Reading Quiz 5.1 PDF

## 8-2 Constant Angular Acceleration In Chapter 2, we derived the useful kinematic equations (Eqs. 2-11) that relate acceleration, velocity, distance, and time for the special case of uniform linear acceleration. Those equations were derived from the definitions of linear velocity and acceleration,...

## 8-2 Constant Angular Acceleration In Chapter 2, we derived the useful kinematic equations (Eqs. 2-11) that relate acceleration, velocity, distance, and time for the special case of uniform linear acceleration. Those equations were derived from the definitions of linear velocity and acceleration, assuming constant acceleration. The definitions of angular velocity and angular acceleration (Eqs. 8-2 and 8-3) are just like those for their linear counterparts, except that θ replaces the linear displacement x, ω replaces v, and α replaces a. Therefore, the angular equations for constant angular acceleration will be analogous to Eqs. 2-11 with x replaced by θ, v by ω, and a by α, and they can be derived in exactly the same way. ### Kinematic equations for constant angular acceleration * **Angular** **Linear** * θ = ω₀t + ½αt² x = v₀t + ½at² [constant α, a] (8-9a) * ω = ω₀ + αt v = v₀ + at [constant α, a] (8-9b) * ω² = ω₀² + 2αθ v² = v₀² + 2ax [constant α, a] (8-9c) * θ = (ω₀ + ω)t/2 x = (v₀ + v)t/2 [constant α, a] (8-9d) Note that ω₀ represents the angular velocity at t = 0, whereas ω and θ represent the angular position and velocity, respectively, at time t. Since the angular acceleration is constant, α = a. ### **Example 8-6 Centrifuge acceleration** A centrifuge rotor is accelerated for 30 s from rest to 20,000 rpm (revolutions per minute). * (a) What is its average angular acceleration? * (b) Through how many revolutions has the centrifuge rotor turned during its acceleration period, assuming constant angular acceleration? #### **Approach** To determine Δω/Δt, we need the initial and final angular velocities. For (b), we use Eqs. 8-9 (recall that one revolution corresponds to 2π rad). #### **Solution** * (a) The initial angular velocity is 0. The final angular velocity is ω = (20,000 rev/min)(2π rad/rev)(1 min/60s) = 2100 rad/s Then, since α = Δω/Δt and Δt = 30 s, we have α = (2100 rad/s - 0)/30s = 70 rad/s². That is, every second the rotor's angular velocity increases by 70 rad/s, or by (70 rad/s)(1 rev/2π rad) = 11 revolutions per second. * (b) To find θ we could use either Eq. 8-9b or 8-9c (or both to check our answer). The former gives θ = ω₀t + ½at² = (0) + ½(70 rad/s²)(30s)² = 3.15 x 10⁴rad Where we have kept an extra digit because this is an intermediate result. To find the total number of revolutions, we divide by 2π rad/rev and obtain (3.15 x 10⁴ rad)/(2π rad/rev) = 5.0 x 10³rev **Note:** Let us calculate θ using Eq. 8-9c: ω² - ω₀² = 2αθ θ = (ω² - ω₀²)/2α = (2100 rad/s)² - 0)/(2(70 rad/s²) )= 3.15 x 10⁴ rad which checks our answer above from Eq. 8-9b perfectly. ## 8-4 Torque We have so far discussed rotational kinematics-the description of rotational motion in terms of angular position, angular velocity, and angular acceleration. Now we discuss the dynamics, or causes, of rotational motion. Just as we found analogies between linear and rotational motion for the description of motion, so rotational equivalents for dynamics exist as well. To make an object start rotating about an axis clearly requires a force. But the direction of this force, and where it is applied, are also important. Take, for example, an ordinary situation such as the overhead view of the door in Figure 8-10. If you apply a force F₁ perpendicular to the door as shown, you will find that the greater the magnitude, F₁, the more quickly the door opens. But now if you apply the same force at a point closer to the hingesay, F₂, in Fig. 8-10-the door will not open so quickly. The effect of the force is less: where the force acts, as well as its magnitude and direction, affects how quickly the door opens. Indeed, if only this one force acts, the angular acceleration of the door is proportional, not only to the magnitude of the force, but is also directly proportional to the perpendicular distance from the axis of rotation to the line along which the force acts. This distance is called the lever arm, or moment arm, of the force, and is labeled r and r₁ for the two forces in Fig. 8-10. Thus, if r₂ in Fig. 8-10 is three times larger than r₁, then the angular acceleration of the door will be three times as great, assuming that the magnitudes of the forces are the same. To say it another way, if r₂ = 3r₁, then F₁ must be three times as large as F₂ to give the same angular acceleration. (Figure 8-11 shows two examples of tools whose long lever arms are very effective.) The angular acceleration, then, is proportional to the product of the force times the lever arm. This product is called the moment of the force about the axis, or, more commonly, it is called the torque, and is represented by τ (Greek lowercase letter tau). Thus, the angular acceleration α of an object is directly proportional to the net applied torque τ: α α τ, we see that it is torque that gives rise to angular acceleration. This is the rotational analog of Newton's second law for linear motion, a α F. - We defined the lever arm as the perpendicular distance from the axis of rotation to the line of action of the force that is, the distance which is perpendicular both to the axis of rotation and to an imaginary line drawn along the direction of the force. We do this to take into account the effect of forces acting at an angle. It is clear that a force applied at an angle, such as F_c in Fig. 8-12, will be less