Rotation of an Object About a Fixed Axis

Chapter 1 Rotation of an Object About a Fixed Axi 1.1 The Important Stuff 1.1.1 Rigid Bodie; Rotation So far in our tudy of phyic we have (with few exception) dealt with particle, object whoe patial dimenion were unimportant for the quetion we were aking. We now deal with the (elementary!) apect of the motion of extended object, object whoe dimenion are important. The object that we deal with are thoe which maintain a rigid hape (the ma point maintain their relative poition) but which can change their orientation in pace. They can have tranlational motion, in which their center of ma move but alo rotational motion, in which we can oberve the change in direction of a et of axe that i glued to the object. Such an object i known a a rigid body. We need only a mall et of angle to decribe the rotation of a rigid body. Still, the general motion of uch an object can be quite complicated. Since thi i uch a complicated ubject, we pecialize further to the cae where a line of point of the object i fixed and the object pin about a rotation axi fixed in pace. When thi happen, every individual point of the object will have a circular path, although the radiu of that circle will depend on which ma point we are talking about. And the orientation of the object i completely pecified by one variable, an angle θ which we can take to be the angle between ome reference line painted on the object and the x axi (meaured counter-clockwie, a uual). Becaue of the nice mathematical propertie of expreing the meaure of an angle in radian, we will uually expre angle in radian all through our tudy of rotation; on occaion, though, we may have to convert to or from degree or revolution. Revolution, degree and radian are related by: 1revolution = 360degree = 2π radian 1

2 CHAPTER 1. ROTATION OF AN OBJECT ABOUT A FIXED AXIS r q Figure 1.1: A point on the rotating object i located a ditance r from the axi; a the object rotate through an angle θ it move a ditance. [Later, becaue of it importance, we will deal with the motion of a (round) object which roll along a urface without lipping. Thi motion involve rotation and tranlation, but it i not much more complicated than rotation about a fixed axi.] 1.1.2 Angular Diplacement A a rotating object move through an angle θ from the tarting poition, a ma point on the object at radiu r will move a ditance ; length of arc of a circle of radiu r, ubtended by the angle θ. When θ i in radian, thee are related by θ = r θ in radian (1.1) If we think about the conitency of the unit in thi equation, we ee that ince and r both have unit of length, θ i really dimenionle; but ince we are auming radian meaure, we will often write rad next to our angle to keep thi in mind. 1.1.3 Angular Velocity The angular poition of a rotating change with time; a with linear motion, we tudy the rate of change of θ with time t. If in a time period t the object ha rotated through an angular diplacement θ then we define the average angular velocity for that period a ω = θ t (1.2) A more intereting quantity i found a we let the time period t be vanihingly mall. Thi give u the intantaneou angular velocity, ω: θ ω = lim t 0 t = dθ dt Angular velocity ha unit of rad, or equivalently, 1 or 1. (1.3)

1.1. THE IMPORTANT STUFF 3 In more advanced tudie of rotational motion, the angular velocity of a rotating object i defined in uch a way that it i a vector quantity. For an object rotating counterclockwie about a fixed axi, thi vector ha magnitude ω and point outward along the axi of rotation. For our purpoe, though, we will treat ω a a number which can be poitive or negative, depending on the direction of rotation. 1.1.4 Angular Acceleration; Contant Angular Acceleration The rate at which the angular velocity change i the angular acceleration of the object. If the object (intantaneou) angular velocity change by ω within a time period t, then the average angular acceleration for thi period i α = ω (1.4) t But a you might expect, much more intereting i the intantaneou angular acceleration, defined a ω α = lim t 0 t = dω (1.5) dt We can derive imple equation for rotational motion if we know that α i contant. (Later we will ee that thi happen if the torque on the object i contant.) Then, if θ 0 i the initial angular diplacement, ω 0 i the initial angular velocity and α i the contant angular acceleration, then we find: ω = ω 0 + αt (1.6) θ = θ 0 + ω 0 t + 1 2 αt2 (1.7) ω 2 = ω 2 0 + 2α(θ θ 0) (1.8) θ = θ 0 + 1 2 (ω 0 + ω)t (1.9) where θ and ω are the angular diplacement and velocity at time t. θ 0 and ω 0 are the value of the angle and angular velocity at t = 0. Thee equation have exactly the ame form a the equation for one dimenional linear motion given in Chapter 2 of Vol. 1. The correpondence of the variable are: x θ v ω a α. It i almot alway implet to et θ 0 = 0 in thee equation, o you will often ee Eq. 1.6 1.9 written with thi ubtitution already made. 1.1.5 Relationhip Between Angular and Linear Quantitie A we wrote in Eq. 1.1, when a rotating object ha an angular diplacement θ, then a point on the object at a radiu r travel a ditance = rθ. Thi i a relation between the angular motion of the point and the linear motion of the point (though here linear i a bit of a minomer becaue the point ha a circular path). The ditance of the point from the axi

4 CHAPTER 1. ROTATION OF AN OBJECT ABOUT A FIXED AXIS doe not change, o taking the time derivative of thi relation give the intantaneou peed of the particle a: v = d dt = rdθ = rω (1.10) dt which we imilarly call the point linear peed (or, tangential peed), v T ) to ditinguih it from the angular peed. Note, all point on the rotating object have the ame angular peed but their linear peed depend on their ditance from the axi. Similarly, the time derivative of the Eq. 1.10 give the linear acceleration of the point: a T = dv dt = rdω dt = rα (1.11) Here it i eential to ditinguih the tangential acceleration from the centripetal acceleration that we recall from our tudy of uniform circular motion. It i till true that a point on the wheel at radiu r will have a centripetal acceleration given by: a c = v2 r = (rω)2 r = rω 2 (1.12) Thee two component pecify the acceleration vector of a point on a rotating object. (Of coure, if α i zero, then a T = 0 and there i only a centripetal component.) 1.1.6 Rotational Kinetic Energy Becaue a rotating object i made of many ma point in motion, it ha kinetic energy; but ince each ma point ha a different peed v our formula from tranlational particle motion, K = 1 2 mv2 no longer applie. If we label the ma point of the rotating object a m i, having individual (different!) linear peed v i, then the total kinetic energy of the rotating object i K rot = i 1 2 m iv 2 i = 1 2 m i vi 2 I If r i i the ditance of the i th ma point form the axi, then v i = r i ω and we then have: ( ) K rot = 1 m 2 i (r i ω) 2 = 1 (m 2 i ri 2 )ω 2 = 1 m 2 i ri 2 ω 2. i i i The um i m i r 2 i i called the moment of inertia for the rotating object (which we dicu further in the next ection), and uually denoted I. (It i alo called the rotational inertia in ome book.) It ha unit of kg m 2 in the SI ytem. With thi implification, our lat equation become K rot = 1 2 Iω2 (1.13)

