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Notes_10 - Vector nature of rotation r ω r ri CHAPTER 10 θ increasing ccw dθ θ CONSERVATION OF ANGULAR MOMENTUM • Vector nature of rotation o

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Unformatted text preview: Vector nature of rotation r ω r ri CHAPTER 10 θ increasing ccw dθ θ CONSERVATION OF ANGULAR MOMENTUM • Vector nature of rotation o the cross product o the right hand rule Because rotation can be clockwise or counter clockwise, • Angular momentum The directions are given by the right-hand-rule. • Conservation of angular momentum o everyday examples o the gyroscope* o Kepler’s 2nd Law* angular velocity and angular acceleration are actually vector quantities, with both direction and magnitude . ccw * You study using the notes provided. cw r F φ r r l = r sin φ RMM02VD2.MOV ω= dθ and v = rω dt Torque is also a vector. Expressed mathematically, the torque is defined as rrr τ = r × F, r which is a vector (or cross) product. The direction of τ r is given by the right-hand-rule and the magnitude of τ is given by RMM02VD3.MOV dω d 2θ α= = and a t = rα dt dt 2 r r rr τ = r F sin φ = l F . r Thus, in the case shown, the direction of τ is out of the page. Note: the value of the torque depends on the rotation axis and the direction of the force. r B θ r A We have already seen that the scalar or dot product of two vectors is defined as: rr rr A • B = A B cos θ r r If A ⇒ ( A x , A y , A z ) and B ⇒ (B x , B y , B z ) , then rr A • B = A xB x + A yB y + A zB z. ~ see revision notes on website ~ Example: r r Work done (W) = F • ∆x However, the cross (vector) product is very different ... for example: the result is a vector. r B ˆ n θ r A The vector or cross product of two vectors is defined as: rrr rr ˆ C = A × B = A B sin θ n r ˆ where n is a unit vector that is perpendicular to both A r and B and in a direction given by the right-hand-rule. ~ See web-site ~ ijˆ Since the unit vectors ˆ , ˆ , k are orthogonal we have: ˆ×ˆ=k ˆ × ˆ = −k ˆ ijˆ ji ˆ×k=ˆ jˆi ˆij k׈=ˆ ˆj k × ˆ = −ˆ i ˆ × k = −ˆ iˆ j and ˆ × ˆ = ˆ × ˆ = k × k = 0, iijjˆˆ ˆ i − ve VTM07AN2.MOV ˆ k The Right hand rule for the cross product rrr C= A×B + ve ˆ j ~ Here’s an easy way to remember it ~ r B ˆ j θ r r i j ij Question 1: If A = 2ˆ + 3ˆ and B = −4ˆ + ˆ , (a) what is rr r r A × B , and (b) what is the angle between A and B? r A r i j (a) Given: A = 2ˆ + 3ˆ r ij and B = −4ˆ + ˆ , ˆ i rrr i j ij then C = A × B = (2ˆ + 3ˆ ) × ( −4ˆ + ˆ ) = −8( ˆ × ˆ ) + 2( ˆ × ˆ ) − 12( ˆ × ˆ ) + 3( ˆ × ˆ ) ii ij ji jj 0 0 ˆ ˆ ˆ = 2 k − 12( − k ) = 10 k . rr rr ˆ (b) We know A × B = A B sin θ k , r but A = 22 + 32 = 13 r and B = ( −4)2 + 12 = 17 rr 10 ˆ = 0.672 . So A × B = 14.87 sin θ k , i.e., sin θ = 14.87 The solution θ = 42.2o is clearly not appropriate, but we remember that sin α ≡ sin(180 − α ). ∴ θ = 180o − 42.2o = 137.8o. Another important concept in rotation about an axis is angular momentum ... y ˆ j r v r r O φ m r r ( p = mv ) ˆ i x r L is outward The angular momentum about O is defined as: rrr L = r × p, r r where p = mv is the linear momentum of the object. The magnitude of the angular momentum is; L = mvr sin φ = pr sin φ , r and the direction of L is given by the right-hand-rule and r r is perpendicular to r and v . Note: a particle does not have to be rotating to have angular momentum. Dimension: [M][L]2 [ T] Units: kg ⋅ m2 /s (or J.s) Question 2: An object of mass 3kg is moving with a r ˆ velocity v = (3.0m/s)i along the line z = 0, y = 5.3m. (a) What is the angular momentum of the object relative to the origin at the point x = 0, y = 5.3m? (b) What is the angular momentum of the