consider now a rigid body rotating around a fixed axis with angular velocity $\omega$. The angular momentum L_i of a point of mass m_i moving at a speed $v_{i} = \omega \ r_{i}$ around the axis is

$$L_{i} = r_{i} p_{i} = r_{i} m_{i} v_{i} = m_{i} r_{i}^{2} \omega.$$

By summing up the agular momentum contributions of all the different points and using the fact that all angular momenta point in the direction of the rotation axis we get for the total angular momentum of the rigid body

$$L = I \omega,$$

where I is the moment of inertia.

Similar to the total momentum, the rate of change of the total angular momentum of a system of point-like particles, ${\bf L} = \sum {\bf L_{i}}$,is related to the net torque ${\bf \tau}_{\rm{ext}} = \sum {\bf \tau}_{i}$ caused by all the external forces as follows:

$${\bf \tau}_{\rm{ext}} = \frac{d}{dt} {\bf L}.$$

All internal forces describing mutual interactions among the particles inside the system cancel each other out due to Newton's third law. For this reason, the net torque arising from all internal forces is always zero.

Angular momentum is especially useful when dealing with rotational motion because, similarly to linear momentum, it is conserved when the net external torque is zero. We can learn a lot about the complex motion of an isolated system by using the fact that the total angular momentum vector remains constant at all times. For example, the angular momentum $L = I \omega$ of a spinning iceskater is conserved because, in the absence of friction with the ice, her net external torque is zero. Iceskaters can change their moment of inertia I quite easily. All they have to do is move their mass closer to the center, e.g. by pulling in their extended hands. Indeed, I depends on the way that the mass is distributed throughout the object. Mass located away from the rotational axis has more rotational inertia than mass located closer to the axis. Michelle Kwan's moment of inertia is larger if she stretches her arms than if she keeps them close to her body. Even though she can change her moment of inertia during her rotational (spinning) motion, her angular momentum $L = I \omega$ must always remain the same. Therefore, the smaller her moment of inertia the larger her rotational speed and vice versa.