1.1. THE IMPORTANT STUFF 5 1.1.7 The Moment of Inertia; The Parallel Axi Theorem For a rotating object compoed of many ma point, the moment of inertia I i given by I = i m i r 2 i (1.14) I ha unit of kg m 2 in the SI ytem, and a we ue it in elementary phyic, it i a calar 1 (i.e. a ingle number which in fact i alway poitive). More frequently we deal with a rotating object which i a continuou ditribution of ma, and for thi cae we have the more general expreion I = r 2 dm (1.15) Here, the integral i performed over the volume of the object and at each point we evaluate r 2, where r i the ditance meaured perpendicularly from the rotation axi. The evaluation of thi integral for variou of cae of interet i a common exercie in multi-variable calculu. In mot of our problem we will only be uing a few baic geometrical hape, and the moment of inertia for thee are given in Figure 1.2 and in Figure 1.3. Suppoe the moment of inertia for an object of ma M with the rotation axi paing through the center of ma i I CM. Now uppoe we diplace the axi parallel to itelf by a ditance D. Thi ituation i hown in Fig. 1.4. The moment of inertia of the object about the new axi will have a new value I, given by I = I CM + MD 2 (1.16) Eq. 1.16 i known a the Parallel Axi Theorem and i ometime handy for computing moment of inertia if we already have a liting for a moment of inertia through the object center of ma. 1.1.8 Torque We can impart an acceleration to a rotating object by exerting a force on it at a particular point. But it turn out that the force i not the implet quantity to ue in tudying rotation; rather it i the torque imparted by the force. Suppoe the force F (whoe direction lie in the plane of rotation) i applied at a point r (relative to the rotation axi which i at the origin O). Suppoe that the (mallet) angle between r and F i φ. Then the magnitude of the torque exerted on the object by thi force i τ = rf in φ (1.17) By ome very imple regrouping, thi equation can be written a τ = r(f inφ) = rf t 1 In advanced mechanic it i treated a a matrix, but we don t need to make thing that complicated!

6 CHAPTER 1. ROTATION OF AN OBJECT ABOUT A FIXED AXIS R R 1 R 2 (a) 2 2 M(R MR 2 1 +R 2 ) 2 1 (b) L R R (c) (d) 1 2 MR 2 1 4 MR 2 + 1 12 ML 2 1 12 L ML 2 (e) 1 3 ML 2 L (f) Figure 1.2: Formulae for the moment of inertia I for imple hape about the axe, a hown. In each cae, the ma of the object i M. (a) Hoop about ymmetry axi. (b) Annular cylinder about ymmetry axi. (c) Solid cylinder (dik) about ymmetry axi. (d) Solid cylinder (dik) about axi through CM, perpendicular to ymmetry axi. (e) Thin rod about axi through CM, perpendicular to length. (f) Thin rod about axi through one end, perpendicular to length.

1.1. THE IMPORTANT STUFF 7 R R (a) 2 5 MR 2 MR 2 2 3 (b) R b (c) a (d) 1 2 MR 2 1 12 M(a 2 +b 2 ) Figure 1.3: More formulae for moment of inertia I for imple hape about the axe, a hown. (a) Solid phere; axi i through a diameter. (b) Spherical hell; axi i through a diameter. (c) Hoop; axi i through a diameter. (d) Rectangular lab; axi i perpendicular, through the center. M D CM CM (a) (b) Figure 1.4: (a) Moment of inertia about an axi through the center of ma of an object i I CM. (b) We diplace the axi o that it i parallel and a ditance D away. New moment of inertia i given by the Parallel Axi Theorem. Thi object look like ome kind of potato, but the theorem will work for any vegetable.

8 CHAPTER 1. ROTATION OF AN OBJECT ABOUT A FIXED AXIS F F F t F O r f P O r f P O r r f P (a) (b) (c) Figure 1.5: (a) Rigid body rotate about point O. A force F i exerted at point P. (b) Torque τ can be viewed a the product of the ditance r and the tangential component of the force, F t. (c) Torque τ can alo be viewed a the product of the force F and the ditance r (often called the moment arm, or lever arm). where F t = F inφ i the component of the force perpendicular to r, or a τ = (r inφ)f = r F where r = r in φ i the ditance between the axi and the line which we get by extending the force vector into a line often called the line of action. The ditance r i called the moment arm of the force F. Thee different way of thinking about the term in Eq. 1.17 are illutrated in Fig. 1.5. Formula 1.17 give u the magnitude of the torque; trictly peaking, torque i a vector and in more advanced tudie of rotation, it mut be treated a uch. The (vector) torque due a force F i defined a τ = r F o we ee that Eq. 1.17 give the magnitude of thi vector. For now we will treat torque a a number which i poitive if the force give a counterclockwie rotation and negative if the force give a clockwie rotation. What we are calling τ here i really the component of the torque vector along the rotation axi. To repeat, τ i poitive or negative depending on whether the rotation which the force (acting alone) induce i counter-clockwie or clockwie. (However on occaion we may want to chooe the other convention.) When a number of individual force act on a rotating object, we can compute the net torque: τ net = τ i 1.1.9 Torque and Angular Acceleration (Newton Second Law for Rotation) The angular acceleration of a rotating object i proportional to the net torque on the object; they are related by: τ net = Iα (1.18)

1.1. THE IMPORTANT STUFF 9 Linear Quantity Angular Quantity x θ v ω a α m I F τ p L Table 1.1: Correpondence between linear and angular quantitie where I i the moment of inertia of the object. Thi equation look a lot like Newton Second Law for one dimenional motion, F x = ma x. 1.1.10 Work, Energy and Power in Rotational Motion A with the (abbreviated) formula for the work done by a force for a mall diplacement, linear motion, W = F x dx, we have a formula for the work done by a torque for a mall angular diplacement dθ: W = τdθ For a finite angular diplacement from θ i to θ f, the work done i W = 1.1.11 The New Equation Look Like the Old Equation θf θ i Although the rotational motion we have begun to tudy in thi chapter i really quite a different thing from the linear motion we have tudied up to now, it i of great help in uing (or at leat remembering) the equation by drawing correpondence between the new rotational quantitie and the old linear quantitie. Thee are ummarized in Table 1.1. For completene, we include the quantity angular momentum, L, which will be dicued in the next chapter. The baic relation between the old linear quantitie and the correponding relation between the new angular quantitie are ummarized in Table 1.2. Of coure, the firt three pair of equation deal with the pecial cae of contant linear or angular acceleration. τ dθ A few pointer for olving problem which involve rotation: The object which rotate are (obviouly) all extended object, that i, they have dimenion (width, height... ) which can t be ignored. In ome problem the center of ma of the rigid body doe move and we may need to undertand how the gravitational potential energy of the object change. The baic anwer i that we compute the gravitational potential energy by treating all the ma of the object a being located at it center of ma.

