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Unformatted text preview: Angular velocity and angular acceleration dθ ds i CHAPTER 9 r ri ROTATION θi • Angular velocity and angular acceleration o equations of rotational motion The arc length moved by the ith element in a rotating • Torque and Moment of Inertia o Newton’s 2nd Law for rotation • Determination of the Moment of Inertia • Rotational kinetic energy o power • Rolling objects (with no slip) rigid, non-deformable disk is: dsi = ri dθ where dθ is in radians. The angular velocity of the rotating disk is defined as: ω= dθ , dt and so the linear velocity of the ith element (in the direction of the tangent) is: vi = dsi = riω . dt 1 What’s the relationship between the angular and linear velocity of two points on the same disk? If the angular velocity changes there is angular acceleration ... dθ ds i r ri θi r a it r ai r a ir r r a ir ⊥ a it The angular acceleration of the disk is: α= RMM02VD2.mov dω d dθ d 2θ =.= , dt dt dt dt 2 and the tangential acceleration of the ith element is: dv dω a it = i = ri = riα . dt dt The angular velocity ω is the same for all points on the But, because the ith element is traveling in a circle, it disk but the linear (or tangential) velocity is not. Look, experiences a radial (centripetal) acceleration: v = rω , so, since r2 > r1, then v 2 > v1. v2 a ir ( = a ic ) = i = riω 2 . ri r The resultant linear acceleration is a i = a ir 2 + a it2 . 2 CONNECTION BETWEEN LINEAR AND ROTATIONAL MOTION Angular velocity ( ω ) (vector) Dimension: ω ⇒ Linear motion (Check: v i = riω ⇒ [L] 1 [L ] = ). [ T] [ T] Units: rad/s Angular acceleration ( α ) (vector) 1 Dimension: α ⇒ [ T]2 Rotational motion a ⇒ constant 1 [ T] α ⇒ constant v = v o + at ω = ω o + αt 1 ( x − x o ) = v o t + at 2 2 1 ( θ − θ o ) = ω o t + αt 2 2 v 2 = v o2 + 2a ( x − x o ) ω 2 = ω o2 + 2α(θ − θo ) You see, they’re very similar Units: rad/s2 3 r = 0.12 m: ω o = 0 : θo = 0 : t = 5s r α = 3.00 rad/s2. ω = ? : θ = ?: a t = ?: a c = ? (a) ω = ω o + αt = 3.00 × 5 = 15.0 rad/s . Question 1: A disk of radius 12cm, initially at rest, begins rotating about its axis with a constant angular 2 acceleration of 3.00 rad/s . After 5s, what are (a) the angular velocity of the disk, and (b) the tangential and centripetal accelerations of a point on the perimeter of the disk? (c) How many revolutions were made by the disk in those 5s? (b) v i = riω = 0.12 × 15.0 = 1.80 m/s (linear). • tangential acceleration: a t = riα = 0.12 × 3.00 = 0.36 m/s2. • centripetal acceleration: a c = riω 2 = 0.12 × 15.02 = 27.0 m/s2. (Check ... a c = v 2 1.802 = = 27.0 m/s2 .) 0.12 r 1 1 (c) (θ − θo ) = ω o t + αt 2 = × 3.00 × 52 = 37.5rad , 2 2 37.5 ⇒n= = 5.97 rev . 2π 4 In many applications a belt or chain is pulled from or wound onto a pulley or gear wheel ... vt vt at DISCUSSION PROBLEM [9.1]: R As the string (chain or belt) is removed (or added), its instantaneous velocity is the same as the tangential velocity at the rim of the wheel, providing there is no slip: i.e., v t = Rω . When you rewind an audio or video tape why does it Also, under the same conditions, the instantaneous wind up fastest when it has almost completed its task, acceleration of the string is the same as the tangential i.e., when the take-up spool is almost full? acceleration at the rim of the wheel: dv dω = Rα . i.e., a t = t = R dt dt 5 As we saw in chapter 4 As we have seen, the magnitude of the torque is: force ⇒ change in motion, the same is true for rotational motion. τ = lF = ( r sin θ)F . The tendancy for a force to cause rotation of a rigid body is called torque ... consider a mass m attached to a The following demonstrations show the properties of this massless rigid rod that rotates around an axis O. A force expression. F will cause the mass to rotate. r O l r F θ m l = r sin θ The magnitude of the torque due to a force F on the mass is: τ = lF = ( r sin θ)F , where l is called the lever arm. RMM05VD1.MOV The lever arm is the perpendicular distance from the axis of rotation (O) to the line of action of the force. 