Lecture8 - Chapter 10 Rota-on of Mo-on of rigid...

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Unformatted text preview: Chapter 10 Rota-on of Mo-on of rigid bodies Rota%onal mo%on: mo-on in which an extended object rotates about an axis A rigid object is one that is non ­deformable. This means that the rela-ve loca-ons of all par-cles of which the object is composed remain constant. We model extended object as a rigid body. In transla-onal mo-on, mo-on occurs when the object changes its posi-on during some -me interval. Rota-on axis Rota-on axis What does an object change during rota-onal mo-on ? Answer: Its angle with respect to some reference line. In terms of rota-on, a rigid object about a fixed axis means every par-cle on the object rotates in the same angle in a given -me interval. Rota-on angle (review) Angle in radians described by the reference line is given by θ= s r Final posi-on aOer rota-on s Tip of the “moving line” follows arc s Recall that 1 radian is the angle subtended at the center of the circle by and arc equal to the radius. How many radians corresponds to one full revolu-on of an object ? Conver-ng from radians to degrees: θ r Fixed reference line θ = 1rev = 360° = x rad = x rad ⋅ s 2π r = = 2π rad r r 360° 180° = x rad ⋅ 2π rad π rad NOTE: Unless specifically men-oned, calcula-ons are to be made in radians. So set your calculator accordingly! Angular displacement and angular velocity By replacing posi-on with angle, we can translate all formulas we learned from transla-on mo-on into rota-onal! Angular displacement: Average angular velocity: Instantaneous angular velocity: Angular accelera-on α avg Δθ = θ f − θ i ω avg = Δθ Δt ω= dθ dt Δω dω d 2θ = , α= =2 Δt dt dt Units: θ is measured in radian. Therefore ω is expressed in rad/s and α in rad/s2. Note that θ is dimensionless since it is ra-o of lengths. The dimensions ω of and α are T ­1 and T ­2 respec-vely. Rota-on with constant angular accelera-on Here, again, we just do “transla-on” of transla-onal mo-on with constant linear accelera-on to rota-onal mo-on with constant angular accelera-on: ω f = ωi + αt 1 θ f = θi + ω it + α t 2 2 ω 2 = ω i2 + 2α θ f − θ i f ( ) Also: θ f = θi + 1 ωi + ω f t 2 ( ) Note that we “translate” my making the following subs-tu-ons: x →θ v→ω a→α Angular displacement, velocity and vectors Although angular displacement Δθ also has a direc-on, it cannot be easily described by a vector. Consider the situa-on described in the figure on the right. For the final orienta-on, it depends on which order you rotate the object! This means that angular displacement does not follow the basic rules of vector addi-on, in par-cular: " Δθ1 " + " Δθ 2 " ≠ " Δθ 2 "+ " θ1 " Δ Therefore angular displacements are generally not considered vectors, unless they are very small (dθ). ω Angular velocity is a vector quan%ty. It is directed along the axis of rota-on. To fix which way it is directed along the axis of rota-on, we use the right hand rule: If your fingers curl around the rota:on axis and point in the direc:on of the rota:on your outstretched thumb will indicate the direc:on of the ω vector. Angular and linear veloci-es In rota-on of a rigid body, every par-cle in it moves in a circle whose center is on the axis of rota-on. The tangen-al velocity a par-cle of the object is given by v= We know, s = rθ. Thus v=r ds dt dθ dt P v θ r s → v = rω Every par-cle of a rigid object rotate about the common axis with the same angular speed. The tangen-al velocity, however, as shown by the last equa-on, depends on r – infact, it is propor-onal to r. A par-cle further from the axis moves faster. . v2 We also know: the centripetal accelera-on a and the tangen-al speed are related by ac = r 2 2 v In terms of