10 CHAPTER 1. ROTATION OF AN OBJECT ABOUT A FIXED AXIS Linear Relation Angular Relation v = v 0 + at ω = ω 0 + αt x = x 0 + v 0 t + 1 2 at2 θ = θ 0 + ω 0 t + 1 2 αt2 v 2 = v0 2 + 2a(x x 0) ω 2 = ω0 2 + 2α(θ θ 0) K = 1 2 mv2 K = 1 2 Iω2 F = ma τ = Iα p = mv L = Iω Table 1.2: Correpondence between baic equation for linear and angular motion. Some problem involve a pulley which turn becaue a tring i tightly wrapped around it or becaue a tring pae over the pulley and doe not lip againt it. In thoe cae the tring exert a force on the pulley which i tangential, and we ue thi fact in computing the torque on the pulley. But ee the next point... When a tring doe pa over a real pulley (i.e. it ha ma even though we might till ignore friction in the bearing) and exert force on the pulley, the tring tenion will not be the ame on both ide of the place where it i in contact. Thi differ from the problem in the chapter on force where pulley were involved; there, we ued the idealization of male pulley, and the tenion wa the ame on both ide. 1.2 Worked Example 1.2.1 Angular Diplacement 1. (a) What angle in radian i ubtended by an arc that ha length 1.80m and i part of a circle of radiu 1.20m? (b) Expre the ame angle in degree. (c) The angle between two radii of a circle i 0.620 rad. What arc length i ubtended if the radiu i 2.40m? [HRW5 11-1] (a) Eq. 1.1 relate arclength, radiu and ubtended angle. We find: θ = r = 1.80m 1.20m = 1.50rad The ubtended angle i 1.50 radian. (b) To expre thi angle in degree ue the relation: 360deg = 2π rad (or, 180deg = π rad). Then we have: ( ) 180deg 1.50rad = (1.50rad) = 85.9 π rad (c) We can find the arc length ubtended by an angle θ by the relation: = θr. Then for an angle of 0.620rad and radiu 2.40m, the arclength i = θr = (0.620)(2.40m) = 1.49m.

1.2. WORKED EXAMPLES 11 1.2.2 Angular Velocity 2. What i the angular peed in radian per econd of (a) the Earth in it orbit about the Sun and (b) the Moon in it orbit about the Earth? [Ser4 10-3] (a) The Earth goe around in a (nearly!) circular path with a period of one year. In econd, thi i: ( ) ( ) (3600 ) 365.25day 24hr 1yr = (1yr) = 3.156 10 7 1yr 1day 1hr In one year it angular diplacement i 2π radian (all the way around) o it angular peed i ω = θ t = 2π rad = 1.99 (3.156 10 7 10 7 ) (b) How long doe it take the moon to go around the earth? That a number we need to look up. You ought to know that it i about a month, but any good reference ource will tell you that it i 27.3day, officially called the idereal period of the moon.. (Thing get confuing becaue of the motion of the earth; full moon occur every 29.5day, a period which i called the ynodic period.) Converting to econd, we have: ( ) (3600 ) 24hr P = 27.3day = (27.3day) = 2.36 10 6 1day 1hr In that length of time the angular diplacement of the moon i 2π o it angular peed i ω = θ t = 2π (2.36 10 6 ) = 2.66 10 6 rad 1.2.3 Angular Acceleration; Contant Angular Acceleration 3. The angular poition of a point on the rim of a rotating wheel i given by θ = 4.0t 3.0t 2 + t 3, where θ i in radian if t i given in econd. (a) What are the angular velocitie at t = 2.0 and t = 4.0? (b) What i the average angular acceleration for the time interval that begin at t = 2.0 and end at t = 4.0? (c) What are the intantaneou angular acceleration at the beginning and end of thi time interval? [HRW5 11-5] (a) In the problem we are given the angular poition θ a a function of time. To find the (intantaneou) angular velocity at any time, ue Eq. 1.3 and find: ω(t) = dθ dt = d ( 4.0t 3.0t 2 + t 3) dt = 4.0 6.0t + 3.0t 2

12 CHAPTER 1. ROTATION OF AN OBJECT ABOUT A FIXED AXIS where, if t i given in econd, ω i given in rad. The angular velocitie at the given time are then ω(2.0) = 4.0 6.0(2.0) + 3.0(2.0) 2 = 4.0 rad ω(4.0) = 4.0 6.0(4.0) + 3.0(4.0) 2 = 28.0 rad (b) Since we have the value of ω and t = 2.0 and t = 4.0, Eq. 1.4 give the average angular acceleration for the interval: α = ω t = 12.0 rad 2 = (28.0 rad 4.0 rad ) (4.0 2.0) The average angular acceleration i 12.0 rad 2. (c) We find the intantaneou angular acceleration from Eq. 1.5: α(t) = dω dt = d ( ) 4.0 6.0t + 3.0t 2 dt = 6.0 + 6.0t where, if t i given in econd, α i given in rad 2. Then at the beginning and end of our time interval the angular acceleration are: α(2.0) = 6.0 + 6.0(2.0) = 6.0 rad 2 α(4.0) = 6.0 + 6.0(4.0) = 18.0 rad 2 4. An electric motor rotating a grinding wheel at 100 rev/min i witched off. Auming contant negative angular acceleration of magnitude 2.00 rad, (a) how 2 long doe it take the wheel to top? (b) Through how many radian doe it turn during the time found in (a)? [Ser4 10-5] (a) Convert the initial rotation rate to radian per econd: 100 rev = ( ) ( ) (1min ) 100 rev 2π rad = 10.5 rad min min 1rev 60 When the wheel ha topped then of coure it angular velocity i zero. Since we know ω 0, ω and α we can ue Eq. 1.6 to get the elaped time: and we get: ω = ω 0 + αt = t = (ω ω 0) α t = (0 10.5 rad ) ( 2.00 rad 2 ) = 5.24

1.2. WORKED EXAMPLES 13 The wheel take 5.24 to top. (b) We want to find the angular diplacement θ during the time of topping. Since we know that the angular acceleration i contant we can ue Eq 1.9, and it might be implet to do o. then we have: θ = 1 (ω 2 0 + ω)t = 1 rad (10.5 + 0)(5.24) = 27.5rad. 2 The wheel turn through 27.5 radian in coming to a halt. 5. A phonograph turntable rotating at 33 1 rev/min low down and top in 30 3 after the motor i turned off. (a) Find it (uniform) angular acceleration in unit of rev/min 2. (b) How many revolution did it make in thi time? [HRW5 11-12] (a) Here we are given the initial angular velocity of the turntable and it final angular velocity (namely zero, when it top) and the time interval between them. We can ue Eq. 1.6 to find α, which we are told i contant. We have: α = ω ω 0 t We don t need to convert the unit of the data to radian and econd; if we watch our unit, we can ue revolution and minute. Noting that the time for the turntable to top i t = 30 = 0.50min, and with ω 0 = 33.3 rev and ω = 0 we find: min α = 0 33.3 rev min 0.50 min = 66.7 rev min 2 The angular acceleration of the turntable during the time of topping wa 66.7 rev min 2. (The minu ign indicate a deceleration, that i, an angular acceleration oppoite to the ene of the angular velocity.) (b) Here we want to find the value of θ at t = 0.50min. To get thi, we can ue either Eq. 1.7 or Eq. 1.9. With θ 0 = 0, Eq. 1.9 give u: θ = θ 0 + 1 2 (ω 0 + ω)t = 1 rev (33.3 + 0)(0.50min) 2 min = 8.33 rev The turntable make 8.33 revolution a it low to a halt. 6. A dik, initially rotating at 120 rad, i lowed down with a contant angular acceleration of magnitude 4.0 rad. (a) How much time elape before the dik 2 top? (b) Through what angle doe the dik rotate in coming to ret? [HRW5 11-13]

