CIRCULAR MOTION AND ROTATION

Transcription

1 1. UNIFORM CIRCULAR MOTION So far we have learned a great deal about linear motion. This section addresses rotational motion. The simplest kind of rotational motion is an object moving in a perfect circle at a constant speed. This is called Uniform Circular Motion. The first step in describing uniform circular motion is finding rotational variables that are analogous to the linear variables of motion. Consider the circle below: Δs Δθ r When an object moves around the circumference of a circle a distance Δs, this distance corresponds to the angle that is swept out by the motion, Δθ, with the following relationship Δs = rδθ. This is the definition of an angle in radians. It is very important that the angle be represented in radians for this relationship to hold. The velocity, or more precisely the speed, is the rate of change of the arc length Δs, and can be expressed as an equation involving the rate of change of the angle, Δθ:

2 v = Δs Δt = r Δθ Δt = rω ω = Δθ Δt KEY CONCEPTS Here ω is the rate at which the angle θ changes and is called the angular velocity or angular frequency. The velocity vector is tangent to the circle and points in the direction of the displacement, and is therefore always changing. The time it takes for the object to complete one complete turn around the circle is called the period and is denoted by T. The period is equal to the total distance traveled in one cycle divided by the velocity. Since we know the total distance around a circle (the circumference), the period can easily be related to either the velocity or the angular velocity T = 2πr v = 2π ω The frequency, denoted as f, is defined to be one over the period, which is the number of rotations the object undergoes in a certain time interval. It is related to the angular velocity through the equations f = 1 T = ω 2π ω = 2πf Recall that acceleration is the change in velocity with respect to elapsed time. Since velocity is a vector, any change in its magnitude (speed) or its direction involves acceleration. Thus far we have focused on cases in which only the magnitude of the velocity changed, but in uniform circular motion, the magnitude is constant and the direction changes constantly. This change results from a type of acceleration known as centripetal acceleration and written as a c. The centripetal acceleration always points inward, toward the center of the circle of motion, and is given by the equation a c = v 2 r. v a c a c v 2

3 Centripetal acceleration is an important concept that is useful in describing many phenomena, such as the orbits of the moon and planets, and the motion of roller coasters. Whenever working with problems involving circular motion, remember that centripetal acceleration is indeed acceleration not a force - and thus belongs on the ma side of the equation F=ma. Students (and occasionally even instructors) are sometimes tempted to include a separate centripetal force on the left side of the equation. However, whatever force it is that holds the object rotating in a circle is in fact a centripetal force, and no additional separate force by that name is involved. Centripetal acceleration is also part of another misconception. Passengers on quickly rotating rides in amusement parks feel like they are being thrown outward. People like to call this apparent push centrifugal force. In fact the force that people interpret as being thrown outward is a direct result of Newton s first law. If at some point in its rotation a rotating body is released from whatever constrains it to rotate, it would move in a straight line, which would take it farther from the center of the circle. Thus, the feeling is one of being thrown outward. In fact force is required to keep a rotating body moving in a circle. This force again, no matter what provides it is the centripetal force. Centrifugal force is a fictitious force and should never appear in calculations. 2. ROTATIONAL KINEMATICS As shown above position and velocity have the angular counterparts of angle θ and angular velocity ω. There is also an angular counterpart for acceleration known as the angular acceleration α, which is simply the rate of change of the angular velocity. The analogy between linear and angular quantities can be summarized as follows. x v = dx dt a = dv dt θ ω = dθ dt = v r α = dω dt = a // r Please note, the angular acceleration relates to the speed changing part of the acceleration a //. (This component of acceleration is parallel to the direction of motion.) The centripetal acceleration a c is always present (even when the angular acceleration is 0), is perpendicular to the velocity, and changes the direction, but not the magnitude of the velocity. Because of the close analogy between translational and rotational motion, we can immediately transfer our constant acceleration equations to describe rotational motion as follows. 3

4 v = v 0 + at x = x 0 + v 0 t at 2 v 2 2 = v 0 +2a ( x - x 0 ) KEY CONCEPTS ω = ω 0 + αt θ = θ 0 + ω 0 t αt 2 ( ) ω 2 = ω α θ θ 0 The rotational kinematics equations are used just like the translational ones. It is worth noting that ω and α are actually vector quantities just like v and a. (θ is not a vector because rotations around different axes do not commute. However, this is not a great concern for our purposes.) The direction of ω and α is just the direction of the axis about which they rotate. (An axis can actually have either of two directions. To be more accurate it is the direction about which they rotate counterclockwise.) 3. TORQUE AND ROTATIONAL STATICS The rotational counterpart to force is torque. Forces cause translational motion and torques produce rotational motion. Newton s Second Law states that the net force acting on an object is equal to the mass (inertia) times the acceleration (change in the object s translational motion). This is the translational version of Newton s Second Law. There is also a rotational version of Newton s Second Law, which is the net torque on an object, written as τ, is equal to the object s rotational inertia I times its angular acceleration α. These equations are shown side by side below for comparison. F = ma Newton s Second Law τ = Iα Just as the mass of an object is a measure of how difficult it is to start an object moving, the rotational inertia I of an object is a measure of how difficult it is to start the object turning or rotating. The value of I depends not only on the mass, but also on how the mass is distributed. We will discuss the computation of I momentarily. Torque depends on three basic quantities: the size of the force applied F, the location where it is applied r (relative to the pivot point) and the direction in which it is applied. F θ r 4