6 Dimension: τ ⇒ [L] [M][L] [M][L]2 = [ T]2 [ T]2 (vector) Ft = F sin θ Unit: N ⋅ m r Form the radial and tangential components of the force: Ft = F sin θ r O θ m Fr = F cos θ l Newton’s 2nd Law tells us that the tangential component r F θ m O r F of the force Ft produces a tangential acceleration a t , Fr = F cos θ l i.e., Ft = ma t . ∴ τ = rFt = mra t . But, from Chapter 8, the tangential (linear) acceleration is related to the angular acceleration α, viz: a t = rα . Then τ = lF = ( r sin θ)F = r (F sin θ) ∴ τ = mra t = mr 2α . i.e., τ = rFt . Note: the radial component Fr , which passes through A rigid object that rotates about a fixed axis is really just the axis of rotation, does not produce a torque and a collection of individual elements of mass ( ∆mi ) that therefore it does not produce rotation; only the each move in a circular path of radius ri, where ri is tangential component Ft produces a torque that results measured from the axis of rotation in rotation. 7 From above, for each mass element, we have τi = ( ∆mi ) ri2α , ri where τi is the net torque on the ith ∆mi O element. Summing over all elements, the total net torque on the object is: ( ) τ net = ∑i τi = ∑i ( ∆mi ) ri2α = ∑i ( ∆mi ) ri2 α RMM05VD2.MOV = Iα , where we call I = ∑i ( ∆mi ) ri2 the MOMENT OF INERTIA. This is Newton’s 2nd Law for rotation, i.e., τ net = Iα . Same torques ( τ1 = τ2 ). Different moments of inertia ( I1 > I2). But τ = Iα ∴ α1 < α 2 “ A net external torque acting on a body produces an angular acceleration, α , of that body given by Iα , where I is the moment of inertia.” (viz: Fnet = ∑i Fi = ma .) 8 Moment of inertia ... so what’s that all about? Dimension: I ⇒ [M][L]2 (scalar) 2 Units: kg ⋅ m Every object has a moment of inertia about an axis of Question 2: Find the moment of inertia of a uniform thin rod of length l and mass M rotating about an axis perpendicular to the rod and through its center. y rotation. Its value depends on how the mass is x distributed around that axis. For a collection of n objects, the moment of inertia about the rotation axis is: l I = ∑ n I n = ∑ n mn rn2. For a “continuous” object: I = Limit ∑i ( ∆mi )ri2 = ∫ r 2dm, ∆mi →0 where m is a function of r. 9 y [1] Show for yourselves that the moment of inertia of a dx x x − l2 l rod of mass M and length l about one end is 2 y Since the rod is a “continuous” object, the moment of inertia is I = ∫ r 2dm = x= l 0 1 I = Ml2. 3 2 the element of length dx is () ∫ x= − l 2 l3 distance from one end is l ( l) l ( − l3 ) 2 x3 2 l3 − l y () x 2 M dx = M 1 = Ml − = Ml2 . 24 l 12 24 −2 () [2] Show for yourselves that the moment of inertia of a rod of mass M and length l about an axis one-third the dm = M l dx . ∴I = l 2 x= − l x x 2 ∫ x dm. The mass per unit length of the rod is M l , so the mass of x= l 2 dx dx 2 − l3 0 x x 2l 3 1 I = Ml2. 9 10 The moment of inertia is given by r 2 I = ∫ r 2dm. r1 Consider a ring of radius r and width dr. dr r Question 3: Find the moment of inertia of the circular If the mass of the object shown below, rotating about an axis perpendicular to the plane and through its center. The mass is 1.50kg. object is M, the mass of the ring is dm = M 2πrdr , π r22 − r12 ( ) where r1 and r2 are the inner and outer radii of the object and M its mass. Then r 2 I = ∫ r 2M 10cm r1 20cm 2πrdr 2M r2 3 =2 ∫ r dr π r22 − r12 r2 − r12 r1 ( )( ) r 2 2M r 4 M = r 4 − r14 =2 2 4 2−r2 2 r2 − r1 r 2 r2 1 1 ( = )( ( ) ) M 1 r22 − r12 r22 + r12 = M r22 + r12 . 