angular speed this becomes ac rω . = ac = = rω 2 r On the other hand the radial accelera-on (nonuniform tangen-al speed ) is obtained from . Radial (centripetal) and tangen-al accelera-on We also know: the centripetal accelera-on a and the tangen-al speed are related by v2 ar = r Here i labels the ith par-cle of the rigid object. ar = rω 2 The tangen-al accelera-on comes from non uniform angular speed. at = d dω (rω ) = r → at = rα dt dt The total accelera-on is then a = a 2 + a t2 = r α 2 + ω 4 r Rota-onal kine-c energy When an object rotates, different parts of the object have different linear veloci-es: vi = riω And they will have different kine-c energies: ω r2 r1 1 K i = mi vi2 2 The total kine-c energy of a rota-ng object is equal to the sum of all the kine-c energies of its parts: ⎞ 1 1 1⎛ K = ∑ K = ∑ m v = ∑ m (ω r ) = ⎜ ∑ m r ⎟ ω 2 2 2⎝ ⎠ i 2 ii 2 2 i i ii i i i i 2 Thus we have What is I ? K= 12 Iω 2 where I = ∑ mi ri2 i I tells us how the mass is distributed about the axis of rota-on, i.e. what quan-ty of mass is what distance from the axis. I is called the moment of iner-a or rota-onal iner-a. I takes the role that mass m takes in transla-onal mo-on: K = 2 mv 2 ↔ K = 2 Iω 2 1 1 Example: Two wheels have the same mass M and radius R. Wheel 1 has the mass distributed uniformly throughout the wheel, wheel 2 has all the mass concentrated around its rim. Which has the higher moment of iner:a about the central axis? 1 2 Answer: Wheel 2, since equal mass is concentrated further from the axis. In wheel 1, some of the mass is close to the center, not contribu:ng much to the rota:onal iner:a (specifically ‘r2’). Example 10.3 (textbook) Four rota-ng masses. Calculate moment of iner-a and the rota-onal kine-c energy for both cases. (Chalk board stuff) Moment of iner-al of extended objects In case of an extended objects, we imagine it to be composed of many small elements, each of mass Δmi, and take the limit Δmi, ­> 0 : I = lim ∑ Δmi ri2 → I = ∫ r 2 dm Δmi → 0 i Examples 10.4 and 10.5 (rigid rod and solid cylinder) (Chalk board stuff) Example: What is the kine:c energy of a wheel which consists of two concentric rings rota:ng at 4 rad/s, one of mass 2 kg and a radius of 0.5 m from the center, the other with a mass of 1 kg and a radius of 0.2 m (assume they are held together by massless spokes)? Since they are rota-ng about the same axis, the moments of iner-a simply adds up: I = I1, circle + I 2, circle = m1r12 + m2 r22 = ( 2kg )(0.5m )2 + (1kg )(0.2m )2 = 0.54kg m 2 Kine-c Energy: Parallel axis theorem Lets take a simple case, in which the original axis is through the center of mass and shiO axis, such that new axis is parallel to the original one. Suppose that the the axis is now shiOed to P(a,a). Then the new distance of a small mass element dm at (x,y) is P x d z O y r 2 = ( x − a )2 + ( y − b )2 The moment of iner-al of the body about the new axis is then I = ∫ r 2 dm = ∫ [( x − a )2 + ( y − b )2 ] dm = ∫ ( x 2 + y 2 ) dm − 2 a ∫ x dm − 2b ∫ y dm + ∫ (a 2 + b 2 ) dm The two terms in the middle gives the coordinates of the CM, which are 0, here. Therefore, we have I = I CM + Md 2 Example: Center of mass of a rod about one of its ends. 1 1 ⎛1 ⎞ I= ML2 + M ⎜ L ⎟ = ML2 ⎝2 ⎠ 12 3 2 I CM Md 2 Torque We all know that it is easier to open the door by applying a force near the outer edge (that where the knob is placed) away from the hinges. We need to apply greater force if we want to open the door by pushing near the hinge. This and many other examples hints that we need a quan-ty that combines a force and the distance at which the force is applied from the axis of rota-ons. This quan-ty is torque. It is a vector quan-ty (denoted by τ) defined as the cross (vector) product of force and the