object relative to the origin at r i the point x = 12 m, y = 5.3m? (c) If a force F = ( −3N)ˆ is applied to the object, what is the torque of this force relative to the origin? ˆ j y (0, 5.3m) r 5.3m r1 r r2 (12m, 5.3m) r v = (3m/s) ˆ i ˆ k x O Take a closer look at a similar scenario ... ˆ j Origin ˆ i d ˆ i r rr (a) Angular momentum about O is L1 = r1 × p, r j where the position vector is: r1 = (5.3ˆ ) m and the r r momentum is p = mv . r ˆ ∴ L = 5.3ˆ × 9ˆ = ( −47.7 k ) kg ⋅ m2 /s . ji 1 (Note the direction) r i j (b) For the position vector r2 = (12ˆ + 5.3ˆ ) m, r rr ˆ L2 = r2 × p = (12ˆ + 5.3ˆ ) × 9ˆ = ( −47.7 k ) kg ⋅ m2 /s, i j i i.e., the same ... very interesting! rrr (c) The torque about O is: τ = r × F ˆ = (12ˆ + 5.3ˆ ) × ( −3ˆ ) = ( +15.9 k ) N ⋅ m. i j i (Note the direction) r r1 r r r2 p θ1 r p r r3 θ2 r p θ3 The angular momentum of the object about the origin is r rr ˆ Li = ri × p = ri p sin θi k (outward) r ∴ Li = Li = ri p sin θi r But ri sin θi = d , which is constant for all pairs of ri , θi, r providing the direction of p is unchanged! r So, the angular momentum Li is constant provided there r are no external forces to change the direction of p, i.e., no torques. ~ Rotation analog of the conservation of linear momentum ~ ˆ j r ri Look at the special case of motion in a circle ... angular momentum is conserved .... r vi ∆mi r L ˆ i O r r r p ... although a centripetal force acts on the object, it is Consider a planar object rotating about O. If ∆mi is a mass element, its angular momentum about the rotation axis is: But v i = riω . directed through the rotation axis; it produces zero torque and so there’s no change in angular momentum! r r r ˆ Li = ri × ( ∆mi ) v i = ∆mi ri v i k v ˆ ∴ Li = ( ∆mi ri2 )ω k . So, the total angular momentum is r r ˆ ˆ L = ∑i Li = ∑i ( ∆mi ri2 )ω k = Iω k ˆr But ω k = ω . r r ∴ L = Iω , r r i.e., in the direction of ω . cf: linear momentum p is r parallel to v . r r L = Iω No net torque ... ω I ∴1= 2 I2 ω1 RMM09VD1.MOV If there is a net external torque acting on an object then, from Newton’s 2nd Law of rotation, we find r r r r r dω d (Iω ) dL τ = Iα = I = =, dt dt dt DISCUSSION PROBLEM [10.1]: i.e., the rate of change of angular momentum of the object is equal to the net external torque. (Compare with r r dp Newton’s 2nd Law F = .) Note: if there is no net dt r r dL = 0 , as before. external torque, i.e., τ ⇒ 0, then dt A rotating cloud of gas begins to collapse under If a torque is applied for time ∆t , then r r (Units of L ⇒ J.s) τ.∆t = ∆L . r r Remember: F.∆t = ∆p, the impulse equation? gravitational forces within. As it collapses, what happens to its: • Moment of inertia? • Angular momentum? Let’s look some situations involving angular momentum ... • Angular velocity? ω1 l l ω2 ⇒ I2 I1 This is a conservation of angular momentum problem ... I1ω1 = I2ω 2 ... but we need to make some assumptions ... • body is a cylinder ( r ≈ 0.2 m, Mb ≈ 50 kg ). Question 3: An ice skater starts a spin with arms outstretched, rotating at 1.5rev/s. Estimate her rotational speed (in rev/s) when she brings her arms against her body. • each arm is a thin rod ( l ≈ 1m, Ma ≈ 5kg). 1 1 With arms “out” I1 ⇒ MbR12 + 2 × Ma (2l)2 2 12 1 1 = × 50 × (0.2)2 + 2 × × 5 × 22 = 4.33 kg ⋅ m2. 2 12 1 With arms “in” I2 ⇒ (Mb + 2Ma )R 22 2 1 = × (50 + 10) × (0.22)2 = 1.45 kg ⋅ m2 . 