Physics A Chapter 10 - Rotational Motion Fall 018 These notes are five pages. A quick summary: The concepts of rotational motion are a direct mirror image of the same concepts in linear motion. Follow the same rules and procedures in Chapter 10 that you used in earlier chapters. A description of Webassign Problems 1 and 13 is included at the end of these notes. Our study of rotational motion directly parallels what we have already studied for linear motion. This included kinematics (from Chapter & 3), force (from Chapters 4 & 5), energy (from Chapters 6 & 7) and momentum (from Chapter 8.) For rotational motion, we use exactly the same ideas and rules but we adapt the definitions of motion to rotation. For example, in kinematics we used velocity and acceleration for linear motion; for rotational motion, we will use angular velocity and angular acceleration. There are four types of problems in Chapter 10: kinematics, force & torque, energy, and momentum. Kinematics In Chapter, we defined position, velocity, acceleration and time and we developed equations that expressed how these concepts are related to each other: v f = v i + at and x f = x i + v i t + ½ a t We used these equations only for intervals of constant acceleration, which were defined by initial and final points of interest. For rotational motion, we make exactly the same definitions and use exactly the same equations: Angular position: θ in radians (instead of x in meters) Angular velocity: ω in rad/s (instead of v in m/s) Angular acceleration: α in rad/s (instead of a in m/s ) And we write the same two equations to express how these are related to each other: ω f = ω i + αt and θ f = θ i + ω i t + ½ α t Solving kinematics problems in rotation follows exactly the same procedure as for kinematics problems in Chapter : Identify the problem as kinematics by the clues: position, velocity, acceleration and time. Organize the given information by labeling points of interest. Write two equations for each interval. Count equations and unknowns. Some problems include information about linear and rotational motion. To connect the linear and rotational information, a very important principle is necessary: every point on a rotating object had rotational and related linear motion. Page 1 of 5
Physics A Chapter 10 - Rotational Motion Fall 018 The connection between the linear and rotational motion for any point on a rotating object can be expressed using: s = θ R v = ω R a = α R where R is the distance from the axis of rotation to the point on the rotating object that we want to refer to. Force & Torque In Chapter 4, we used Newton s Second Law to define the effect of forces on an object: force causes acceleration. For rotational motion, torque is the equivalent of force. That is, Newton s Second Law in rotation tells us that torque causes angular acceleration. How do we apply a torque to a rotating object? We have to apply a force (i.e. we have to touch the object to get it to rotate), but the torque created by that force depends on the direction of the force and where on the object that force is applied. We make a definition for a torque that is created by a particular force on an object: τ = r F sinφ where: F is the force applied to the object; r is the distance from the axis of rotation to the point where the force is applied; φ is the angle between r and F. In Chapter 10, force problems are torque problems. Which means that after you have identified a problem as a force problem, follow the same procedure as in Chapters 4 and 5: Draw a picture, organize information Draw a FBD (in the approximate shape of the object) Label forces in the direction and location that they act on the object Define x and y axes; write force equation for each dimension The steps above are directly from Chapters 4 and 5; the additional steps you will need to consider torques acting on the object are: Define the axis of rotation (or point of rotation) and the positive direction of rotation. Create a torque table, with columns for force r sinφ + or - and τ Fill in the first column with the list of forces from your free body diagram. Fill in the second column by finding r for each force. Fill in the third column by recognizing the angle between each force and its r. Determine which direction each force would rotate the object; fill in the fourth column. Combine the entries in the first four columns to create the torque in the final column. Page of 5
Physics A Chapter 10 - Rotational Motion Fall 018 After completing the torque table, you can write Newton s Second Law in rotation: sum of the torques = moment of inertia angular acceleration or: τ = I α If the object is not accelerating, then the angular acceleration is zero and the sum of torques is simply equal to zero. Note that this is exactly what we do with forces for an object that is not accelerating. What is moment of inertia? Just as the torque produced by a given force depends on where that force is applied to the object, the moment of inertia of a rotating object is a measure of its mass and how far the mass is distributed from the axis of rotation. We define the moment of inertia of one tiny bit of mass, sometimes referred to as a point mass, as mr, where r is the distance from the axis of rotation to the location of the bit of mass. The moment of inertia of any object is the sum of all the moments of inertia of the bits of mass that make up the object. An integral (i.e. adding all the tiny bits) is required to derive the expression for the moment of inertia of any object. The moments of inertia for many common objects (disk, sphere, long thin rod, etc) are already well known (i.e. the integral has already been done and the expressions created.) All expressions for moment of inertia will include the mass of the object, some linear distance squared, and a numerical (i.e. a fraction) coefficient. A sample of moments of inertia of common objects that are rotated about an axis that passes through the center of the object: solid uniform disk solid uniform sphere long thin rod rectangular plate 1 MR I = 1 1 I = MR I = ML ( I = M a + b ) 5 1 1 Work & Energy In Chapter 6, we created the work-energy theorem, which gave us definitions for work and kinetic energy, as well as a simple connection between the two: the work done on an object is equal to the change in the object s kinetic energy. The work-energy theorem is valid in rotation also. We simply modify our definitions of work and kinetic energy to make use of the rotational equivalent definitions of motion: In linear motion: Work = F d cosφ Kinetic energy = ½ m v In rotational motion: Work = τ θ Kinetic energy = ½ I ω Page 3 of 5
