Rotation Torque Moment of Inertia Lana Sheridan De Anza College Mar 9, 2020 Last time • rotational quantities • torque • the cross product Overview • net torque • Newton's second law for rotation • moment of inertia • calculating moments of inertia Quick review of Vector Expressions #» #» #» Let a , b , and c be (non-null) vectors. Could this possibly be a valid equation? #» #» #» a = b · c (A) yes (B) no Quick review of Vector Expressions #» #» #» Let a , b , and c be (non-null) vectors. Could this possibly be a valid equation? #» #» #» a = b · c (A) yes (B) no Quick review of Vector Expressions #» #» #» Let a , b , and c be vectors. Let m and n be non-zero scalars. Could this possibly be a valid equation? #» #» m = b × c (A) yes (B) no Quick review of Vector Expressions #» #» #» Let a , b , and c be vectors. Let m and n be non-zero scalars. Could this possibly be a valid equation? #» #» m = b × c (A) yes (B) no Quick review of Vector Expressions #» #» #» Let a , b , and c be vectors. Let m and n be non-zero scalars. Suppose #» • a 6= 0 and #» • b 6= 0 and #» #» • a 6= n b for any value of n. Could this possibly be a true equation? #» #» #» #» a × b = b × a (A) yes (B) no Quick review of Vector Expressions #» #» #» Let a , b , and c be vectors. Let m and n be non-zero scalars. Suppose #» • a 6= 0 and #» • b 6= 0 and #» #» • a 6= n b for any value of n. Could this possibly be a true equation? #» #» #» #» a × b = b × a (A) yes (B) no Torque Torque is a measure of force-causing-rotation. It is not a force, but it is related. It depends on a force vector and its point of application relative to an axis of rotation. Torque is given by: #» #» #» τ = r × F That is: the cross product between #» • a vector r , the displacement of the point of application of the force from the axis of rotation, and #» • an the force vector F 300 Chapter 10 Rotation of a Rigid Object About a Fixed Axis ▸ 10.2 continued v 1.3 m/s 2 Do the same for the outer track: vf 5 5 22 5 22 rad/s 5 2.1 3 10 rev/min rf 5.8 3 10 m The CD player adjusts the angular speed v of the disc within this range so that information moves past the objective lens at a constant rate. (B) The maximum playing time of a standard music disc is 74 min and 33 s. How many revolutions does the disc make during that time? SOL U TION Categorize From part (A), the angular speed decreases as the disc plays. Let us assume it decreases steadily, with a constant. We can then apply the rigid object under constant angular acceleration model to the disc. Analyze If t 5 0 is the instant the disc begins rotating, with angular speed of 57 rad/s, the final value of the time t is (74 min)(60 s/min) 1 33 s 5 4 473 s. We are looking for the angular displacement Du during this time interval. 1 Use Equation 10.9 to find the angular displacement of Du 5 uf 2ui 5 2 vi 1vf t the disc at t 5 4 473 s: 1 5 5 2 57 rad/s11 22 rad2 /s 4 473 s 5 1.8 3 10 rad 1 1 rev21 2 Convert this angular displacement to revolutions: Du 5 1.8 3 105 rad 5 2.8 3 104 rev 2p rad 1 2 a b (C) What is the angular acceleration of the compact disc over the 4 473-s time interval? SOL U TION Categorize We again model the disc as a rigid object under constant angular acceleration. In this case, Equation 10.6 gives the value of the constant angular acceleration. Another approach is to use Equation 10.4 to find the average angular acceleration. In this case, we are not assuming the angular acceleration is constant. The answer is the same from both equations; only the interpretation of the result is different. Analyze Use Equation 10.6 to find the angular vf 2vi 22 rad/s 2 57 rad/s a5 5 5 27.6 3 1023 rad/s2 acceleration: t 4 473 s Finalize The disc experiences a very gradual decrease in its rotation rate, as expected from the long time interval required for the angular speed to change from the initial value to the final value. In reality, the angular acceleration of the disc is not constant. Problem 90 allows you to explore the actual time behavior of the angular acceleration. The component F sin f 10.4 Torque Torque tends to rotate the wrench about an axis through O. In our study of translational motion, after investigating the description of motion, we studied the cause of changes in motion: force. We follow