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Note 10 Rotational Motion I Sections Covered in the Text: Chapter 13 Motion is classified as being of one of three types: translational, rotational or vibrational. Translational motion is the motion executed by the center of mass of an object modelled as a particle (Notes 03 and 06). The motion of a baseball hit in a line drive is largely translational. But a baseball can also rotate, so the motion of a baseball can in general possess both translational and rotational components. A baseball does not usually vibrate. We shall study vibrational motion in Note 12. At first thought the motion of a baseball might seem impossibly complicated to describe mathematically. Physics shows, however, that the motion of any object can be separated into its translational and rotational components and those components solved for separately. This is possible because translational and rotational components of the motion of a rigid body do not interact with one another.1 We have already studied a particularly simple form of rotational motion in Note 05: a particle rotating about a point external to it. This motion is called circular motion. Here we survey a few aspects of the rotational motion of a rigid extended body rotating about a point internal to itself. We begin by laying down the kinematics of a rotating body as we did for a body in translational motion. We derive the kinematic equations of rotational motion and deduce the relationships between rotational and translational quantities. We introduce the construct of torque. Angular Speed and Angular Acceleration For convenience we consider a rigid extended object whose mass is confined to a plane and whose translational motion is zero (Figure 10-1). We suppose this object is rotating in a counterclockwise direction about an axis perpendicular to its plane passing through a point O. In the coordinate system of the figure, this axis can be thought of as the z-axis. We assume that the object is a rigid, extended body. By this we mean it cannot be modelled as a single particle. It can, however, be modelled (approximately) as a collection of particles whose positions are fixed relative to one another. A CD is a rigid body whereas a soap bubble is not. 1 But the body must be rigid. Examples of non-rigid bodies whose modes of translation and rotation interact are typically studied in a higher-level course in classical mechanics. Figure 10-1. A particle P in a rigid body is located with respect to O by the polar coordinates (r,θ). Any point P in the object can be located relative to the point O with polar coordinates (r, θ). As the object rotates, P follows a circle of radius r. Every other point in the object also follows a circular path, but a path with a different radius. Let us suppose that in some elapsed time ∆t, P moves from a position on the positive x-axis to the point where it is shown in the figure. Then the subtended angle θ is called the angular displacement of P measured relative to the positive x-axis. Because the object is rigid, the angular displacement of every particle in the object is the same as the angular displacement of P. Angular displacement is measured in the dimensionless unit called radian (abbreviated rad) and in the following way. The arc length of the circle on which P moves (the distance travelled by P) is related to the radius r and angle θ by and therefore s = rθ , …[10-1] s θ= . r …[10-2] (rad). If s = r then θ = 1 radian. If the object executes one complete revolution, then the angular displacement of P is 2π radians. 2π radians is equivalent to 360˚. If the object executes two revolutions, then its angular displacement is 4π radians. By its nature, angular displacement is a cumulative quantity. 10-1 Note 10 We now have the tools we need to define the rotational equivalents of the translational quantities we defined in Note 03. We begin by making Figure 10-1 more general (Figure 10-2) by supposing that in some elapsed time ∆t = tf – ti any arbitrary point P in the object moves from a position i to a position f. These positions are shown as [A] and [B] in the figure. The corresponding angular positions are θ i and θ f respectively. Δθ dθ = . Δt →0 Δt dt ω ≡ lim …[10-5] We now extend the math to allow for changes in the instantaneous angular velocity.3 If the instantaneous angular velocity is changing, then the object is by definition, undergoing an angular acceleration. Let the instantaneous angular velocity of the point P at positions i and f be ω i and ω f respectively. The change in the instantaneous angular velocity divided by the corresponding elapsed time is defined as the average angular acceleration: α≡ ω f – ω i Δω = . t f – ti Δt …[10-6] Average angular acceleration has dimension T–2 and units s–2. The limit of the average angular acceleration as ∆t → 0 is defined as the instantaneous angular acceleration: Δω dω = . Δt → 0 Δt dt α ≡ lim Figure 10-2. A more general representation of a rotating body than that shown in Figure 10-1. In some elapsed time ∆t a particle P in the body moves counterclockwise between two arbitrary angular positions. The angular displacement of P in the interval chosen is defined as the difference between the angular positions that define the interval: ∆θ = θf – θi …[10-7] ω and α are, in fact, vector quantities (pseudo-vectors) whose complete vector nature is beyond the scope