2 2 r22 − r12 ( )( )( ) ( ) 11 Sure, but, what’s the significance of I? 1 1 ∴ I = M r22 + r12 = × 1.50 × 0.202 + 0.102 2 2 ( ( ) ) = 3.75 × 10 −2 kg ⋅ m2 . 1 [1] If r1 = 0 , then Idisk = MR 2 , where R is the radius of 2 the disk ( = r2 ) . R Remember, from chapter 4 ... Mass ⇒ a measure of resistance to a change in linear motion, e.g., how difficult it is to start or stop linear motion. Moment of Inertia ⇒ a measure of resistance to a change in rotational motion, i.e., how difficult it is [2] For a thin hoop, r1 ≈ r2 = R , then I hoop = MR 2. to start or stop rotational motion. R Note that providing the thicknesses of the disks are uniform, the moments of inertia do not depend on RMM04VD1.MOV thickness. Same torques Different moments of inertia 12 Values of the moment of inertia for “simple” shapes ... Thin rod Slab 1 I = ML2 12 1 I = M(a 2 + b 2 ) 12 b b L L a a Slab 1 I = Ma 2 3 Thin rod 1 2 I = ML 3 Question 4: A 1m ruler has a mass of 0.25kg. A 5kg Hollow cylinder 1 I = M( R 2 + r 2 ) Hollow sphere 2 I = MR 2 2 3 mass is attached to the 100cm end of the rule. What is its moment of inertia about the 0cm end? Thin-walled cylinder 2 I = MR Solid cylinder 1 2 I = MR 2 Solid sphere 2 2 I = MR 5 13 Parallel axis theorem: O 5kg b Usually, the moment of inertia D a is given for an axis that passes O′ Since a >> b we assume the meter rule is a slab with 1 I = Ma 2 about the O − O′ rotation axis. 3 1 1 I ruler → Ma 2 = × 0.25 × 12 = 0.083kg ⋅ m2 . 3 3 The moment of inertia of the 5kg mass about O − O′ is 2 2 2 I mass = ma = 5 × 1 = 5kg ⋅ m . Thus, the total moment of inertia of the ruler and mass is • cm through the center of mass (cm) of the object. The moment of inertia about a general (parallel) axis is given by: I = Icm + MD2 where Icm is the moment of inertia about the center of mass, M is the mass of the object and D is the distance between the axes. 2 I ruler + I mass = 5.083kg ⋅ m . Note: the two rotation axes must be parallel 14 y y 2m 2m 3kg 4kg 2m Question 5: Four masses at the corners of a square with 2m side length L = 2 m are conncted by massless rods. The masses are m1 = m3 = 3kg and m2 = m4 = 4 kg. Find (a) the moment of inertia about the z-axis, (b) the moment 4kg •cm D 3kg x z 2m z x of inertia about an axis that is perpendicular to the plane of the ensemble and passes through the center of mass of the system, (c) the moment of inertia about the x-axis, (a) Moment of inertia about the z-axis: I z = ∑i mi ri2 = 3 × 22 + 4 × (2 2 )2 + 3 × 22 + 4 × 0 = 56 kg ⋅ m2. which passes through m3 and m4, and (d) the moment of inertia about the y-axis, which passes through m1 and m2. (b) By symmetry, the center of mass is at the center of y the square. L m1 ∴ Icm = ∑i mi ri2 m2 = 3 × ( 2 )2 + 4 × ( 2 )2 + 3 × ( 2 )2 + 4 × ( 2 )2 L = 28 kg ⋅ m2. m4 z m3 x • Check, using the parallel axis-theorem I z = Icm + MD2 ∴ Icm = I z − MD2 = 56 − 14 × ( 2 )2 = 28 kg ⋅ m2 . 15 We saw in chapter 6 that a linearly moving object has y 2m 2m 3kg 4kg 4kg an axis has rotational kinetic energy ... 2m 2m z translational kinetic energy ... an object rotating about y •cm If a rigid object is rotating 2m 3kg x z (c) Moment of inertia about the x-axis: I x = ∑i mi ri2 = 3 × 22 + 4 × 22 = 28 kg ⋅ m2. (d) Moment of inertia about the y-axis: I y = ∑i mi ri2 = 4 × 22 + 3 × 22 = 28 kg ⋅ m2. x ω O ri vi ∆mi with angular velocity ω , the kinetic energy of the ith element is: 1 Ki = ( ∆mi ) v i2 2 1 = ( ∆mi )( ri2ω 2 ) , since v i = riω . 2 So, the total rotational kinetic energy is: 1 K = ∑i Ki = ∑i ( ∆mi )ri2ω 2 2 1 = Iω 2. 2 1 1 * K rot = Iω 2 is the analog of K trans = mv 2 . 