14 CHAPTER 1. ROTATION OF AN OBJECT ABOUT A FIXED AXIS (a) We are given the initial angular velocity of the dik, ω 0 = 120 rad. (We let the poitive ene of rotation be the ame a that of the initial motion.) We are given the magnitude of the dik angular acceleration a it low, but then we mut write α = 4.0 rad 2. The final angular velocity (when the dik ha topped!) i ω = 0. Then from Eq. 1.6 we can olve for the time t: ω = ω 0 + αt = t = ω ω 0 α and we get: rad (0 120 t = ) = 30.0 ( 4.0 rad ) 2 (b) We ll let the initial angle be θ 0 = 0. We can now ue any of the contant α equation containing θ to olve for it; let chooe Eq. 1.8, which give u: and we get: ω 2 = ω 2 0 + 2α(θ) = θ = (ω2 ω 2 0 ) 2α θ = (02 (120 rad )2 ) 2( 4.0 rad 2 ) = 1800rad The dik turn through an angle of 1800 radian before coming to ret. 7. A wheel, tarting from ret, rotate with a contant angular acceleration of 2.00 rad. During a certain 3.00 interval, it turn through 90.0rad. (a) How long 2 had the wheel been turning before the tart of the 3.00 interval? (b) What wa the angular velocity of the wheel at the tart of the 3.00 interval? [HRW5 11-19] (a) We are told that ome time after the wheel tart from ret we meaure the angular diplacement for ome 3.00 interval and it i 9.00 rad. Suppoe we tart meauring time at the beginning of thi interval; ince thi time meaurement in t from the beginning of the wheel motion, we ll call it t. Now, with the uual choice θ 0 = 0 we know that at t = 3.00 we have θ = 90.0rad. Alo α = 2.00 rad 2. Uing Eq. 1.7 get: which we can ue to olve for ω 0 : o that 90.0rad = ω 0 (3.00) + 1 rad (2.00 )(3.00) 2 2 2 ω 0 (3.00) = 90.0rad 1 rad (2.00 )(3.00) 2 = 81.0rad 2 2 ω 0 = 27.0 rad (Looking ahead, we can ee that we ve already anwered part (b)!)

1.2. WORKED EXAMPLES 15 Now uppoe we meaure time from the beginning of the wheel motion with the variable t. We want to find the length of time required for ω to get up to the value 27.0 rad. For thi period the initial angular velocity i ω 0 = 0 and the final angular velocity i 27.0 rad. Since we have α we can ue Eq. 1.6 to get t: ω = ω 0 + αt = t = ω ω 0 α which give rad (27.0 0 rad t = ) = 13.5 2.00 rad 2 Thi tell u that the wheel had been turning for 13.5 before the tart of the 3.00 interval. (b) In part (a) we found that at the beginning of the 3.00 interval the angular velocity wa 27.0 rad. 1.2.4 Relationhip Between Angular and Linear Quantitie 8. What i the angular peed of a car travelling at 50 km hr and rounding a circular turn of radiu 110m? [HRW5 11-27] To work conitently in SI unit, convert the peed of the car: ( v = 50 km km 10 3 ) ( ) = (50 ) m 1hr = 13.9 m hr hr 1km 3600 The relation between the car linear peed v and it angular peed ω a it goe around the track i v = rω. Thi give: ω = v r = 13.9 m 110m = 0.126 rad 9. An atronaut i being teted in a centrifuge. The centrifuge ha a radiu of 10m and, in tarting, rotate according to θ = 0.30t 2, where t in econd give θ in radian. When t = 5.0, what are the atronaut (a) angular velocity, (b) linear peed, (c) tangential acceleration (magnitude only) and (d) radial acceleration (magnitude only)? [HRW5 11-32] (a) We are given θ a a function of time. We get the angular velocity from it definition, Eq. 1.3, ω = dθ dt = d dt (0.30t2 ) = 0.60t where we mean that when t i in econd, ω i given in rad. When t = 5.0 thi i ω = (0.60)(5.0) m = 3.0 rad

16 CHAPTER 1. ROTATION OF AN OBJECT ABOUT A FIXED AXIS (b) The linear peed of the atronaut i found from Eq. 1.10 (the linear or tangential peed of a ma point): v = Rω = (10m)(0.60t) = 6.0t where we mean that when t i given in econd, v i given in m. When t = 5.0 thi i v = (6.0)(5.0) m = 30.0 m (c) The (magnitude of the) tangential acceleration of a ma point i given by Eq. 1.11. We will need the angular acceleration α, which i o that α = dω dt = d (0.60t) = 0.60 dt a T = Rα = (10m)(0.60) = 6.0 m 2. Here, ince a T i contant we have written in the appropriate unit, which are mp. (Since a T i contant, the anwer i the ame at t = 5.0 a at any other time.) (d) The radial acceleration of the atronaut i our old friend (?) the centripetal acceleration. From Eq. 1.12 we can get the magnitude of a c from: a c = Rω 2 = (10m)(0.60t) 2 = 3.6t 2 where we mean that if t i given in econd, a c i given in mp. When t = 5.0 thi i a c = (3.6)(5.0) 2 m 2 = 90 m 2 1.2.5 Rotational Kinetic Energy 10. Calculate the rotational inertia of a wheel that ha a kinetic energy of 24, 400J when rotating at 602rev/min. [HRW5 11-45] Find the angular peed ω of the wheel, in rad ω = 602 rev min ( 2π rad 1rev : ) (1min ) 60 = 63.0 rad We have ω and the kinetic energy of rotation, K rot, o we can find the rotational kinetic energy from K rot = 1 2 Iω2 = I = 2K rot ω 2 We get: 2(24, 400J) I = (63.0 rad = 12.3kg m2 )2 The moment of inertia of the wheel i 12.3kg m 2.