Physics A Chapter 10 - Rotational Motion Fall 018 Note that the work in rotational motion does not include the cosine of the angle between the torque and the angular displacement (i.e. between τ and θ.) This is because our treatment of rotational motion only considers rotation in one dimension ; i.e. we only have a positive and negative direction of rotation, much like we had for motion in one dimension in Chapter. For this reason, the work done by a torque will always be either positive, if the torque is in the same direction the object rotates, or negative, if the torque is opposite the direction the object rotates. But there is no angle between the direction of the torque and direction of angular displacement. For an energy problem in rotational motion, write the work-energy theorem as before. The only difference is in the expressions used for work and kinetic energy for the rotating object. An important consideration for the kinetic energy of a rotating object: If a rotating object is held in place, then its kinetic energy is simply that associated with its rotation. That is, K = ½ I ω. But if a rotating object rolls, i.e. it is traveling along a surface as it rotates (like a tire rolling along a road), then the object has both rotational and translational (i.e. linear ) kinetic energy. In this case, the kinetic energy of the object is the sum of ½ I ω and ½ m v. Since there is linear and rotational motion involved in this scenario, a connection (see the kinematics section) between the two types of motion is usually necessary to simplify the resulting algebra. Angular Momentum In Chapter 8, we defined impulse, momentum and collisions. We now do the same thing for the corresponding concepts in rotational motion: Linear: Impulse = force time Momentum = m v Rotational: Impulse = torque time Angular momentum = I ω We used linear momentum primarily for problems that included a collision. Using Newton s Third Law, we found that two objects that apply equal and opposite impulses to each other when they apply equal and opposite forces to each other for the same time. This gave us the principle that the total momentum of a system of objects before a collision is equal to the total momentum of that system after the collision. We apply the same idea to rotational motion: the total angular momentum of a system of objects remains constant, before and after a collision. We can apply this principle to Problems 1 and 13 in the Chapter 10 Webassign assignment. Problem 1: You are given the moment inertia of a merry-go-round and its initial angular velocity. You are also told a child, who will jump on the mgr is initially stationary. The system described is the mgr and the child; you are given its initial angular momentum. After the child jumps on the mgr, the total moment of inertia of the rotating system can be expressed as that of the mgr plus that of the child. If we Page 4 of 5
Physics A Chapter 10 - Rotational Motion Fall 018 approximate the child as a point mass, i.e. all of his mass is at a given distance from the axis of rotation, then the child s moment of inertia is simply his mass times R, where R is the given radius of the mgr. We can then write: initial angular momentum = final angular momentum initial I ω = final I ω I mgr ω 1 = (I mgr + MR ) ω The only unknown in this expression is ω, the final angular velocity of the system. Problem 13: This problem presents a variation on how we used momentum in a collision. Since the mass of a given object is typically fixed, changing the linear momentum of an object usually involves changing its velocity. This can only occur if a force is applied to the object, and so linear momentum problems necessarily involve two objects applying a force to each other... i.e. a collision. But angular momentum includes the moment of inertia of an object, and the moment of inertia is a measure of where the mass of the object is distributed. This means that if the distribution of an object s mass changes, its moment of inertia changes: bring the mass closer to the center, and the moment of inertia decreases; move the mass further from the center and the moment of inertia increases. However, according to the definition of impulse, an object s angular momentum can only change if a torque (from an external source) is applied to the object. If the object is isolated, then its angular momentum must remain constant. The person in Problem 13 changes his moment of inertia by moving the weights in his hands closer to the axis of rotation. But his angular momentum cannot change. The net effect is that when he decreases his moment of inertia (by bringing the mass closer to the axis), his angular velocity must increase in order to keep the same angular momentum. For this problem, define the moment of inertia of the student and stool as I ss. Assume each of the two weights he holds can be treated as a point mass, so that the moment of inertia of each is mr. Then you can write: initial angular momentum = final angular momentum initial I ω = final I ω (I ss + mr 1 + mr 1 ) ω 1 = (I ss + mr + mr ) ω The only unknown in this equation is the final angular velocity. After finding the value for ω, calculate the initial and final kinetic energies by using the definition, ½ I ω. Page 5 of 5