2 2 16 A torque is required to rotate (or slow down) an object ... but torque involves force ... r r r F ds Ft r F γ θ dθ O Fr γ + θ = 90o Ft = F cos γ Question 6: An engine develops 400N ⋅ m of torque at 3700rev/min. What is the power developed by the When force is applied over a distance, work is done, given by: engine? rr dW = F • ds = F.ds cos γ = F cos γ .ds = Ft .ds = Ft .rdθ = τ.dθ (J or N.m). Power is the rate at which the torque does work, dW dθ = τ = τω i.e., P = (watts). dt dt • dW = τ.dθ is the analog of dW = F.ds . • P = τω is the analog of P = Fv . 17 Torque τ = 400N ⋅ m. Angular velocity ω = 3700 rev/min 3700 × 2π = = 387.5rad/s. 60 From earlier, power P = τω = 400 × 387.5 = 1.55 × 105 watts. But 746 watts = 1HP 10N ⋅ m to the shaft of a grindstone. If the moment of inertia of the grindstone is 2 kg ⋅ m2 and the system starts from rest, find (a) the rotational kinetic energy of the grindstone after 8.0s, (b) the work done by the motor during this time, and (c) the average power delivered by 5 ∴P = Question 7: An electric motor exerts a constant torque of 1.55 × 10 = 208HP . 746 the motor. 18 (a) To find the rotational kinetic energy we need to know the angular velocity ω . τ 10 Since τ = Iα , α = = = 5.0 rad/s2. I2 The grindstone starts from rest so ω = αt = 5.0 × 8.0 = 40 rad/s. 1 1 ∴ K = Iω 2 = × 2 × 402 = 1600J . 2 2 (b) There are two ways to determine the work done by the motor. (i) By the work-kinetic energy theorem we would expect the motor to have done 1600J of work. (ii) We can use the expression W = τθ , but we need (c) The average power is the total work done divided by the time interval, i.e., ∆W 1600 Pav = = = 200 W . ∆t 8.0 Note: the expression P = τω we derived earlier is actually the instantaneous power, as ω is not constant. The instantaneous power increases linearly from zero at t = 0 to (10 × 40) = 400 W at t = 8.0s. to find θ , i.e., the angle through which the grindstone has turned in 8.0s. 1 1 θ = αt 2 = × 5.0 × (8.0)2 = 160 rad 2 2 (25.46 rev). ∴ W = τθ = 10 × 160 = 1600J . 19 2v cm If a rigid object is suspended from an arbitrary point O and is free to rotate about that point, it will turn until the v cm ω center of mass is vertically beneath the suspension point. v=0 Consider a ball (cylinder or wheel, etc) rolling on a surface without slipping. y • The point in contact with the surface has zero instantaneous velocity relative to the surface. rcm × cm O xcm x Mg • The velocity of the cm is v cm ( = rω = v ), • The velocity of a point at the top is 2 v cm ( = 2 v ). Since the object has the same linear velocity as the cm, If the y-direction is vertical and the suspension point is not at the center of mass, the object will experience a net i.e., v, the total kinetic energy is: 1 1 K tot = mv 2 + Iω 2 2 2 torque given by τ = Mgx cm, where x cm is the x component of the center of mass. Therefore, the object will rotate until x cm = 0, i.e., the suspension point is direction above the center of mass. translational + rotational where ω = v r . So, as a ball rolls down a hill ... v, ω Gravitational potential energy ⇒ translational energy + rotational energy 20 Consider rolling a sphere, cylinder and a ring down an incline. Do they travel with the same velocity? h v For each object, conservation of energy gives: 1 1 mgh = mv 2 + Iω 2 2 2 and with no slip v = Rω , where R is the radius of the object. After manipulation we find: 2 gh v2 = . 1+ I mR 2 ∴ For maximum velocity: I ⇒ smallest. RMA07VD2.MOV v2 = 2 gh 1+ I mR 2 2 For a solid sphere I = mR 2 . 5 10 gh 2 gh 2 ∴v = , i.e., v = . 