Rotation Kinematics, Moment of Inertia, and Torque

Transcription

1 Rotaton Knematcs, Moment of Inerta, and Torque Mathematcally, rotaton of a rgd body about a fxed axs s analogous to a lnear moton n one dmenson. Although the physcal quanttes nvolved n rotaton are qute dstnct from ther counter for the lnear moton, the formulae look very smlar and may be manpulated n smlar ways. Here s the correspondence table: Lnear Moton Lnear dsplacement x Lnear velocty v = x Lnear acceleraton a = v Moton wth constant acceleraton: v(t) = v 0 +at, x(t) = x 0 +v 0 t+ 2 at2 Mass m Knetc energy K = 2 mv2 Force F Equaton of Moton F = ma Work W = F x Lnear momentum P = mv Angular Moton Angular dsplacement ϕ Angular velocty ω = ϕ Angular acceleraton α = ω Rotaton wth constant angular acceleraton: ω(t) = ω 0 +αt, ϕ(t) = ϕ 0 +ω 0 t+ 2 αt2 Moment of nerta I Knetc energy K = 2 Iω2 Torque τ Equaton of Moton τ = Iα Work W = τ ϕ Angular Momentum L = Iω Lnear Moton of the Parts of a Rotatng Body When a rgd body rotates around a fxed axs, each part of the body moves n a crcle, or n a crcular arc: L 2 = r 2 r 2 r r 2 L = r r

2 Each part moves n an arc of dfferent radus r equal to the dstance of that part to the axs but they all span the same angle equal to the angular dsplacement of the body. The dstance traveled by each part s L = r for n radans. () Dvdng ths dstance by the tme nterval, we fnd the lnear speed of the part: v = L = r = ω r (2) where ω s the angular velocty of the rotaton. Agan, ω should be measured n radans per second. The velocty vector v of the part s tangent to the crcle centered on the axs as shown below: a a t a c v The acceleraton vector of the part has two components: a = a t + a c (3) the tangental acceleraton and the centrpetal acceleraton. The tangental acceleraton has magntude a t = r α for α n radan/s 2 (4) and drecton parallel to the velocty vector v. That s, t has the same drecton as v f the rotatonal speed ncreases or opposte drecton to v f the rotatonal speed decreases. 2

3 The centrpetal acceleraton has magntude a c = v2 r = (ω r)2 r = ω 2 r for ω n radan/s (5) regardless of the angular acceleraton α and ts drecton s towards the axs of rotaton. Snce the tangental and the centrpetal acceleratons are always perpendcular to each other, the net acceleraton vector a has magntude a = a 2 t + a2 c = r α 2 + ω 4. (6) Knetc Energy of Rotaton The net knetc energy of a rotatng body s smply the sum of knetc energes of all ts. Thus, K = = = ω2 2 m v 2 2 m (r ω) 2 2 m r 2 = ω2 2 I (7) where I = m r 2 (8) s the moment of nerta of the body the rotatonal analogue of the mass. The moment of nerta of a body depends on ts mass, sze, and shape, and also on a partcular axs around whch the body s rotated. Indeed, n eq. (8), r s the dstance of part 3

4 # from the axs of rotaton. For example, for the rotaton about the z axs of the 3D body, r 2 = x 2 + y 2 + x 2, (9) so the moment of nerta around ths axs would be I wrtzaxs = m (x 2 +y2 +z2 ). (0) But f the same body rotates around the x axs, the moment of rotaton would be gven by a dfferent sum I wrtxaxs = Also, the locaton of the axs s just as mportant as ts drecton. m (y 2 +z 2 +x 2 ). () For a contnuous body, the sum n eq. (8) for the moment of nerta becomes a volume ntegral, but calculatng such ntegrals s way beyond the scope of ths class. Instead, let me gve you a few examples of moments of nerta for bodes of partcularly smple shapes: Sold rod of length L and unform densty; axs to the rod. For axs though the end of the rod, I = 3 ML2. For axs through the mddle of the rod, I = 2 ML2. A thn cylndrcal shell of radus R and unform densty. For axs=cylnder s axs, I = MR 2. Same formula for a thn hoop and an axs through the center, to the hoops plane. But for axs=dameter of the hoop, I = 2 MR2. A sold cylnder or dsk of radus R and unform densty. For axs=cylnder s axs, I = 2 MR2. A thn sphercal shell of radus R and unform densty and thckness. For any axs through the sphere s center, I = 2 3 MR2. A sold ball of radus R and unform densty. For any axs through the sphere s center, I = 2 5 MR2. Any body whose mass s concentrated n several ponts (or rather of relatvely small szes) use eq. (8). 4

5 Torque The rotatonal analogue of the Newton s Second Law ma = F s Iα = τ (2) where τ s the torque the rotatonal analogue of the force. The torque of a force F depends on the force s magntude and drecton, and also on locaton of the pont where the force s appled. It s defned as a product of a force (actng on a rotatng body) and ts lever arm, τ = F l, (3) where the lever arm l s the dstance between the axs of rotaton and the lne of force lne through the pont P where the force acts n the drecton of the force vector F. In a 2D plane perpendcular to the axs of rotaton, θ F l P r lever arm l = rsnθ torque τ = Fl = Frsnθ In components, force F = (F x,f y,f z ) actng at pont r = (x,y,z) has torque (around the z axs) τ = yf x xf y. (4) Note that the F z component of the force parallel to the axs does not contrbute to the torque. 5

6 The work of a torque on a rotatng body s smply W = τ for n radans. (5) Ths s the rotatonal analogy of the work W = F x of a lnear force. To see how ths works, consder a rotaton through a small angle as shown below: r r θ F 90 θ angle between r and F = 90 θ, r = r, W = r F cos(90 θ) = (r ) F snθ = (r F snθ) = τ. (6) When a body does not have a fxed axs of rotaton but can turn n all three dmensons, we need to generalze the torque to a vector τ. The torque of a force F actng at pont r s the vector product τ = r F. (7) In components, τ x = yf z zf y, τ y = zf x xf z, (8) τ z = xf y yf x. The magntude of the torque vector s τ = r F sn(angle between r and F) (9) and ts drecton s to both the force and the radus vector, τ F, τ r. (20) But n ths class, I wll mostly focus on torques around a fxed axs. 6