1.2. WORKED EXAMPLES 17 20 40 60 80 CM Figure 1.6: Rotating ytem for Example 11. 1.2.6 The Moment of Inertia (and More Rotational Kinetic Energy) 11. Calculate the rotational inertia of a meter tick with ma 0.56kg, about an axi perpendicular to the tick and located at the 20cm mark. [HRW5 11-53] A picture of thi rotating ytem i given in Fig. 1.6. The tick i one meter long (being a meter tick and all that) and we take it to be uniform o that it center of ma i at the 50cm mark. But the axi of rotation goe through the 20cm mark. Now if the axi did pa through the center of ma (perpendicular to the tick), we would know how to find the rotational inertia; from Figure 1.2 we ee that it would be I CM = 1 12 ML2 = 1 12 (0.56kg)(1.00m)2 = 4.7 10 2 kg m 2 The rotational inertia about our axi will not be the ame. We note that our axi i diplaced from the one through the CM by 30cm. Then the Parallel Axi Theorem (Eq. 1.16) tell u that the moment of inertia about our axi i given by I = I CM + MD 2 where I CM = 4.7 10 2 kg m 2, a we ve already found, M i the ma of the rod and D i the ditance the axi i diplaced (parallel to itelf), namely 30cm. We get: I = 4.7 10 2 kg m 2 + (0.56kg)(0.30m) 2 = 9.7 10 2 kg m 2 So the rotational inertia of the tick about the given axi i 0.097kg m 2. 12. Two mae M and m are connected by a rigid rod of length L and negligible ma a in Fig. 1.7. For an axi perpendicular to the rod, how that the ytem ha the minimum moment of inertia when the axi pae through the center of ma. Show that thi moment of inertia i I = µl 2, where µ = mm/(m + M). [Ser4 10-21]

18 CHAPTER 1. ROTATION OF AN OBJECT ABOUT A FIXED AXIS L M m x L-x Figure 1.7: Rotating ytem for Example 12. A noted in the figure, let the ditance of the axi from the ma M be x; then it ditance from ma m mut be L x. The definition of the moment of inertia tell u that for thi ytem I = i m i r 2 i = Mx 2 + m(l x) 2 = Mx 2 + m(l 2 2Lx + x 2 ) = (M + m)x 2 2mLx + ml 2 The final expreion how that a a function of x, I i quadratic with a poitive coefficient in front of the x 2 term. The graph of thi function i a parabola which face up, and it ha a minimum at the value of x for which di/dx = 0. Solving for thi value, we get: m di dx = 2(M + m)x 2mL = 0 = x = ml (M + m) (Note, ince i a number between zero and 1 thi i a point between the two mae.) (M+m) Now, taking the origin to be at ma M, the coordinate of the center of ma of thi ytem i located at (M 0 + m L) ml x CM = = (m + M) (m + M) o the axi of minimum I doe indeed pa through the center of ma. Subtituting thi value of x into the expreion for I we find: I min = M ( ) 2 ( ml + m L (M + m) = Mm2 L 2 (M + m) 2 + m = Mm2 L 2 (M + m) 2 + m ml (M + m) ( ) 2 L(M + m) ml (M + m) ( ML (M + m) = Mm2 L 2 + mm 2 L 2 (M + m) 2 = ) 2 ) 2 mm(m + M) (M + m) 2 L 2

1.2. WORKED EXAMPLES 19 30 o 1.25 m T 0.75 kg (a) (b) mg 30 o Figure 1.8: The pendulum of Example 13, at a poition of 30 from the vertical. The econd picture how the force acting on the ball at the end of the rod. = mm (M + m) L2 Now if we let µ = mm/(m + m) we can write thi a 1.2.7 Torque I min = µl 2. 13. A mall 0.75kg ball i attached to one end of a 1.25m long male rod, and the other end of the rod i hung from a pivot. When the reulting pendulum i 30 from the vertical, what i the magnitude of the torque about the pivot? [HRW5 11-61] Draw a picture of the ytem! In Fig. 1.8 we how the baic geometry and alo the force which are acting on the ball at the end of the rod. (The rod i male o no external force act on it; we only need to worry about the torque produced by the force acting on the ball.) To calculate the torque due to each force, we need the magnitude of the force, the ditance at which it act from the pivot (here, it i 1.25m) and the angle between the force and the lever arm (that i, the line joining the pivot to the place where the force act). You might think we would want to find the tenion T in the rod before calculating the torque, but it i not neceary. The force of the rod on the ball point along the vector r. So φ = 0 and there i no torque. The force of gravity make an angle of 30 from the vector r and ince it ha magnitude mg, the magnitude of it torque i τ = rf inφ = (1.25m)(0.75kg)(9.80 m 2 )in 30 = 4.6N m The net torque ha magnitude 4.6N m.

20 CHAPTER 1. ROTATION OF AN OBJECT ABOUT A FIXED AXIS F A F C C 160 o O 135 o A F B 90 o B Figure 1.9: Force acting on a rotating body in Example 14. 14. The body in Fig. 1.9 i pivoted at O. Three force act on it in the direction hown on the figure: F A = 10N at point A, 8.0m from O; F B = 16N at point B, 4.0m from O; and F C = 19N at point C, 3.0m from O. What i the net torque about O? [HRW5 11-64] Start with the force applied at point A. Here r A = 8.0m and for our purpoe φ A can be taken a 135 or 45. The magnitude of the torque imparted by F A i τ A = r A F A inφ A = (8.0m)(10.0N)in(135 ) = 56.6N m Clearly F A i a force which give a torque in the counterclockwie (poitive, uually) ene, o we write: τ A = +56.6N m Next, the force applied at B. With r B = 4.0m and φ B = 90 we find: τ B = r B F B inφ B = (4.0m)(16.0N)in(90 ) = 64.0N m but F B i clearly a force which give a clockwie (i.e. negative) torque, and o τ B = 64.0N m And finally, the force applied at C. With r C = 3.0m and φ C = 160, we have τ C = r C F C inφ C = (3.0m)(19.0N)in(160 ) = 19.5N m. One can ee that F C give a counterclockwie torque and o Now find the total torque! τ C = +19.5N m τ net = τ i = (56.6N m) + ( 64.0N m) + (19.5N m) = 12.1N m The net (counterclockwie) torque on the object i 12.1N m.

1.2. WORKED EXAMPLES 21 1.2.8 Torque and Angular Acceleration (Newton Second Law for Rotation) 15. When a torque of 32.0N m i applied to a certain wheel, the wheel acquire an angular acceleration of 25.0 rad 2. What i the rotational inertia of the wheel? [HRW5 11-65] Torque, angular acceleration and the moment of inertia are related by τ = Iα. Solving for I, we find: I = τ α = 32.0N m 25.0 rad = 1.28kg m 2 2 16. A thin pherical hell ha a radiu of 1.90m. An applied torque of 960N m impart to the hell an angular acceleration equal to 6.20 rad about an axi through 2 the center of the hell. (a) What i the rotational inertia of the hell about the axi of rotation? (b) Calculate the ma of the hell. [HRW5 11-69] (a) From the relation between (net) torque, moment of inertia and angular acceleration, τ = Iα we have I = τ α Uing the given torque on the hell and it angular acceleration we find that the rotational inertia of the hell i (960N m) I = (6.20 rad = 155kg m 2. ) 2 (b) From Figure 1.3 the moment of inertia for a pherical hell of ma M and radiu R rotating about any diameter i I = 2 3 MR2 We know I and R o we can olve for the ma of the hell: M = 3I 2R = 3(155kg m2 ) 2 2(1.90m) 2 = 64.4kg 17. A wheel of radiu 0.20 M i mounted on a frictionle horizontal axi. A male cord i wrapped around the wheel and attached to a 2.0kg object that lide on a frictionle urface inclined at 20 with the horizontal, a hown in Fig. 1.10. The object accelerate down the incline at 2.0 m 2. What i the rotational inertia of the wheel about it axi of rotation? [HRW5 11-72] A in mot problem of thi type, we will find a olution by diagramming the force which act on each ma and then uing Newton law to et up equation. Here there are two mae we need to iolate: The block and the pulley.