vector posi%on of the force: τ =r×F We will elaborate on this defini-on later. Right now we focus on the magnitude of torque: d = rsinθ θ r τ = rF sin θ = Fd Here d is the perpendicular distance of the force from the axis. SI unit of torque is newton ­meter (Nm). Note that torque and work done are dimensionally same though they are quite different quan--es. F Rigid body under net torque Consider a par-cle of mass m rota-ng in a circle under the influence of a net tangen-al force ∑ F and a net radial force Fr . The radial force provides the centripetal t ∑ accelera-on while the tangen-al force provides the tangen-al accelera-on. ∑ F = ma t t ∑F 2 t The net torque on this force is m ∑ τ = ∑ F r = ma r = mr α t t ∑F r r But mr2 is the moment of iner-a of the par-cle about the axis of rota-on (at center of the circle). Thus: ∑ τ = Iα It can be shown that this formula also holds for an extended object of arbitrary shape rota-ng about a fixed axis (see the textbook, page 284). Thus net torque ac-ng on a par-cle or a rigid body about an axis is equal to the product of its moment of iner-a about that axis and angular accelera-on. Note that this formula is similar to, Newton’s second law ∑ F = ma , where τ, I and α are like F, m and a . Example: Problem 36. Chalk board stuff Work, power & Rota-onal Kine-c Energy Work: dW = Fds = F r dθ = τ dθ W = ∫ dW = ∫ τ dθ θi θf F dθ ds Power (for const. torque) P= dW dθ =τ → P = τω dt dt Compare this with P = F.v, of transla-onal mo-on. Rolling mo-on of a rigid object When an object rolls, it rotates and translates at the same -me. Assume the object rolls without slipping: Then the length of the arc s through which the wheel has moved is equal to the distance the wheel has been translated. The center of mass then moves through a distance of B A A B s = θR ds dθ = =R → vCM = Rω dt dt B A Taking the -me deriva-ve, we get the transla-onal speed of the center of mass: vCM And the accelera-on of the CM for this pure rolling mo-on is: aCM = R dω → aCM = Rα dt Rolling (contd…) We can see Rolling = Transla-on + Rota-on Note: the instantaneous velocity of the point on the wheel that is in contact with the ground is zero. This means that the wheel does not slip. The rolling mo-on can thus be seen as pure rota-on of the wheel about the contact point P at the bokom. This means that the total kine-c energy of rolling mo-on can be wriken as: 1 K = I Pω 2 2 Using parallel axis theorem: I P = I CM + Md 2 = I CM + MR 2 Thus, K= 1 1 1 2 I CM + MR 2 ω 2 ⇒ K = I CM ω 2 + MvCM 2 2 2 ( ) Total kine%c energy of a rolling object is the sum of the rota%onal kine%c energy about the center of mass and the transla%onal kine%c energy of the center of mass. Example Three objects, a ring, a disk and a sphere roll down a hill. The three objects have the same mass M and radius R. The hill makes an angle of 30o with the horizontal. If the hill has a height of 2 m, which of the three objects will reach the bottom first, and how fast will they be going ? How does this compare to a block sliding down the same hill without friction? 2 Block sliding down a fric-onless hill: Mgh = Mv ⇒ v = 2 gh = 6.3m/s 1 2 h 30o Object rolling down the hill: Mgh = Mv 2 + Iω 2 2 2 v I = nMR 2 n = 1 for ring ω= R n = ½ for disk n = 2/5 for sphere 1 1 1 1 1+ n ⎛ v⎞ ⇒ Mgh = Mv 2 + nMR 2 ⎜ ⎟ = Mv 2 ⎝ R⎠ 2 2 2 2 ⇒v= 2 gh < 2 gh 1+ n Ring v = 4.4 mls Disk v = 5.1 mls Sphere v = 5.3 mls So, the sphere will make it down first, followed by the disk, and the ring will be last. ...
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This note was uploaded on 11/16/2010 for the course PHY 2170 taught by Professor Blank during the Spring '08 term at Wayne State University.

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