2 Iω 4.33 × 1.5 ∴ ω2 = 1 1 = = 4.5rev/s. I2 1.45 (a) The angular momenta ωo ωo ωo 2r are r 2r ωo r L1 = I1ω o and L2 = I2ω o, but they are in opposite directions. Since there are no external torques angular momentum is conserved. Question 4: Two disks, with identical masses but The initial angular momentum is different radii (r and 2r) are spinning on frictionless L1 + ( −L2 ) = I1ω o − I2ω o = (I1 − I2 )ω o, and the final angular momentum is bearings with the same angular speed ω o but in opposite (I1 + I2 )ω . directions. The two disks are brought slowly together and the resulting frictional force between their surfaces eventually brings them to a common angular velocity. (a) What is the magnitude of the final angular speed in terms of ω o ? (b) What is the change in the rotational kinetic energy? 1 1 But I1 = M(2 r )2 = 2Mr 2 and I2 = Mr 2. 2 2 1 1 ∴ 2 − Mr 2ω o = 2 + Mr 2ω , 2 2 i.e., 32 5 Mr ω o = Mr 2ω 2 2 3 ∴ ω = ω o. 5 The total initial rotational kinetic energy: 1 1 1 Ki = I1ω o2 + I2ω o2 = (I1 + I2 )ω o2 2 2 2 The total final rotational kinetic energy: 1 K f = (I1 + I2 )ω 2. 2 ∴ ∆K K f − Ki ω 2 − ω o2 = = , K Ki ω o2 3 but ω = ω o, so (ω 2 − ω o2 ) < 0, i.e., ∆K is always < 0. 5 ∆K ∴ = −0.64, K i.e., 64% of the kinetic energy is “lost”. • What’s happened to it? Note: this is the analog of the inelastic collision we studied in chapter 8. Question 5: (a) Assuming the Earth to be a homogeneous sphere of radius R and mass M, show that the period of rotation about its axis is 4 πM 2 T= R, L where L is the angular momentum of the Earth about its axis. (b) Suppose the radius changes by a small amount ∆R . Show that the fractional change in period is ∆T ∆R ≈2 . T R (c) By how many kilometers would the Earth have to expand for the period to change by 0.25 days/year so that leap years would no longer be necessary? (a) If the angular velocity is ω we have • angular momentum L = Iω , 2 • moment of inertia I = MR 2. 5 2π The period of rotation is T = ω ω = 2 π 2 πI 4 π M 2 R. = = L L 5L I QED. DISCUSSION PROBLEM [10.2]: A student sits on a stool that is free to rotate. She is handed a wheel that is spinning in the horizontal plane as shown below. (b) Let R ⇒ R + ∆R so T ⇒ T + ∆T , then 4 πM ∆T dT T ≈ = 2 R = 2 2 R . 5L R ∆R dR ∴ ∆T ∆R ≈2 . T R Assuming she (and the stool) are stationary to begin (c) ∆R ≈ R ∆T ∆T 0.25 day 0.25 day = = and we need . 2T T 1 year 365 day ∴ ∆R = 6.37 × 103 × 0.25 = 2.18 km. 2 × 365 with, what happens if she inverts the wheel (through 180o) ? We have a perfectly inelastic collision and no external torques. So, angular momentum is conserved at the collision. Question 6: A uniform rod of length 1.20m and mass 0.80kg is pivoted at its top. The rod, initially at rest, is struck by a lump of putty of mass 0.30kg at a point x cos θ d θ x x x(1 − cos θ) 0.96m below the pivot point. If the putty sticks to the d rod and the maximum deflection of the rod is 60o from θ 2 + d× cm the vertical, what was the speed of the putty at impact? d (1 − cos θ) 2 Also, mechanical energy is conserved, i.e., 0.96m 1.20m 0.30kg v M = 0.80kg Ki + Ui = K f + Uf . But K f = Ui = 0, so Ki = Uf . The rotational kinetic 1 energy of the rod and putty at impact is Ki = Iω 2, 2 where I is the moment of inertia of the rod plus putty. An increase in potential energy occurs for both the rod and the putty. The increase in potential energy at the top of the swing is d Uf = Mg (1 − cos θ) + mgx (1 − cos θ) , 2 where M is the mass of the rod and m is the mass of putty. Conservation of mechanical energy gives 12 d Iω = g M + mx (1 − cos θ). 2 2 Also, at impact angular momentum is conserved, i.e., the angular momentum of the putty before the impact equals the angular momentum of the rod and putty after the impact. mvx 0.288 v ∴ mvx = Iω ⇒ ω = = , I I 1 where I = Md 2 + mx 2 3 0.80 × (1.20)2 = + 0.30 × (0.96)2 = 0.660 kg ⋅ m2. 3 ∴ω = 0.288 v = 0.436 v , 0.660 which gives ω in terms of v, the velocity of the putty. Hence, substituting for ω gives 1 × 0.660 × (0.436 v )2 2 0.80 × 1.20 = 9.81 + 0.30 × 0.96 (1 − cos 60o ). 