of these notes to describe adequately.4 But you can find the direction of the vectors with the help of the righthand rule as was introduced for the vector cross product (Figure 10-3). …[10-3] (rad).2 The angular displacement divided by the corresponding elapsed time is defined as the average angular velocity: ω≡ θ f – θi Δθ = . t f – ti Δt …[10-4] Average angular velocity has dimension T –1 and units rad.s–1 (or just s–1 since rad has no dimension). The limit of the average angular velocity as ∆t → 0 is defined as the instantaneous angular velocity: Figure 10-3. How to use the right hand rule to find the direction of the vector ω of a rotating body. 3 2 The alert reader will notice the word displacement here implying a vector quantity. Angular displacement is, in fact, a vector quantity, or more correctly, a pseudo-vector. This aspect of rotational motion is somewhat advanced and best left to a second year course in classical mechanics. 10-2 Of course, in order for the object’s angular velocity to change, the object must be subject to a force applied to it in a special way. For the moment we ignore this force, as we did in Notes 02 and 03, and stay within the area of kinematics. 4 We leave this description to a higher-level course in classical mechanics. Note 10 To find the direction of the ω vector, extend your right hand, curl your fingers as if you are to grip something and extend your thumb. Now curl your fingers in the direction of the angular displacement of the object (the direction the object is rotating). Then your thumb points in the direction of the ω vector. By convention, the ω vector is placed on a diagram along the body’s axis of rotation. Later in this note we shall see other uses of the right hand rule. Rotational Kinematics As implied in the previous section, a set of kinematic equations exist for rotational motion just as they do for translational motion. They have a similar form and are derived in a similar fashion. We shall therefore just list them (Table 10-1). Table 10-1. Comparison of translational and rotational kinematic equations. Translational Motion Rotational Motion v f = vi + at ω f = ω i + αt 1 x f = xi + vi t + at 2 2 1 x f = xi + (vi + v f )t 2 2 2 v f = vi + 2a(x f – xi ) 1 θ f = θi + ω it + αt –1 The final angular speed is 10.0 s or 10.0 rad.s . (b) Using the second equation in Table 10-1 we have for the angular displacement 1 2 θ f = θ i + (ω i + ω f )t 1 2 = 25.0 radians. = 0 + (0 + 10.0rad.s –1 )(5.00s) The number n of revolutions is this number divided by the number of radians per revolution (i.e., 2π): n= 25.0(rad) = 3.98 rev. rad 2π rev Thus nearly 4 revolutions are required for the wheel to accelerate to the final angular speed of 10.0 s–1. € 2 2 1 θ f = θi + (ω i + ω f )t 2 2 –1 2 ω f = ω i + 2α (θ f – θi ) Problems in rotational kinematics can be solved much like problems in translational kinematics. Assuming you have memorized the translational equations and know the rotational equivalents, you can easily reconstruct the rotational equations. Let us consider an example. Relations exist between the angular and tangential speeds of a particle in a rotating rigid body, and between the angular and tangential accelerations. Since these relations are useful in solving rotational motion problems we consider them next. Relations Between Rotational and Translational Variables Suppose that a rigid extended body rotates about an axis that passes through an internal point O as shown in Figure 10-4. Consider a point P in this body. Example Problem 10-1 A Problem in Rotational Kinematics Starting from rest a wheel is rotated with a constant angular acceleration of 2.00 rad.s –2 for 5.00 s. (a) What is the final angular speed of the wheel? (b) How many complete revolutions does the wheel execute in the elapsed time of 5.00 s? Solution: (a) Using the first equation in Table 10-1 we have for the final angular speed ω f = ω i + α t = 0 + (2.00s–2 )(5.00s) = 10.0 s–1. Figure 10-4. A particle P in a rotating rigid body. 10-3 Note 10 The magnitude of the tangential velocity of P is, from eq[10-1], v= ds dθ =r , dt dt since r is constant. Thus using the definition of ω in eq[10-5] v = rω . …[10-8] Table 10-2. Relationships between the magnitudes of translational and rotational variables. Translational Rotational Relationship x v a θ ω α x = θr v = ωr a = αr = ω 2r Let us consider an example. The tangential acceleration of P is, using eq[10-8], at = dv dω =r , dt dt again since r is constant, or using the definition of α in eq[10-7], at = rα . …[10-9] Now since P is moving in a circle it is undergoing a centripetal acceleration. The magnitude of this centripetal or radial component of the acceleration is ac = v2 = rω z2 . r using eq[10-8]. The relationship between tangential and radial components of the acceleration of P can be seen with the help of Figure 10-5. The total acceleration of P is the sum of the at and ar vectors. The complete vector nature of at and ac is beyond the scope of these notes to describe. The relationships between the magnitudes of these quantities are summarized in Table 10-2. Example Problem 10-2 Tangential and Angular Speeds A bicycle wheel of diameter 1.00 m spins freely on its axis at an angular speed of 2.00 rad.s–1. (a) What is the tangential speed of a point on the rim of the wheel? (b) What is the tangential speed of a point halfway between the axis and the rim? Solution: (a) Using eq[10-8] the tangential speed of a point on the rim of the wheel is 1.00m v = rω = (2.00rad.s –1 ) = 1.00 m.s–1. 