22 CHAPTER 1. ROTATION OF AN OBJECT ABOUT A FIXED AXIS 2.0 kg 20 o Figure 1.10: Sliding ma on tring and pulley, a decribed in Example 17. N T T R mg in q mg mg co q (a) (b) Figure 1.11: (a) Force on the liding ma in Example 17. (b) Force (and it poition of application on the pulley) in Example 17.

1.2. WORKED EXAMPLES 23 We tart with the liding ma; the force acting on it are hown in Fig. 1.11(a). There i the downward force of gravity mg on the ma, which we eparate into it component: mg inθ down the lope and mg coθ into the plane. There i the normal force N of the urface (outward and perpendicular to the lope) and the tenion of the tring T which point up the lope. Since in thi problem we take the urface to be frictionle, that all. The block can only move along the lope o the force perpendicular to the lope mut cancel. Thi give N = mg co θ, which we don t really need! If the acceleration down the lope i a, then adding up the down the lope force and uing Newton 2 nd law give: mg inθ T = ma (1.19) Now conider the pulley. With the tring wrapped around it at a radiu R, the action of the tring on the wheel i that of a tangential force of magnitude T applied at a ditance R, hown in Fig. 1.11(b). The force i perpendicular to the line joining the axle and the point of application o that it give a (counterclockwie) torque of magnitude rf in φ = RT. Then the relation between torque and angular acceleration give: τ = RT = Iα where I i the moment of inertia of the wheel. Now if the tring i wrapped around the wheel then a it roll off the edge of the wheel it mut be true that the linear acceleration of any piece of the tring i the ame a the tangential acceleration of the wheel edge, and thi i alo the ame a the acceleration of the ma down the lope. Since the tangential acceleration of the edge of the wheel i a t = Rα, we have a = Rα, or α = a/r. Putting thi into the lat equation give RT = I a R or T = Ia R 2 and thi give an expreion for T that we can put into Eq. 1.19. Doing thi, we get Solving for I we find mg inθ Ia R 2 = ma Ia/R 2 = m(g inθ a) = I = mr 2(g in θ a) a Now we can plug in the number and we get: I = (2.0kg)(0.20m) 2((9.80 m 2 )in20 2.0 m 2 ) (2.0 m 2 ) = 5.4 10 2 kg m 2 The rotational inertia of the wheel i 0.054kg m 2. 18. A block of ma m 1 and one of ma m 2 are connected by a male tring over a pulley that i in the hape of a dik having radiu R and ma M. In

24 CHAPTER 1. ROTATION OF AN OBJECT ABOUT A FIXED AXIS m 1 M, R m 2 q Figure 1.12: Mae joined by tring and moving on wedge; pulley (a dik) ha ma! addition, the block are allowed to move on a fixed block wedge of angle θ a in Fig. 1.12. The coefficient of kinetic friction for the motion of either ma on the wedge urface i µ k. Determine the acceleration of the two block. [Ser4 10-29] Thi problem i imilar to one you may have een in the chapter on force problem. We will olve it in the ame way, by drawing a free body diagram for each ma, writing out Newton 2 nd Law for each ma and olving the equation. The difference come from the fact that one of the ma element in thi problem (the pulley!) undergoe rotational motion becaue of the torque that are exerted on it. It will undergo an angular acceleration which will be related to the common linear acceleration of the two block. We tart by noting that for a problem where a tring pae over a pulley which ha ma there will be a different tenion for each ection of the tring. (Thi differ from the cae of the ideal male pulley where the tring tenion wa the ame on both ide.) We will ee better why thi ha to be true when we look at the torque acting on the pulley. The part of the tring which i connected to m 1 will have a tenion T 1 and the part of the tring connected to m 2 will have a tenion T 2. Then we think about how the block are going to move in thi ituation. A they are connected by a tring, the magnitude of their acceleration will be the ame; we can call it a. We know that regardle of the value of the mae, m 1 mut move to the right and m 2 mut move down the lope. (Thi aume that friction i not o trong a to prohibit any motion!) And the pulley will be turning clockwie. Now we can draw the force diagram for the two block; we get the diagram given in Fig. 1.13. On m 1 we have the downward force of gravity m 1 g and the upward normal force N 1 of the urface; the tring pull to the right with a force of magnitude T 1 while there i a leftward friction force on the block which we denote f k,1. We know that m 1 ha an acceleration to right of magnitude a. The vertical force on m 1 mut cancel out, giving N 1 = m 1 g. The kinetic friction force i then f k,1 = µ k N 1 = µ k m 1 g

1.2. WORKED EXAMPLES 25 N 2 T 2 N 1 f k,1 m 1 T 1 f k,2 m 2 q m 1 g m 2 g Figure 1.13: Free body diagram for the mae m 1 and m 2. Applying Newton 2 nd law for the horizontal force give giving u our firt take home equation, T 1 f k,1 = T 1 µ k m 1 g = m 1 a T 1 µ k m 1 g = m 1 a (1.20) We move on to m 2. On thi ma we have the downward force of gravity, m 2 g, (which we how a the um of it component along the lope and perpendicular to the lope) the normal force from the urface, N 2, the force from the tring (magnitude T 2, directed up the lope) and the force of kinetic friction, which ha magnitude f k,2 and i directed up the lope becaue we know that block m 2 i moving down the lope. The force (that i, their component) perpendicular to the lope mut cancel out, and from our experience with imilar problem we can ee that thi will give N 2 = m 2 g co θ. Thi give the magnitude of the force of kinetic friction on m 2, f k,2 = µ k N 2 = µ k m 2 g co θ Now look at the force component on m 2 in the down the lope direction. By Newton 2 nd law they um to give m 2 a, and o we write: m 2 g inθ T 2 f k,2 = m 2 a We ubtitute for f k,2 and get our next take home equation, m 2 g inθ T 2 µ k m 2 g coθ = m 2 a (1.21) The two equation we have found have three unknown: T 1, T 2 and a. We will never olve the problem if we don t get a third equation, and it i the equation of (rotational) motion for the pulley that give u the lat equation.