5 The formula for torque is τ = rf sinθ and its units are Newton-meters or N m. (Note: One Joule is equal to one N m, but it is not appropriate to use Joules for units of torque.) The formula for torque can be best understood by considering an example. Consider turning a bolt, door, or merry-go-round. The object moves farther when a larger force is applied or when it is applied farther from the pivot point. The angular dependence of torque is perhaps the hardest to understand. Imagine exerting a force on a bolt or a merry-go-round. If the force is parallel to the radius vector, then it just pushes or pulls on the pivot point and no rotation results. By contrast a force applied at an angle that is perpendicular to the radius vector has a great turning effect and thus creates a large torque. It is often useful to associate all the angular dependence with either the force or the position vector, either by finding the component of force perpendicular to the radius, or by finding the component of radius perpendicular to the force as illustrated below. Then the torque can be calculated by using the perpendicular component, which simplifies calculations. F // F F r r F = F sinθ r // τ = rf sinθ = rf = r F r = rsinθ The perpendicular part of the radius is used so often that it has a special name, the lever arm. Torque is actually a vector quantity and is simply the cross product of the position and force vectors. From our previous study of cross products we know that this 5

6 operation yields exactly the equation we have been looking at for the magnitude of the torque. τ = r F τ = rfsinθ The simplest kind of rotational problem is one in which the sum of the torques, and thus the angular acceleration, is 0: τ = Iα = 0 α = 0 Such problems are called rigid body or equilibrium problems. 4. PROBLEM SOLVING STRATEGY ROTATIONAL STATICS When solving equilibrium problems using the following steps is helpful. 1. Double check that it is a static problem, one with no accelerations or angular accelerations. 2. Choose a pivot point. Often this is obvious because the problem involves a hinge or a fixed point. If the choice is not obvious pick the pivot point as being where you have the most unknowns. (This simplifies things because forces at the pivot point create no torque.) 3. Write an equation for the sum of torques, and equations for the sum of forces in both the x and y directions. Set these totals equal to 0. Be careful with your signs. 4. Solve the equations for your unknowns algebraically. 5. Insert numbers to find the final answer. 5. MOMENT OF INERTIA AND ROTATIONAL DYNAMICS If we want to study the dynamics of a rotating object we need to be able to calculate I, the moment of inertia. As stated before I depends on both the mass and its distribution. For a point mass, I is just the mass times its rotational radius squared that is. I = mr 2 For several masses this becomes a sum over the various masses and for a continuous rigid body it becomes an integral. 6

7 I = m i r i 2 I = i r 2 dm KEY CONCEPTS In each of these equations r is the distance between the point under consideration and the rotational axis. (This is particularly easy to confuse when working with spheres where r is often used to describe the distance to the center of the sphere.) Because I depends on the distance to the axis of rotation, the value of I changes when the axis of rotation changes. Generally it is most convenient to calculate the moment of inertia around the center of mass and then use the parallel axis theorem to find the moment of inertia around the axis of interest. The parallel axis theorem states that given the moment of inertia I about an axis that goes through the center of mass of an object, you can find the moment of inertia about any parallel axis a distance R away, with the formula. I axis = I C.M. + MR 2 Here M is the mass of the entire object. A brief list of useful moments of inertia is listed below. Object Moment of Inertia I Thin Ring MR 2 Disk 1 2 MR2 Ring of Finite width 1 2 MR 2 2 in + R out Spherical Shell 2 3 MR2 Solid Sphere 2 5 MR2 Thin Rod About the Center 1 12 M 2 Thin Rod about one Edge 1 3 M 2 ( ) Once we know the moment of inertia, torque problems proceed almost identically to force problems. The same steps used force problems and for static torque problems should be applied here, but instead of setting the angular acceleration to 0, we leave it as a variable to be solved for. 7

8 Many rotational dynamics problems can be solved in multiple ways. For example, objects rolling down a hill can be treated as both translational motion and rotational around the center of mass, or as purely rotational motion around the point of contact. 7. ROTATIONAL KINETIC ENERGY AND ANGULAR MOMENTUM There are also rotational counterparts of kinetic energy and momentum. The counterpart of translational kinetic energy is rotational kinetic energy. The two equations are written below for comparison. K = 1 2 mv 2 K rot = 1 2 Iω 2 The applications of rotational kinetic energy are fairly straight forward. It is simply another form of energy. The rotational counterpart of momentum is angular momentum (L). It can be written in a variety of ways, and provides a variety of new applications. For a point object, the angular momentum is the cross product of the radius and the momentum. L = r p = mr v L = rpsinθ = mrv sinθ Notice that angular momentum is a vector quantity, though in introductory courses its vector nature is rarely important. For a rotating rigid body angular momentum can be written in a form analogous to the formula for momentum. p = mv L = Iω The vector nature of angular momentum can get very complex for a rigid body, so we are suppressing it for both translational and rotational situations here. Just as force is the rate of change of momentum, torque is the rate of change of angular momentum. F = dp dt τ = dl dt 8

9 Thus, just as momentum is conserved in the absence of external forces, angular momentum is conserved in the absence of external torques. This is known as the conservation of angular momentum. Like the conservation of linear momentum, the conservation of angular momentum can be used in collisions (including cases where there are external forces, but no external torques). It can also be applied in a variety of other examples. For example both an object moving in a straight line at a constant velocity and an object in uniform circular momentum have constant angular momentum. A variety of other motions have constant angular momentum and this can be useful in describing the orbits of planets. In another example, and ice-skater pulling her hands in reduces her moment of inertia and thus her angular velocity increases to maintain constant angular momentum. Notice this results in an increase in rotational kinetic energy. This energy comes from the work the skater had to do to pull her hand inward. 9