7 1 + 25 Since this is independent of m and R, a bowling ball and a pool ball would have the same speeds at the bottom of So, in a race between objects rolling down a slope, the an incline, despite their different masses and radii! order would be (1) sphere, (2) cylinder, (3) hoop, and is completely independent of m and R! 21 Draw the free body diagram of all the forces acting on the sphere. Use Newton’s 2nd Law along the incline: Mg sin 30o − f = Ma cm , N f R horizontal. Express the frictional force acting at the point of contact with the surface in terms of M, R and g. where a cm is the acceleration of the center of mass. To get Mg Question 8: A uniform sphere of mass M and radius R rolls without slipping down an incline at 30o to the 30o O an expression for a cm, we take torques about O, the center of the disk. τ = fR = Iα , where I is the moment of inertia of the sphere and α its angular acceleration. (Note that the normal force N and the weight force do not contribute to the torque as their 30 o lines of action pass through O.) With no slip a cm = Rα = fR 2 . I Substituting for a cm in the first equation, we get Mg sin 30o − f = M fR 2 fR 2 =M I 2MR 2 5 5 = f. 2 22 Mg 7 1 . ∴ f = Mg ⇒ f = 2 2 7 F You can show yourselves that if we put I = βMR 2, where β = 1 for a hoop, β = 12 for a disk, etc., and the slope of the incline is θ, then a general expression for the x R frictional force acting on an object, with no slip, is Mg sin θ f= . 1 + β −1 Question 9: A cue ball is struck by a horizontal cue a You can also show that since in terms of the radius R of the ball. a cm = fR 2 , I distance x above the center of the ball. If the cue ball is to roll without slipping, what is x? Express your answer You can assume that the frictional force of the table on the ball is negligible compared with the applied force F. (from earlier) then a cm = g sin θ . 1+ β 23 F x R • The force produce a torque about the center τ = Fx . From earlier, if there is no-slip then v cm = Rω , dv dω = Rα . i.e., a cm = cm = R dt dt Using Newton’s 2nd Law F = ma cm and τ = Iα = Fx . F Fx I ∴ = a cm = Rα = R , i.e., x = . m I mR 2 For a sphere I = mR 2 . 5 2 ∴ x = R. 5 2 • x > R ⇒ top spin 5 “SLIP” 2 • x < R ⇒ back spin 5 m2 30kg 20kg m1 2m Question 10: The system shown above is released from rest, with the 30kg mass 2m above the ground. Modeling the pulley as a uniform disk with a radius of 10cm and mass 5kg, find (a) the speed of the 30kg mass just before it strikes the ground, (b) the angular velocity of the pulley at that instant, (c) the tensions in the two strings, and (d) the time it takes for the 30kg block to reach the ground. Assume the bearings in the pulley are frictionless and there is no slip between the string and the pulley. 24 • ω • m2 30kg m1 2m m2 ω • m2 30kg v 20kg m1 • m1 v v 2m 20kg m1 We have both translational and rotational motion. The moment of inertia of the pulley is: 1 1 I = mR 2 = × 5 × 0.102 = 0.025kg ⋅ m2. 2 2 (a) Conservation of energy ... as m2 hits the ground 1 1 1 m2 gh = m1v 2 + m1gh + m2 v 2 + Iω 2 . 2 2 2 v The angular velocity of the pulley ω = , so R 1 30 × 9.81 × 2 = × 20 v 2 + 20 × 9.81 × 2 2 2 1 1 v + × 30 v 2 + × 0.025 × , 2 2 (0.10)2 (b) ω = (c) m2 v v 2.73 = = 27.3rad/s. R 0.10 T1 m1g T2 a Draw the free-body a m2 g diagrams for the masses. What is the upward acceleration of m1? v 2 = 2ah . ∴ a = v 2 2.742 = = 1.88 m/s2. 2h 2 × 2 Then T1 − m1g = m1a , i.e., T1 = m1 ( g + a ) = 234 N. Also T2 − m2 g = m2 ( −a ), i.e., T2 = m2 ( g − a ) = 238 N. i.e., 589 = 10 v 2 + 392 + 15v 2 + 1.3v 2 ∴v = 197 = 2.74 m/s. 26.3 1 2h 2×2 = = 1.46s. (d) ( y − y o ) = h = at 2. ∴ t = a 1.88 2 25 ...
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This note was uploaded on 07/13/2011 for the course PHY 2048 taught by Professor Guzman during the Spring '08 term at FAU.

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