Rotational Dynamics. Physics 1425 Lecture 19. Michael Fowler, UVa


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1 Rotatonal Dynamcs Physcs 1425 Lecture 19 Mchael Fowler, UVa
2 Rotatonal Dynamcs Newton s Frst Law: a rotatng body wll contnue to rotate at constant angular velocty as long as there s no torque actng on t. Pcture a grndstone on a smooth axle. BUT the axle must be exactly at the center of gravty otherwse gravty wll provde a torque, and the rotaton wll not be at constant velocty! A
3 How s Angular Acceleraton Related to Torque? Thnk about a tangental force F appled to a mass m attached to a lght dsk whch can rotate about a fxed axs. (A radally drected force has zero torque, does nothng.) The relevant equatons are: F = ma, a = rα, τ = rf. Therefore F = ma becomes τ = mr 2 α Vhas zero Lght dsk axle r m F
4 Newton s Second Law for Rotatons For the specal case of a mass m constraned by a lght dsk to crcle around an axle, the angular acceleraton α s proportonal to the torque τ exactly as n the lnear case the acceleraton a s proportonal to the force F: τ = mr 2 α F = ma The angular equvalent of nertal mass m s the moment of nerta mr 2.
5 More Complcated Rotatng Bodes Suppose now a lght dsk has A several dfferent masses attached at dfferent places, and varous forces act on them. As before, radal components cause no rotaton, we have a sum of torques. BUT the rgdty of the dsk ensures that a force appled to one mass wll cause a torque on the others! How do we handle that? m 2 F 2 r 1 m 1 F 1
6 Newton s Thrd Law for a Rgd Rotatng Body If a rgd body s made up of many masses m connected by rgd rods, the force exerted along the rod of m on m j s equal n magntude, opposte n drecton and along the same lne as that of m j on m, therefore the nternal torques come n equal and opposte pars, and cannot contrbute to the body s angular acceleraton. It follows that the angular acceleraton s generated by the sum of the external torques.
7 Moment of Inerta of a Sold Body Consder a flat square plate rotatng about a perpendcular axs wth angular acceleraton α. One small part of t, Δm, dstance r from the axle, has equaton of moton τ = τ + τ = mr α ext nt 2 Addng contrbutons from all parts of the wheel ext 2 τ = τ = mr α = Iα I s the Moment of Inerta. Z Δm
8 Calculatng Moments of Inerta A thn hoop of radus R (thnk a bcycle wheel) has all the mass dstance R from a perpendcular axle through ts center, so ts moment of nerta s I = m r = MR 2 2 A unform rod of mass M, length L, has moment of nerta about one end L I = x ( M / L) dx = ML v x dx L R Mass of length dx of rod s (M/L)dx
9 Dsks and Cylnders A dsk: mass M, radus R, s a sum of nested rngs. The red rng, radus r and thckness dr, has area 2πrdr, hence mass dm = M(2πrdr/ πr 2 ). Addng up rngs to make a dsk, R R 2 2( 2) / I r dm r M R rdr MR = = = A cylnder s just a stack of dsks, so t s also ½MR 2 about the axle. c
10 Parallel Axs Theorem If we already know I CM about some lne through the CM (we take t as the z axs), then I about a parallel lne at a dstance h s I = I CM + Mh 2 A r h y dm r CM at O x Here s the proof: Moment 2 I = mr = m r + h ( ) 2 mr = + 2h mr + Mh 2 2 = + (Snce = 0.) 2 ICM Mh mr of nerta I about perpendcular axs through A We prove t for a 2D object the proof n 3D s exactly the same, takng the lne through the CM as the zaxs.
11 Clcker Queston We found the moment of nerta of a rod about a 1 perpendcular lne through one end was ML 2 3. Use the parallel axs theorem to fgure out what t s about a perpendcular lne through the center of the rod. A B C D E 1 ML ML ML ML 14 ML 2 12
12 Perpendcular Axs Theorem For a 2D object (a thn plate) the moment of nerta I z about a perpendcular axs equals the sum of the moments of nerta about any two axes at rght angles through the same pont n the plane, a x z y I z = I x + I y Proof: ( ) = = + = + I mr m x y I I z x y
13 Clcker Queston Gven that the moment of nerta of a dsk about ts 1 axle s MR 2, use the perpendcular axs theorem to fnd 2 the moment of nerta of a dsk about a lne through ts center and n ts plane. A B C 1 MR MR 2 MR
14 Rotatonal Knetc Energy Imagne a rotatng body as composed of many small masses m at dstances r from the axs of rotaton. The mass m has speed v = ωr, so KE = ½m r 2 ω 2. The total KE of the rotatng body (assumng the axs s at rest) s ( ) K = 2mr ω = 2 Iω
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