2 (b) A point halfway between axis and rim will have a tangential speed one half of this value, or v = 0.50 m.s –1. Clearly, the further a point on the wheel is from the axis of rotation the greater is its tangential speed. Though the two points have different tangential speeds they have the same angular speed. We stated in Note 07 without proof that the centre of mass of a rigid body can be taken to be the body’s geometric centre. We are now ready to extend the idea of the centre of mass to a system of discrete bodies, and to define the centre of mass in a proper mathematical fashion. Figure 10-5. The resultant acceleration of any particle P in a rotating rigid body is the vector sum of the tangential and radial acceleration vectors. 10-4 Note 10 Center of Mass Revisited Example Problem 10-3 Finding the Centre of Mass of a System of Particles We have all seen pictures of a body rotating in the weightless environment of a space capsule. If the body is a flat plate (Figure 10-6) we would see that a point in the body doesn’t rotate at all. That point is the body’s centre of mass. If we model the body as a collection of particles in a coordinate system (Figure 10-6b) then the coordinates of the centre of mass can be shown to be A 0.50 kg ball and a 2.00 kg ball are connected by a massless rod of length 0.50 m. Where is the centre of mass of the system located relative to the 2.00 kg ball? Solution: The system is shown in Figure 10-7. We model the two balls as particles. For convenience we put the large ball at the origin. Using eqs[10-10] (and taking ycm = 0) we have Figure 10-7. Finding the centre of mass of a system. x cm = = € 1 m x + m2 x 2 mi x i = 1 1 ∑ M i m1 + m2 (2.0 kg)(0.0 m) + (0.50 kg)(0.50 m) 3.0 kg + 0.50 kg = 0.10 m € Figure 10-6. An unconstrained body rotating freely in space rotates about its centre of mass. x cm = 1 ∑ mi x i M i …[10-10] y cm € 1 = ∑ mi y i . M i where M = m 1 + m2 + … is the body’s total mass. In a moment we shall put this definition into more general form. (In € the event that the body has an appreciable thickness then we would have to add an equation in z to eqs[10-10].) Let us consider an example. € The centre of mass of the system lies between the two balls and 0.10 m from the 2.0 kg ball. The centre of mass of a system of particles lies at its “weighted” centre. A solid rigid body does not consist of a collection of discrete particles, but rather of a continuous distribution of mass. We must therefore generalize eqs[10-10] for a continuous distribution. To do this we imagine the body divided up into many small cells of boxes, each with the small mass ∆m (Figure 10-8). Eqs[10-10] thus become x cm = 1 1 x iΔmi and y cm = ∑ y iΔmi . ∑ M i M i If we now let the number of cells become larger and larger and the size of the cells ∆mi become smaller and smaller these equations go over to 10-5 Note 10 Eq[10-11a] becomes x cm 1 M = M L ∫ 1 xdx = L L ∫ xdx . 0 Evaluating the integral we get € Figure 10-8. The division of a solid body into cells. x cm 1 = M ∫ xdm and y cm 1 = M ∫ ydm …[10-11a and b] € Eqs[10-11] is the most general definition of the centre of mass. When evaluating eqs[10-11] you must remember to express dm in terms of dx, dy or both and then calculate the integrals over coordinates. Let us consider an example. Example Problem 10-4 Finding the Centre of Mass of a Uniform Rod Find the centre of mass of a thin, uniform rod of length L and mass M relative to one end. x cm L 1 x2 1 L2 1 = = − 0 = L . L 2 0 L 2 2 As expected, the rod’s centre of mass is located midway along the rod, or at its geometric centre. € Thus far we have considered the kinematics of rotational motion. We are now ready to broaden our study to the dynamics of rotational motion. We know that a body’s translational motion can only be changed by the application of a net force to the body. By the same token, a body’s rotational motion can only be changed by the application of a net force. But the force must be applied in a special way. This warrants the definition of another tool of physics called torque. Torque We shall see in what follows that the rotational equivalent of force is a construct called torque. Torque may be thought of as the “turning action” that produces a rotation in the same sense that force is the “straightline action” that produces a translation. Torque can be understood with the help of Figure 10-10. Figure 10-9. Finding the centre of mass of a long thin rod. Solution: The rod is an example of a body with a continuous distribution of mass. Since it is very thin we can set ycm = 0 (and of course zcm = 0). We put one end of the rod at the origin and imagine the length of the rod to be divided into increments dx. Since the rod is uniform each increment has mass dm = (M/L)dx. 10-6 Figure 10-10. How the torque produced by a force is defined. The figure shows a wrench whose jaws fit onto a nut centered at the point O. We assume that the nut-bolt Note 10 combination is a right-handed one, meaning that the nut must be rotated counterclockwise to loosen it (so that its translational motion along the axis of the screw is out of the plane of the page). A force must therefore be applied to the wrench roughly as shown. Though the ultimate cause of rotational motion is a force, the physical “action” of loosening the nut cannot be adequately described by force alone. As anyone knows, if you apply the force in a line through O no loosening of the nut occurs at all. Moreover, the further away from O you apply the force, the larger is the turning effect for the same magnitude of force. We shall see in what follows that this rotational “action” is best described by a new construct called torque. The torque about an axis O is defined to have the magnitude τ = Fd , where F is the magnitude of the force applied and d is the perpendicular distance from O to the line along which the force acts. According to Figure 10-10 we can also write τ ≡ rFsin φ , …[10-12] where r is the distance from O to the point on the body at which the force is applied and φ is the smallest angle between the vectors r and F . The dimension of torque is M.L2.T –2. Its units are N.m. WARNING! Torque is not the same as Work Torque and work have the same units, namely N.m. But torque is not the same as work. For one thing, torque is a (pseudo) vector whereas work is a scalar. Any number of forces may act on a body at the same time. For example, Figure 10-11 shows two forces with magnitudes, F 1 and F 2, acting on the same body. The force F 1 tends to produce a rotation of the body counterclockwise about O. For this reason the torque produced by this force is arbitrarily given a positive sign. The force F 2 tends to produce a rotation of the body clockwise about O. For this reason the torque produced by this force is given a negative sign. The resultant torque has the magnitude ∑ τ = Fd 1 1 clockwise about O. If the result is negative, then the body will tend to rotate clockwise about O. Thus far we have considered the magnitude of torque. To understand the direction of torque we must work with its vector definition. This brings us to the major reason for introducing the vector cross product: Figure 10-11. Two forces applied to a body in arbitrary directions. τ = r × F, …[10-13] where r is the vector locating the point at which the force F is applied relative to the axis of rotation. The directions of the three vectors in relation to one another can be seen with the help of Figure 10-12. Let us suppose for convenience that a force is applied to a point P in the xy-plane of the body. The vector r locates the point P relative to the point O about which the object rotates. The torque vector τ is then given by the right hand rule. With reference to the figure, the magnitude of the torque as given by eq[10-13] can be seen to be equivalent to eq[10-12]. We now consider an example of two forces applied to a body. – F2 d2 . According to our sign convention, if this result is positive, then the body will tend to rotate counter- 10-7 Note 10 (given a positive sign because it tends to produce a counterclockwise rotation). The magnitude of the net torque about the rotation axis is the sum Figure 10-12. The relationahip between the vectors τ , r and F. Figure 10-13. Two torques acting on a cylinder. Example Problem 10-5 The Net Torque on a Cylinder A one-piece cylinder is shaped as in Figure 10-13 with a core section protruding from a larger drum. The cylinder is free to rotate around the central (z-) axis shown in the drawing. Two ropes are wound, one around the core, the other around the drum, and exert forces T1 and T 2 on the cylinder in the directions shown. Calculate the net torque acting on the cylinder about the rotation axis. Solution: The magnitude of the torque produced by T1 is –R 1T1 (given a negative sign because it tends to produce a clockwise rotation about the axis through O). The magnitude of the torque produced by T 2 is +R 2T2 10-8 τ net = τ 1 + τ 2 = R2 T2 – R1 T1 , This result can be positive or negative depending on the relative magnitudes of the forces and radii. Clearly, if the sum is zero, then the body undergoes no rotation at all. Question. Does the body in Figure 10-13 tend to move with translational motion? Explain. To Be Mastered • • Definitions: angular displacement, radian, average angular velocity, average angular speed, instantaneous angular velocity, instantaneous angular acceleration, average angular acceleration equations of rotational kinematics (to be memorized): ω f = ω i + αt • 1 θ f = θi + ω it + αt 2 2 1 θ f = θi + (ω i + ω f )t 2 2 2 ω f = ω i + 2α (θ f – θi ) relations between the magnitudes of translational and rotational quantities x = θr a = αr = ω 2r v = ωr Typical Quiz/Test/Exam Questions 1. (a) Define torque (b) What are the units of torque? 2. State the following as being a vector or a scalar. Give the units of each. (a) coefficient of friction (b) potential energy (c) linear momentum (d) angular velocity 3. A body is subject to non-concurrent forces. Write down the conditions for the body to be in a state of mechanical equilibrium. Explain the meaning of each quantity in your expressions. 4. A uniform plank of length 2.00 m and weight 100.0 N is to be balanced on a fulcrum or support point (see the figure). A 500.0 N weight is suspended from the right end of the plank and a 200.0 N weight suspended from the left end. Answer the following questions. right left fulcrum (a) Describe the state of the plank when it is balanced. (b) What are the conditions for this state? (c) How far from the left end of the plank must the fulcrum be placed? (d) What is the reaction force exerted by the fulcrum on the plank? 10-9