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Unformatted text preview: Chapter 10 Rotaon of Moon of rigid bodies Rota%onal mo%on: moon in which an extended object rotates about an axis A rigid object is one that is non
deformable. This means that the relave locaons of all parcles of which the object is composed remain constant. We model extended object as a rigid body. In translaonal moon, moon occurs when the object changes its posion during some me interval. Rotaon axis Rotaon axis What does an object change during rotaonal moon ? Answer: Its angle with respect to some reference line. In terms of rotaon, a rigid object about a ﬁxed axis means every parcle on the object rotates in the same angle in a given me interval. Rotaon angle (review) Angle in radians described by the reference line is given by θ=
s r Final posion aOer rotaon 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 revoluon of an object ? Converng 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 menoned, calculaons are to be made in radians. So set your calculator accordingly! Angular displacement and angular velocity By replacing posion with angle, we can translate all formulas we learned from translaon moon into rotaonal! Angular displacement: Average angular velocity: Instantaneous angular velocity: Angular acceleraon 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 rao of lengths. The dimensions ω of and α are T
1 and T
2 respecvely. Rotaon with constant angular acceleraon Here, again, we just do “translaon” of translaonal moon with constant linear acceleraon to rotaonal moon with constant angular acceleraon: 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 substuons: x →θ v→ω a→α Angular displacement, velocity and vectors Although angular displacement Δθ also has a direcon, it cannot be easily described by a vector. Consider the situaon described in the ﬁgure on the right. For the ﬁnal orientaon, it depends on which order you rotate the object! This means that angular displacement does not follow the basic rules of vector addion, in parcular: " 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 rotaon. To ﬁx which way it is directed along the axis of rotaon, 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 velocies In rotaon of a rigid body, every parcle in it moves in a circle whose center is on the axis of rotaon. The tangenal velocity a parcle 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 parcle of a rigid object rotate about the common axis with the same angular speed. The tangenal velocity, however, as shown by the last equaon, depends on r – infact, it is proporonal to r. A parcle further from the axis moves faster. . v2 We also know: the centripetal acceleraon a and the tangenal 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 acceleraon (nonuniform tangenal speed ) is obtained from . Radial (centripetal) and tangenal acceleraon We also know: the centripetal acceleraon a and the tangenal speed are related by v2 ar = r
Here i labels the ith parcle of the rigid object. ar = rω 2
The tangenal acceleraon comes from non uniform angular speed. at = d dω (rω ) = r → at = rα dt dt The total acceleraon is then a = a 2 + a t2 = r α 2 + ω 4 r Rotaonal kinec energy When an object rotates, diﬀerent parts of the object have diﬀerent linear velocies: vi = riω
And they will have diﬀerent kinec energies: ω r2 r1 1 K i = mi vi2 2 The total kinec energy of a rotang object is equal to the sum of all the kinec 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 rotaon, i.e. what quanty of mass is what distance from the axis. I is called the moment of inera or rotaonal inera. I takes the role that mass m takes in translaonal moon: 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 rotang masses. Calculate moment of inera and the rotaonal kinec energy for both cases. (Chalk board stuﬀ) Moment of ineral 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 rotang about the same axis, the moments of inera simply adds up: I = I1, circle + I 2, circle = m1r12 + m2 r22 = ( 2kg )(0.5m )2 + (1kg )(0.2m )2 = 0.54kg m 2 Kinec 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 ineral 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 quanty that combines a force and the distance at which the force is applied from the axis of rotaons. This quanty is torque. It is a vector quanty (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 deﬁnion 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 quanes. F Rigid body under net torque Consider a parcle of mass m rotang in a circle under the inﬂuence of a net tangenal force ∑ F and a net radial force Fr . The radial force provides the centripetal t ∑ acceleraon while the tangenal force provides the tangenal acceleraon. ∑ 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 inera of the parcle about the axis of rotaon (at center of the circle). Thus: ∑ τ = Iα
It can be shown that this formula also holds for an extended object of arbitrary shape rotang about a ﬁxed axis (see the textbook, page 284). Thus net torque acng on a parcle or a rigid body about an axis is equal to the product of its moment of inera about that axis and angular acceleraon. 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 & Rotaonal Kinec 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 translaonal moon. Rolling moon 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 derivave, we get the translaonal speed of the center of mass: vCM And the acceleraon of the CM for this pure rolling moon is: aCM = R dω → aCM = Rα dt Rolling (contd…) We can see Rolling = Translaon + Rotaon 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 moon can thus be seen as pure rotaon of the wheel about the contact point P at the bokom. This means that the total kinec energy of rolling moon 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 friconless 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.
 Spring '08
 BLANK

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