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Rotation Moment of Inertia of a Rotating Body: W Rotation Moment of inertia of a rotating body: 2 I =w r dm Usually reasonably easy to calculate when ● Body has symmetries ● Rotation axis goes through Center of mass Exams: All moment of inertia will be given! No need to copy the table from the book. Rotation Parallel axis theorem: Assume the body rotates dm around an axis through P. P Let the COM be the center b . of our coordinate system. COM . P has the coordinates (a,b) a 2 2 2 I =w r dm =w (x-a) +(y-b) dm 2 2 2 2 = w (x +y )dm – 2aw xdm – 2bw ydm + (a +b )w dm 2 = ICOM – 0 – 0 + h M 2 I = ICOM + h M This is something you might need Rotation Parallel axis theorem: Assume the body rotates dm around an axis through P. P Let the COM be the center b . of our coordinate system. COM . P has the coordinates (a,b) a 2 I = ICOM+Mh The moment of inertia of a body rotating around an arbitrary axis is equal to the moment of inertia of a body rotating around a parallel axis through the center of mass plus the mass times the perpe ndicular distance between the axes h squared. Example: Moment of inertia Solid sphere of radius R rotating around symmetry axis: I = 2MR2/5 2 I = ICOM+Mh 2 If R<<h ICOM <<< Mh h=1m 36kg All have essentially the same ICOM,A moment of inertia: > 2 h=2m 9kg I ~ 36 kg m ICOM,B > h=3m 4kg ICOM,C Example: Moment of inertia Lets calculate the moment of inertia for P an annular homogeneous cylinder rotating . around the central axis: Parameters: Mass M, Length L Outer and Inner Radii R1, R2 2 I =w r dm Step1: Replace dm with an integration over a volume element dV. Step 2: Express the volume element in useful coordinates and find the boundaries for the integration. Step 3: Integrate Example: Moment of inertia Lets calculate the moment of inertia for P an annular homogeneous cylinder rotating . around the central axis: Parameters: Mass M, Length L Outer and Inner Radii R1, R2 2 I =w r dm Step1: Replace dm with an integration 2 2 over a volume element dV. Vcyl = p L (R1 -R2 ) Homogeneous: I = r2 dV rw Density r = M/V dm = r dV and r is independent of the coordinates (As long as we only integrate over the body itself and not over empty space) Example: Moment of inertia Lets calculate the moment of inertia for P an annular homogeneous cylinder rotating . around the central axis: Parameters: Mass M, Length L Outer and Inner Radii R1, R2 Step 2: Express the volume element in 2 I =rw r dV useful coordinates and find the boundaries for the integration. dV = dxdydz in cartesian coordinates ds y Better coordinates: cylindrical coordinates dr x Volume element in cylindrical coordinates 2 I =rw r dV dV = dxdydz in cartesian coordinates Better coordinates for this problem: Cylindrical Coordinates ● dz is the same in both Area element in x-y plane ● only change how we ds y describe things in the x-y plane dr dsdr = rdfdr x dV = dxdydz = rdfdrdz Example: Moment of inertia Step 2: Express the volume element in P useful coordinates and find the boundaries . for the integration. 2 I =rw r dV Integration boundaries: z: -L/2 to L/2, f: 0 to 2p 2 r: R to R I =rw r rdfdrdz 2 1 4 4 4 4 2 2 2 2 I =r L 2p (R1 -R2 )/4 Use (R1 -R2 ) = (R1 +R2 ) (R1 -R2 ) z-integration r-integration 2 2 Recall: Vcyl = p L (R1 -R2 ) f-integration 2 2 I = M (R1 +R2 )/2 See Table in book, also Sample Problem 10-7 Kinetic Energy Kinetic energy of a rotating body: ● Lets split up the body into a collection of particles 2 2 K = 0.5 (Smi ri ) w ri Rotational Inertia: . 