12. Rolling, Torque, and Angular Momentum

Transcription

1 12. olling, Toque, and Angula Momentum 1 olling Motion: A motion that is a combination of otational and tanslational motion, e.g. a wheel olling down the oad. Will only conside olling with out slipping. Fo a disk o sphee olling along a hoizontal suface, the motion can be consideed in two ways: I. Combination of otational and tanslational motion: Cente of mass moves in a tanslational motion. The est of the body is otating aound the cente of mass. v el cm v cm II. Pue otational Motion:

2 The whole object is evolving aound a point on the object in contact with the suface. The point of contact changes with time. Most people find method I simple to undestand. v el gnd 2 axis of otation Use method I to analyze olling without slipping: θ d = 2π When the object makes one complete evolution, the object has moved a distance equal to the cicumfeence, and each point on the exteio has touched the gound once. When the object otates though an angle θ, the distance that the cente of mass has moved is:

3 s = θ v cm = dθ dt = ω whee ω is the angula velocity of one object otating about its cente of mass. This looks vey simila to the elationship between angula velocity and the tanslational velocity of a point on a otating object: v = ω v cm is the velocity of the cente of mass with espect to the gound fo the olling motion. v is the velocity of a point on the object with espect to the axis of otation. The velocity of any point on the disk as seen by an obseve on the gound is the vecto sum of the velocity with espect to the cente of mass and the velocity of the cente of mass with espect to the gound: v gnd = v el cm + v cm (1) 3

4 v el cm v cm v el gnd 4 v cm Conside the point on the top of the wheel: v el cm v cm v cm v el cm = +ω v cm = ω (1): v gnd = ω + v cm = v cm + v cm = 2v cm

5 The point on the top of the wheel has a speed (elative to the gound) that is twice the velocity of the cente of mass. 5 Conside the point in contact with the gound: v cm v el cm v el cm = ω v cm = ω v cm v gnd = ω + ω = 0 The point in contact with the gound has a speed of zeo, i.e. momentaily at est. If you ca is taveling down the highway at 70 mph, the tops of you wheels ae going 140 mph while the bottoms of the wheels ae going 0 mph. Conside a disk olling down a amp without slipping:

6 6 h θ Assuming the disk is initially at est: What makes the disk stat olling? What is the tanslational speed of its cente of mass when it eaches the bottom of the amp? What is its angula velocity when it eaches the bottom of the amp? f N mg We need to find the toque that makes the disk stat olling. Conside the otation aound the cente of mass: The foce of gavity acts at the cente of mass and hence poduces no toque. The nomal foce has zeo level am and hence poduces no toque. The foce of fiction povides the toque!

7 If the object is olling without slipping, the fiction foce is static fiction. If the amp is fictionless, the disk will slide down without otation. The velocity and angula velocity at the bottom of the amp can be calculated using enegy consevation. The kinetic enegy can be witten as a sum of tanslational and otational kinetic enegy: 7 K tot = K tancm + K ot eltocm = 1 2 mv cm I cm ω 2 whee ω is the angula speed of the otation elative to the cente of mass and I cm is the moment of inetia aound an axis passing though the cente of mass. Consevation of Enegy: E i = E f mgh = 1 2 mv cm I cm ω 2 (1) Since the disk olls without slipping:

8 v cm = ω 8 ω = v cm ( ) v cm (1): mgh = 1 2 mv cm m2 v cm = = mv cm 4gh 3 2 (2) (2): ω = 4gh 3 2 The speed at the bottom does not depend on the adius o the mass of the disk. The speed at the bottom is less than when the disk slides down a fictionless amp: v = 2gh The angula speed depends on the adius but not the mass. We can still apply consevation of enegy even though thee is a fiction foce. The fiction foce cannot dissipate mechanical enegy because it is a static fiction. The point

9 in contact with the suface is at est while in contact with the suface. A sphee o a hollow cylinde will each the bottom with diffeent speeds due to diffeent centes of mass: E sphee f = mv cm mgh = mv cm v cm = 10 7 gh ( ) v cm m 2 E hollowcyl f = 1 2 mv cm m2 2 mgh = mv cm ( ) v cm v cm = gh The sphee will each the bottom fist, followed by the disk and hollow cylinde. The esult is independent of the mass and adius! Example: A sphee olls down a amp as shown. The amp and the floo has fiction to keep the ball 2 2 9