26 CHAPTER 1. ROTATION OF AN OBJECT ABOUT A FIXED AXIS T 1 R M T 2 Figure 1.14: Force acting on the pulley in Example 18. The force diagram for the pulley i given in Fig. 1.14. When the tring i wrapped over the pulley and doe not lip, it act to exert a tangential force T 2 at the rim of the dik in one direction and a tangential force T 1 in the other direction, a hown. Thee force act at a ditance R from the pivot point (the pulley axle) and each i perpendicular to the line joining the pivot and the point of application. If we take the clockwie torque a being poitive, then the net torque on the dik i τ net = T 2 R T 1 R = (T 2 T 1 )R. From the rotational form of the econd law, we then have τ net = (T 2 T 1 )R = Iα where I i the moment of inertia of the pulley and α i it angular acceleration in the clockwie ene, ince that i how we defined poitive torque τ. We can ue a couple other fact before finihing with thi equation; we aume the pulley i a uniform dik, and o I i given by I = 1 2 MR2. Alo, we know that the linear tangential acceleration of the rim of the dik mut be the ame a the linear acceleration of the tring and alo the mae m 1 and m 2. Thi give u: αr = a = α = a/r Putting both of thee relation into the τ = Iα equation, we get ( ) a (T 2 T 1 )R = Iα = 1 2 MR2 = 1 2 R MRa and o if we cancel a factor of R we have a third equation relating the three unknown, (T 2 T 1 ) = 1 Ma (1.22) 2 Interetingly enough, the radiu R doe not appear. Our final anwer will not depend on it!

1.2. WORKED EXAMPLES 27 R M w R M m 50 cm m Figure 1.15: Before and after picture for a ma pulley ytem which tart from ret. Take the pulley to be a uniform dik. At thi point, we have three equation for three unknown (Eq. 1.20, 1.21 and 1.22.) The phyic part of the problem i done. What remain i the algebra involved in olving for the unknown T 1, T 2 and a. Eq. 1.20 and 1.21 give T 1 = m 1 a + µ k m 1 g and T 2 = m 2 a + m 2 g inθ µ k m 2 g co θ When we put thee into Eq. 1.22 we get: (m 2 + m 1 )a µ k g(m 2 co θ + m 1 ) + m 2 g inθ = 1 2 Ma Thi contain only the acceleration a and o we can quickly olve for it. A little rearranging give (m 1 + m 2 + M/2)a = m 2 g inθ µ k g(m 1 + m 2 co θ) Iolating a and factoring out g give a = [m 2 inθ µ k (m 1 + m 2 co θ)] g (m 1 + m 2 + M/2) for the acceleration of the block. Thi expreion can only make ene if µ k i not o big that it make a negative in thi expreion. 1.2.9 Work, Energy and Power in Rotational Motion v 19. (a) If R = 12cm, M = 400g and m = 50g in Fig. 1.15, find the peed of the block after it ha decended 50cm tarting from ret. Solve the problem uing energy conervation principle. (b) Repeat (a) with R = 5.0 cm. (Aume the pulley i a uniform dik.) [HRW5 11-77] (a) If there i no friction in the axle of the pulley then mechanical energy will be conerved a the block decend. The change in total energy between the before and after picture in Fig. 1.15 will be zero: E = K + U = 0

28 CHAPTER 1. ROTATION OF AN OBJECT ABOUT A FIXED AXIS Initially, the block i at ret and o i the pulley, o there i no kinetic energy in the ytem. In the final picture, the block ha ome peed v o it ha kinetic energy (namely 1 2 mv2 ) and the pulley i turning at ome angular peed ω o that it too ha kinetic energy. The kinetic energy of the pulley i K rot = 1 2 Iω2, o that the total kinetic energy in the after picture i K f = 1 2 mv2 + 1 2 Iω2 We can implify thi expreion by realizing that if the tring doe not lip on the pulley (and it i implied in the et-up that it i wrapped around it, o it doen t) then the tangential peed of the edge of the dik i the ame a that of the falling ma. Thi give ωr = v, or ω = v/r. Alo, auming that the pulley i a uniform dik, we have I = 1 2 MR2. Uing thee relation we get: ) 2 ( ( K f = 1 2 mv2 + 1 1 2 2 MR2) v R ( m = 1 2 mv2 + 1 4 Mv2 = 2 + M 4 ) v 2 The potential energy of the ytem change only from the change in height of the upended ma m. It change in height i y = 0.50m o that the change in potential energy of the ytem i U = mg y = mg( 0.50m) Putting everything into the energy conervation equation we get ( m K + U = 2 + M 4 ) v 2 + mg( 0.50m) = 0 which we can ue to olve for v ince we know all the other quantitie: ( m 2 + M 4 ) v 2 = mg(0.50m) = (0.050kg)(9.80 m 2 )(0.50m) = 0.245J ( (0.050kg) 2 + (0.400kg) ) v 2 = (0.125kg)v 2 = 0.245J 4 v 2 = (0.245J) (0.125kg) The final peed of the block i v = 1.4 m. v = 1.4 m = 1.96 m2 2 (b) In thi part we are aked to find the final peed v if R ha a different value. But if we look back at our olution for part (a) we ee that we never ued the given value of R! (It cancelled out when we wrote out the energy conervation condition in term of m, M and v.) So we get the ame anwer: v = 1.4 m.

1.2. WORKED EXAMPLES 29 10 cm CM 5 cm Axi Axi w CM (a) (b) Figure 1.16: Cylinder rotating about an off center axi, in Example 20. (a) i the initial poition of the cylinder; (b) how the cylinder a it move through it lowet poition. 20. A uniform cylinder of radiu 10 cm and ma 20 kg i mounted o a to rotate freely about a horizontal axi that i parallel to and 5.0cm from the central longitudinal axi of the cylinder. (a) What i the rotational inertia of the cylinder about the axi of rotation? (b) If the cylinder i releaed from ret with it central longitudinal axi at the ame height a the axi about which the cylinder rotate, what i the angular peed of the cylinder a it pae through it lower poition? (Hint: Ue the principle of conervation of energy.) [HRW5 11-84] (a) Firt, draw a picture of thi ytem; Fig. 1.16 (a) how an end on view of the rotating cylinder. It ymmetry axi i labelled CM but it rotational axi i marked Axi. The cylinder doe not turn about it center! (If it did, it moment of inertia would be I = 1 2 MR2, but that will not be the cae here.) We need ome other way of getting I. There i one pecial feature about thi rotation which will give u I. The rotation take place about an axi which i parallel to an axi for which we do know I, namely the ymmetry axi. For that axi, we have I CM = 1 2 MR2 (where M and R are the ma and radiu of the cylinder). Our axi i diplaced from that one by a ditance D = 5.0cm = 0.050m. Then the Parallel Axi Theorem, Eq. 1.16, give u: I = I CM + MD 2 = 1 2 MR2 + MD 2 = 1 2 (20kg)(0.10m)2 + (20kg)(0.050m) 2 = 0.15kg m 2 So the rotational (moment of) inertia of the cylinder about the given axi i 0.15kg m 2. (b) The cylinder tart in the poition hown Fig. 1.16 (a), that i, with the ymmetry axi at the ame height a the rotational axi. We know that the force of gravity act on the cylinder to give it an angular velocity which increae a the cylinder wing downward. We want to know the angular velocity when the center of ma of the cylinder i at it lowet point.