effective than the same magnitude force applied perpendicular to the door, such as F_A (Fig. 8-12a). And if you push on the end of the door so that the force is directed at the hinge (the axis of rotation), as indicated by F_D, the door will not rotate at all. The lever arm for a force such as F_c is found by drawing a line along the direction of F_c (this is the "line of action" of F_c). Then we draw another line, perpendicular to this line of action, that goes to the axis of rotation and is perpendicular also to it. The length of this second line is the lever arm for F_c and is labeled r_c in Fig. 8-12b. The lever arm for F_A is the full distance from the hinge to the doorknob, r (just as in Fig. 8-10). Thus r_c is much smaller than r. The magnitude of the torque associated with F_c is then r_c F_c. This short lever arm r_c and the corresponding smaller torque associated with F_c are consistent with the observation that F_c is less effective in accelerating the door than is F_A with its larger lever arm. When the lever arm is defined in this way, experiment shows that the relation τ α rF is valid in general. Notice in Fig. 8-12 that the line of action of the force F_D passes through the hinge, and hence its lever arm is zero. Consequently, zero torque is associated with F_D, and it gives rise to no angular acceleration, in accord with everyday experience (you can't get a door to start moving by pushing directly at the hinge). In general, then, we can write the magnitude of the torque about a given axis as τ = r⊥F (8-10a) where r is the lever arm, and the perpendicular symbol (⊥) reminds us that we must use the distance from the axis of rotation that is perpendicular to the line of action of the force (Fig. 8-13a). An equivalent way of determining the torque associated with a force is to resolve the force into components parallel and perpendicular to the line that connects the axis to the point of application of the force, as shown in Fig. 8-13b. The component F exerts no torque since it is directed at the rotation axis (its lever arm is zero). Hence the torque will be equal to F₁ times the distance from the axis to the point of application of the force: τ = rF₁ This gives the same result as Eq. 8-10a because F₁ = F sinθ and r = r sinθ. Thus τ = rF sinθ (8-10b) = F (rsinθ ) (8-10c) = F⊥r in either case. [Note that θ is the angle between the directions of F and r (radial line from the axis to the point where F acts).] We can use any of Eqs. 8-10 to calculate the torque, whichever is easiest. Because torque is a distance times a force, it is measured in units of m · N in SI units, cm · dyne in the cgs system, and ft · lb in the English system. ### Example 8-8 Biceps torque The biceps muscle exerts a vertical force on the lower arm, bent as shown in Figs. 8-14a and b. For each case, calculate the torque about the axis of rotation through the elbow joint, assuming the muscle is attached 5.0 cm from the elbow as shown. #### Approach The force is given, and the lever arm in (a) is given. In (b) we have to take into account the angle to get the lever arm. #### Solution * (a) F = 700 N and r₁ = 0.050 m, so: τ = rF = (0.050 m)(700 N) = 35 m·N * (b) Because the arm is at an angle below the horizontal, the lever arm is shorter (Fig. 8-14c) than in part (a): r = (0.050 m)(sin 60°), where θ = 60° is the angle between Fand r. F is still 700 N, so: τ = rF = (0.050 m)(0.866)(700 N) = 30 m·N The arm can exert less torque at this angle than when it is at 90°. Weight machines at gyms are often designed to take this variation with angle into account. **Note:** In (b), we could instead have used τ = rF. As shown in Fig. 8-14d, Fι = F sin 60°. Then τ = (0.050 m)(700 N)(0.866) gives the same result. *Note that the units for torque are the same as those for energy. We write the unit for torque here as m.N (in SI) to distinguish it from energy (N.m) because the two quantities are very different. The special name joule (1 J = 1 N.m) is used only for energy (and for work), never for torque. **Exercise B** Two forces (F_A= 20 N and F_B= 30 N) are applied to a meter stick which can rotate about its left end, Fig. 8-15. Force F_A is applied perpendicularly at the midpoint. Which force exerts the greater torque: F_A, F_B, or both the same? When more than one torque acts on an object, the angular acceleration α is found to be proportional to the net torque. If all the torques acting on an object tend to rotate it in the same direction about a fixed axis of rotation, the net torque is the sum of the torques. But if, say, one torque acts to rotate an object in one direction, and a second torque acts to rotate the object in the opposite direction, the net torque is the difference of the two torques. We normally assign a positive sign to torques that act to rotate the object counterclockwise (just as θ is usually positive counterclockwise), and a negative sign to torques that act to rotate the object clockwise. ### Forces that act to tilt the axis We have been considering only rotation about a fixed axis, and so we considered only forces that act in a plane perpendicular to the axis of rotation. If there is a force (or component of a force) acting parallel to the axis of rotation, it will tend to tilt the axis of rotation-the component F in Fig. 8-16 is an example. Since we are assuming the axis remains fixed in direction, either there can be no such forces or else the axis must be mounted in bearings or hinges that hold the axis fixed. Thus, only a force, or component of a force (F, in Fig. 8-16), in a plane perpendicular to the axis will give rise to rotational acceleration about the axis.