2 ∴ 0.0627 v 2 = 9.81(0.480 + 0.288) × Hence v = 1 = 3.767. 2 3.767 = 7.75m/s. 0.0627 Physics of the Gyroscope The Gyroscope The axle of a wheel is supported by a pivot at O and held A spinning wheel exhibits a remarkable property: horizontal by a hand, say. The weight of the wheel r produces a torque τ in the horizontal plane, so, if the end r τ r τ d rr ∆L = τ∆t r L ω O r F ω r mg r L ∆φ gyro.mov To download a copy of this movie go to the “Useful notes ...” link on the homepage and look under Chapter 10. r ∆L r L of the axle is released the wheel falls. If the wheel is given a very large angular velocity ω before the end is released, it does not fall; rather, the axle rotates slowly in the horizontal plane about O. Why? The action is explained by Newton’s 2nd Law for rotation. r τ rr ∆L = τ∆t r L r L ∆φ r ∆L r τ r L rr ∆L = τ∆t r L r r The wheel has large angular momentum L = Iω along its spin axis. When the end is released the weight of the r wheel applies a torque τ on the wheel. Therefore, according to Newton’s 2nd Law for rotation, there will ∴ωp = r L ∆φ r ∆L r L ∆φ mgd mgd = = . ∆t L Iω be a change in angular momentum, given by r rr r ∆L τ= ⇒ ∆L = τ∆ t , ∆t r in the direction of τ, i.e., in the horizontal plane. All the If the angular momentum of the spinning wheel is large, time the wheel is spinning, the “free” end will rotate the axle would point downward resulting in a large about the pivot support at O. This motion is called vertical component of angular angular momentum; but precession. If the angular velocity of the precession is there is no torque in the vertical direction! So, the end ω p, then (working with magnitudes), does not fall. ∆L = τ∆t = ( mgd )∆t , and ∆L = L∆φ = ( mgd )∆t . the precession can be very slow. Note that if the “free” end were to fall when released, Question 7: A bicycle wheel of radius 28cm has an axle that is 10cm long. The total mass of the wheel and tire is 0.95kg and it is rotating at 15rev/s. If one end of the axle is supported by a pivot, (a) what is the angular momentum of the spinnng wheel? (Assume the wheel is (a) The angular momentum of the spinning wheel is L = Iω, 1 1 where I = MR 2 = × 0.95 × (0.28)2 = 0.0372 kg ⋅ m2 . 2 2 ∴ L = 0.0372 × (15 × 2π ) = 3.51J ⋅ s. a hoop.) (b) What is the angular velocity of precession, and (c) how long does it take the axle to rotate through 360o about the supporting pivot? (b) From before ω p = = mgd L 0.95 × 9.81 × 0.05 = 0.133rad/s. 3.51 (c) Time for one revolution is 2π 2π T= = = 47.2s. ω p 0.133 Kepler’s 2nd Law: the Law of equal areas Kepler’s 2nd Law of planetary motion states that if a line is drawn from a planet to the Sun, the areas swept by the line in the same time intervals are equal throughout the orbit. r vdt dA dt dA dt dA Sun r r vp dA dt The shaded area is one-half the area of the parallelogram, 1r r 1r 1r r r ∴ dA = r × vdt = r × mv dt = r × p dt . 2 2m 2m i.e., Since the mass of the planet is constant dA L = = constant . dt 2 m dA L = , dt 2 m rr where L = r × p is the magnitude of the angular momentum of the planet about the Sun. The gravitational r force between the Sun and planet is along r so it has no torque about the Sun. Thus, L is conserved, i.e., constant. Aphelion v a ra rp Perihelion Also, since L is constant, rv sin φ is constant, where φ is r r the angle between r and v . At aphelion and perihelion, φ = 90o. ∴ ra v a = rp v p. ...
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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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