30 CHAPTER 1. ROTATION OF AN OBJECT ABOUT A FIXED AXIS It would be very hard to find the final ω uing torque, becaue t he force of gravity doe not act on the cylinder in the ame way through the wing. A with imilar problem in particle (point ma) motion, it i eaier to ue conervation of evergy. If there i no friction in the axi, then total mechanical energy i conerved, which we can write a: E = K + U = 0 Now, the cylinder tart from ret when it i at the poition hown in (a) o that K i = 0. In poition (b), the cylinder ha angular peed ω about the axi o that the final kinetic energy i K f = 1 2 Iω2 where I i the moment of inertia we found in part (a) of thi problem. There i a change in potential energy of the cylinder becaue there i a change in height of it center of ma. For an extended object we find the gravitational potential energy by imagining that all the ma i concentrated at the center of ma. Then for the cylinder, ince it center of ma ha fallen by a ditance 5.0cm (compare picture (a) and picture (b)) we ee that the change in U i: U = Mg y = (20kg)(9.80 m 2 )( 0.050m) = 9.80J Plugging everything into our energy conervation equation, we find for which we olve for ω: K + U = 1 2 Iω2 9.80J = 0 ω 2 = 2(9.80J) I = 131 2 = 2(9.80J) (0.15kg m 2 ) which give u ω = 11.4 rad 21. A tall, cylinder haped chimney fall over when it bae i ruptured. Treating the chimney a a thin rod with height h, expre the (a) radial and (b) tangential component of the linear acceleration of the top of the chimney a function of the angle θ made by the chimney with the vertical. (c) At what angle θ doe the linear acceleration equal g? [HRW5 11-87] (a) A uggeted in the problem we model the falling chimney a a thin rod of height h and ma M which turn about a pivot which i fixed to the ground at one place. (A real falling chimney may behave differently... it might break a it fall and the bae may move along the ground. Real life i lot more complicated.) The model i hown in Fig. 1.17 (a). The angle θ meaure the (intantaneou) angle of the chimney with the vertical. We know that

1.2. WORKED EXAMPLES 31 M q h q CM h/2 (h/2) co q (a) (b) Figure 1.17: (a) The falling chimney decribed in Example 21. (b) Poition of the center of ma of the chimney. a the chimney fall over (tarting from ret) it will loe gravitational potential energy and pick up rotational kinetic energy from it rotation about the pivot. It angular peed will increae a it fall. In part (a) we are aked for the radial acceleration of the end of the chimney when the chimney ha rotated through an angle θ. That jut the centripetal acceleration a c of the top of the chimney, which i given by Eq. 1.12 and ince the radiu of the circle in which the top move i h, thi i given by a c = v2 h = hω2. All that we need to know to compute a c i the angular peed of the rod, and for thi we can ue energy conervation for the rod a it fall. (Energy conervation i good for finding final peed!) When the rod i in the initial (vertical) poition, it i at ret o it ha no kinetic energy. Suppoe we meaure height from the ground level; then the center of ma of the rod i at a height h/2 and o the initial gravitational potential energy i Mg h. So the 2 initial total energy i E i = Mgh 2 Now uppoe the rod ha fallen through an angle θ, a hown in Fig. 1.17 (b). If it ha angular peed ω at that time, then it kinetic energy i 1 2 Iω2, where I i the moment of inertia of the rotating chimney. For a uniform rod of length h rotating about one end, thi i I = 1 3 Mh2, o we have: ( K f = 1 2 Iω2 = 1 1 2 3 Mh2) ω 2 = 1 6 Mh2 ω 2 A for the potential energy in thi poition, we can ee from geometry that the center of ma of the rod i at a height of h coθ. So the rod final potential energy i 2 and it final total mechanical energy i U f = Mg h 2 coθ. E f = K f + U f = 1 6 Mh2 ω 2 + Mg h 2 co θ

32 CHAPTER 1. ROTATION OF AN OBJECT ABOUT A FIXED AXIS CM q q F pivot Mg Figure 1.18: Force acting on falling rod (chimney) Energy conervation (we aume that thi pivot i frictionle... ), E i = E f give Mgh 2 which can olve for ω 2. Rearranging, we get: which we can ue in expreion for a c : = 1 6 Mh2 ω 2 + Mg h 2 coθ 1 6 Mh2 ω 2 = Mgh (1 co θ) 2 ω 2 = 3g (1 co θ) h a c = hω 2 = h 3g (1 coθ) = 3g(1 coθ). h The radial part of the acceleration of the tip of the chimney ha thi magnitude and i of coure directed inward toward the pivot. (b) From Eq. 1.11, the tangential component of the acceleration for a point on a rotating object a ditance h away from the pivot i a T = hα where α i the object angular acceleration. So we need to find the angular acceleration α of the chimney when it ha fallen through an angle θ. We can get α by finding the net torque acting on the rod and then uing τ = Iα. In Fig. 1.18 we how the force that act on the rod a it fall, and their point of application. Gravity (effectively) pull downward at the center of the rod with a force Mg, and the pivot alo exert a force on the rod. It not o clear which direction the latter force point, but in the end it doe not matter becaue we only want the torque that thee force exert about the pivot point and force of the pivot itelf will give no torque. The force of gravity i applied at a ditance h/2 from the pivot. It make an angle θ with the line which join the pivot and application point. Here, it give a torque in the counter-clockwie ene. So the torque due to gravity i τ = h Mg inθ 2

1.2. WORKED EXAMPLES 33 and thi i alo the total (counter-clockwie) torque becaue the force of the pivot give none. Then τ net = Iα give u h Mg inθ = Iα 2 where I i the moment of inertia of the rod. For a uniform rod of length h rotating about one end, it i I = 1 3 Mh2 o we get: h 2 Mg in θ = 1 3 Mh2 α Some algebra give u: α = 3g 2h inθ And at lat we find the tangential acceleration of the end of the chimney: a T = hα = h 3g 2h in θ = 3 2 g in θ. The anwer to (a) and (b) do not depend on the ma M of the chimney. (c) We want to know at what angle the linear (i.e. the tangential) acceleration of the chimney top i equal to g. Uing our anwer from part (b), we have the condition: Solve for θ: 3 g inθ = g 2 inθ = 2 = θ = in ( ) 1 2 3 3 = 41.8 The chimney will have fallen by 41.8 when the linear acceleration of it top i equal to g.

34 CHAPTER 1. ROTATION OF AN OBJECT ABOUT A FIXED AXIS