10 olling without slipping until it eaches the second amp which is fictionless. How high on the second amp does the sphee ise? 10 h1 h2 fictionless The sphee olls down the 1st amp and eaches some angula speed at the bottom. The sphee olls acoss the floo with the same angula speed. The sphee slides up the amp with the same angula speed because thee is no toque (fiction) acting on the sphee. E i = mgh 1 (1) E floo = 1 2 mv2 + Iω 2 = 1 2 mω m2 ω 2 = 7 10 mω 2 2 (2) E f = mgh Iω 2 = mgh m2 ω 2 = mgh mω 2 2 (3) Apply consevation of enegy to (1) and (2):

11 mgh 1 = 7 10 mω 2 2 ω 2 2 = 10 7 gh 1 (4) Apply consevation of enegy to (1) and (3): mgh 1 = mgh mω gh 1 = gh ω 2 2 (4): gh 1 = gh gh 1 h 2 = 5 7 h 1 Angula Momentum: otational analogue of linea momentum must be defined with espect to some point φ φ p p simila to toque, two calculation methods: L = p = psin φ = mvsinφ L = p = ( sinφ)p = mvsinφ

12 A vecto quantity: L = p L is pependicula to and p diection is defined by the ight hand ule unit: kg m 2 / s A paticle does not have to tavel in a cicle to have angula momentum. A paticle taveling in a staight line has angula momentum elative to a paticula point. If a point that the angula momentum is defined elative to is along the momentum vecto, then the angula momentum is zeo ( = 0). Fo a system of paticles: L = = L 1 + L L i L n 12 Newton s Second Law: Tanslational: F ext = d p dt otational: τ ext = d L dt

13 The vecto sum of all the toques acting on a paticle is equal to the ate of change of the angula momentum. Both τ and L must be defined elative to the same oigin. Angula Momentum aound a Fixed Axis: z 13 v θ θ L Angula momentum of a small element of mass dm on the object: dl z = dlsinθ = (dm)vsinθ = (dm)v = ω 2 dm L z = ω 2 dm L z = Iω

14 The component of the angula momentum along a fixed axis of otation is just the moment of inetia times the angula velocity. We often just wite: L = Iω Othe components often cancel by symmety Consevation of Angula Momentum: If any component of the net extenal toque on a system is zeo, then the component of the angula momentum of the system along that axis is conseved. If a otating object can some how changes its moment of inetia by intenal foces, then the object will spin faste o slowe depending on whethe the moment of inetia deceases o inceases: L i = L f I i ω i = I f ω f ω f = I i I f ω i 14

15 15 Example: An ice skate would stat spinning with thei ams extended away fom the cente of he body (the axis of otation). As the ice skate pulls he ams tight to he body, the mass is now close to the axis of otation, theefoe the moment of inetia has been educed and the skate spins faste in ode to conseve angula momentum. Example: A unifom thin od of length 0.5 m and mass 4.0 kg can otate in a hoizontal plane about a vetical axis though its cente. The od is at est when a 3.0 g bullet taveling the hoizontal plane is fied into one end of the od. As viewed fom above, the diection of the bullet s velocity makes an angle of 60 o with the od. If the bullet lodges in the od and the angula velocity of the od is 10 ad/s immediately afte the collision, what is the magnitude of the bullet s velocity just befoe the impact?

16 L 16 axis θ Consevation of angula momentum: L i = L f m b v b L 2 sinθ = (I + I b )ω f ω f = 12 m L 2 L + m b 2 L m b v b 2 sinθ = 1 2 ω f 1 12 m L 2 2 L + m = 1290 m / s b 2 Example: Conside a peson standing on a platfom that can otate. The peson is holding a wheel that is spinning such that its angula momentum is pointing upwads: L w

17 L i = L w If the peson tuns the wheel ove so that the angula momentum of the wheel is pointing down, what happen to the motion of the man? L w 17 The peson must stat otating because the angula momentum of the peson-wheel system must be conseved: L f = L w + L p L w = L w + L p L p = 2L w