the same plan here: F sin f S F What is the cause of changes in rotational motion? Imagine trying to rotate a door by applying a force of magnitude F perpendic- r f ular to the door surface near the hinges and then at various distances from the hinges. You will achieve a more rapid rate of rotation for the door by applying the O f F cos f Sr force near the doorknob than by applying it near the hinges. d Line of When a force is exerted on a rigid object pivoted about an axis, the object tends action to rotate about that axis. The tendency of a force to rotate an object about some axis is measured by a quantity called torque St(Greek letter tau). Torque is a vector, S but we will consider only its magnitude here; we will explore its vector nature in Figure 10.7 The force F has a Chapter 11. greater rotating#» #»tendency about an axis throughτ = r ×O asF F= increasesrF sin φ andn^ Consider the wrench in Figure 10.7 that we wish to rotate around an axis that is as the moment arm d increases. perpendicular to the page and passes through the center of the bolt. The applied #» #» where φ is the angle between r and F, and n^ is the unit vector #» #» perpendicular to r and F, as determined by the right-hand rule. Question B A A torque is supplied by applying a force at point A. To produce the same torque, the force applied at point B must be: (A) greater (B) less (C) the same 1Image from Harbor Freight Tools, www.harborfreight.com Question B A A torque is supplied by applying a force at point A. To produce the same torque, the force applied at point B must be: (A) greater (B) less (C) the same 1Image from Harbor Freight Tools, www.harborfreight.com 268 Chapter 9 Linear Momentum and Collisions is applied at the center of mass, the system moves in the direction of the force with- The system rotates clockwise 268when a force is appliedChapter 9 Linear outMomentum rotating and (see Collisions Fig. 9.13c). The center of mass of an object can be located with above the center of mass. this procedure. 268 Chapter 9 Linear is Momentum appliedThe center at and the Collisionsof center mass of themass, pair the of system particles moves described in the direction in Figure of 9.14 the is force located with on- The system rotates clockwise the x axis and lies somewhere between the particles. Its x coordinate is given by when a force is applied out rotating (see Fig. 9.13c). The center of mass of an object can be located with m x 1 m x CMabove the center of mass. isthis applied procedure. at the center of mass, the system1 moves1 in2 2the direction of the force with- Torque The system rotates clockwise xCM ; (9.28) out The rotating center (see of massFig. 9.13c).of the pairThe ofcenter particles ofm 1mass 1describedm of2 an objectin Figure can 9.14 be locatedis located with on Torque: when a force is applied the x axis and lies somewhere between the particles. Its x coordinate is given by above the center of mass. thisFor procedure.example, if x 5 0, x 5 d, and m 5 2m , we find that x 5 2d. That is, the a 1 2 2 1 1 CM 3 CM centerThe centerof mass of lies mass closer of the to pairthe moreof particles mmassive1x1 describedm particle.2x2 in If Figure the two 9.14 masses is located are equal, on xCM ; (9.28) thethe x center axis and of masslies somewhere lies midway between between the mthe particles.1 1 particles.m2 Its x coordinate is given by The system rotates counter- 1 2 For We example, can extend if x this5 0, concept x 5 d, toand a system mm 15x1 2ofm manym, 2wex2 findparticles that xwith5 massesd. That mi inis, threethe clockwiseaCM when a force is applied 1 2 x ; 2 1 CM 3 (9.28) dimensions. The x coordinate ofCM the center of mass of n particles is defined to be below the center of mass. center of mass lies closer to the more massivem1 1 mparticle.2 If the two masses are equal, Forthe example,center of massif x 5lies 0, midway x 5 d ,between and m 5the 2 mparticles., we find that x 5 2d. That is, the Thea system rotates counter- 1 2 2 1 mixi CMmixi3 We can extendm1x1 1 thism2 xconcept2 1 m3x to3 1 a systemc1 m ofn xmanyn a particlesi withai masses1 m in three clockwise when a force is applied center x of ;mass lies closer to the more massive particle.5 If the5 two masses5 arei equal,m x Torque: thedimensions.