2 I = (Smi ri ) K = 0.5 I w2 Compare with translation: 2 v K = 0.5 m v m Correspondences: ● v <-> w ● I <-> m Example: Kinetic Energy Energy storage in an electric flywheel: 2 Krot = 0.5 I w To reduce mass w/o reducing I it might be useful to use an annular cylinder. But material stress favors solid wheels for high end applications. http://en.wikipedia.org/ wiki/Flywheel#Physics Example: Kinetic Energy Energy storage in an electric flywheel: 2 Krot = 0.5 I w Example (solid wheel): M = 600kg, R=0.5m I = 75kg m2 Rotates 30000rpm: f = 30000rpm/60s/min = 500Hz w = 2pf = p * 1000 rad/s Krot = 370MJ A lot of energy which can be released very fast in fusion reactors, in braking systems used in trains etc. Torque We have ● discussed the equations describing the dynamical variables of rotation (q, w, a) ● calculated the kinetic energy associated with a rotating body ● pointed out the similarities with translations What is missing: The 'force' of rotation, the torque Note: Nm = J but J is only used for energies and Nm only for torques Example Pivot. Point L/4 3L/4 What is the angular acceleration? F Use a = t/I need to get I and t 2 2 2 Moment of inertia: I = I com+M(L/4) = ML /12+ML /16 I = 7ML2/48 Torque: t = F L/4 (Note F is perpendicular to lever arm) a = t/I = (FL/ML2) (48/(7*4)) = 12 * F/(7*M*L) Example Pivot. Point L/4 3L/4 F2 F1 Assume a second force is now applied at the other end of the stick. What would be its magnitude and direction if it prevents the stick from rotating? No rotation: tNet = 0 t1 = F1L/4 = – t2 = F2 3L/4 Dire ction of the force will be the same (both down) Magnitude: F2=F1/3 Example A uniform disk with mass M = 2.5kg and radius R=20cm is mounted on a . horizontal axle. A block of mass m=1.2kg hangs from a massless cord that is wrapped around the rim of the disk. Q1: Find the acceleration of the falling block. Notice: ● We have two forces acting on mass m: Gravity and tension from the string ● We have one torque caused by the tension in the string acting on the disk ● The linear motion of the mass is linked to the circular motion of the disk via the cord. Example A uniform disk with mass M = 2.5kg and radius R=20cm is mounted on a . horizontal axle. A block of mass m=1.2kg hangs from a massless cord that is wrapped around the rim of the disk. Q1: Find the acceleration of the falling block. Notice: ● We have two forces acting on mass m: Gravity and tension from the string ma = T-mg (equation of motion of the falling block) (Unknowns: a and T) Example A uniform disk with mass M = 2.5kg and radius R=20cm is mounted on a . horizontal axle. A block of mass m=1.2kg hangs from a massless cord that is wrapped around the rim of the disk. Q1: Notice: Find the acceleration of the falling block. ● We have one torque caused by the tension in the string acting on the disk t = -TR (force T perpendicular to lever arm negative, because of clockwise rotation) = Ia =0.5 MR2a T = -0.5MRa Example A uniform disk with mass M = 2.5kg and radius R=20cm is mounted on a . horizontal axle. A block of mass m=1.2kg hangs from a massless cord that is wrapped around the rim of the disk. Q1: Find the acceleration of the falling block. Notice: ma = T-mg and T = -0.5MRa ● The linear motion of the mass is linked to the circular motion of the disk via the cord. a = Ra 3 equa tions, 3 unknowns: T, a, a Example A uniform disk with mass M = 2.5kg and radius R=20cm is mounted on a . horizontal axle. A block of mass m=1.2kg hangs from a massless cord that is wrapped around the rim of the disk. ma = T-mg, T = -0.5MRa, a = Ra Some math: ma = -0.5Ma – mg a = -2mg/(M+2m) = -4.8m/s2 Angular accel.: a = a/R = -4.8/0.2rad/s2= -24rad/s2 Tension in the cord: T = -0.5MRa = 6.0N Che ck out Table 10.3 for analogy with translation Work and kinetic Energy y F Ft Fr r m Rod x The radial component of F does not do any work as it does not displace the mass m The tangential component of F does displace the mass by ds and does work: dW = Ftds = Ft rdq = tdq Work and kinetic Energy qf W= dW =w t dq = t (qf -qi) Work done q i during displacement for constant torque Recall Work for constant force and translation: xf W= dW = Fdx = F(x -x ) w w f i xi Power: P=dW/dt = tw (rotation about fixed axis) Work and kinetic Energy y F Ft Fr r m Rod x The tangential component of F does increase the velocity of the mass: 2 2 Kf-Ki = 0.5mvf -0.5mvi =W 2 2 2 2 use v=wr Kf-Ki = 0.5mr wf -0.5mr wi 2 2 Kf-Ki = 0.5Iwf -0.5Iwi =W Work and kinetic Energy y We just derived the F relation between Ft work done by Fr a net torque and the m r change in kinetic Rod energy for a single particle.
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