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Summary Lecture 11 Rotational Motion 10.5 Relation between angular and linear variables 10.6 Kinetic Energy of Rotation 10.7 Rotational Inertia 10.8 Torque 10.9 Newton 2 for rotation 10.10 Work and Power Problems:Chap.10: 6, 7, 16, 21, 28 , 33, 39, 49 This Friday Tomorrow 12 – 2 pm 20-minute test material in PPPon “ExtEnsion” lectures 1-7 lecture. during Room 211lecture podium level Turn up any time Relating Linear and Angular variables Need to relate the linear motion of a point inq the rotating body with the angular variables q r s s = qr and s Relating Linear and Angular variables w and v w s q r Not quite true. V, r, and w are all vectors. Although magnitude of v = wr. The true relation is v = w x r s = qr ds v dt d v (qr) dt dθ v r dt v ωr Direction of vectors Grab first vector (w) with right hand. Turn to second vector (r) . v=wx r w Direction of screw is direction of third vector (v). r Vector Product A C =Ax B A = iAx + jAy B = iBx + jBy So C = (iAx + jAy) x (iBx + jBy) Ay = Asinq q Ax = Acosq B = iAx x (iBx + jBy) + jAy x (iBx + jBy) = ixi AxBx + ixj AxBy + jxi AyBx + jxj AyBy now So C= 0 + k AxBy - kAyBx + 0 = 0 - k ABsinq ixi = 0 jxj = 0 ixj = k jxi = -k C= ABsinq Is q a vector? Rule for adding vectors: The sum of the vectors must not depend on the order in which they were added. However q is a vector! Relating Linear and Angular variables a and a v w a r vwxr a dv d (ω x r ) dt dt dr dω a ωx x r dt dt Since w = v/r this term = v2/r (or w2r) The centripetal acceleration of circular motion. Direction to centre a ωxv αxr This term is the tangential accel atan. (or the rate of increase of v) Relating Linear and Angular variables a and a Total linear acceleration a a The acceleration “a” of a point distance “r” from axis consists of 2 terms: a a = ar & v2/r r Central Tangential acceleration even when acceleration Present a is zero! (how fast v is changing) The Falling Chimney The whole rigid body has an angular acceleration a CM gcosq g The tangential acceleration atan distance r from the base is L atan ar q At the CM: atan aL/2, and at the end: atan = aL But at the CM, atan= g cosq (determined by gravity) The tangential acceleration at the end is twice this, but the acceleration due to gravity of any mass point is only g cosq. The rod only falls as a body because it is rigid ………..the chimney is NOT. Kinetic Energy of a rotating body What is the KE of the Rotating body? w cm 1/2 MVcm2 ?? It is clearly NOT ½ MV2cm since Vcm = 0 Kinetic Energy of Rotation Krot = ½m1v12 + ½m2v2 2 +½m3v3 2 + w V1 V 2 m1 m2 V3 But all these values of v are different, since the masses are at different distances from the axis. m3 However w (angular vel.) is the same for all. We know that v = wr. So that Krot=½m1(wr1)2+½m2(wr2)2+½m3(wr3)2+ = ½miri2w2 Krot = ½ w2 miri2 Krot=½m(wr1)2+½m(wr2)2+½m(wr3)2+ . . . . . . = ½miri2w2 = ½ w2 miri2 Krot=½ I w2 I= miri2 Where I is Rotational Inertia or Moment of Inertia of the rotating body So Krot = ½ I w2 (compare Ktrans= ½ m v2) Rotational Inertia I= mi ri2 “I” is the rotational analogue of inertial mass “m” For rotational motion it is not just the value of “m”, but how far it is from the axis of rotation. The effect of each mass element is weighted by the square of its distance from the axis The further from the axis, the greater is its effect. Krot = ½ I w2 The bigger I , the more KE is stored in the rotating object for a given angular velocity A flywheel has (essentially) all its mass at the largest distance from the axis. Some values of rotational inertia for mass M M R Mass M on end of (weightless) rod of length R I= mi ri2 = MR2 Some values of rotational inertia for mass M M/2 M/2 R 2 Masses M/2 on ends of (weightless) rod of length 2R (dumbell of mass M) I= mi ri2 =1/2 MR2 + 1/2 MR2 = MR2 Same as mass M on end of rod of length R ...MR2 Some values of rotational inertia for mass M Mass M in a ring of radius R R I= mi ri2 = mi R2 = MR2 Same as mass M on end of rod, Same as dumbell...MR2 Rotational Inertia of a thin rod about its centre For finite bodies x Rotation axis L I = mi ri2 I r 2dm Linear density (kg/m) L/2 I 2 x 2dm dm ρ dx 0 thickness dx L/2 mass M mass of the rod M=L I 2ρ x 2dx 0 L/2 2 ρx 3 0 3 2 L3 ρ 3 8 1 1 2 L ( L ) 12 I 12 ML2 Some Rotational Inertia Parallel-axis Theorem The rotational inertia of a body about any parallel axis, is equal to its R.I. about an axis through its CM, PLUS R.I. of its CM about a parallel axis through the point of rotation CM h Axis of Rotation I = ICM + Mh2 Proof of Parallel-axis Theorem One rotation about yellow axis involves one rotation of CM about this axis plus one rotation of body about CM. I = Icm + Mh2 What is it about here? Example R RI of ring of mass M about CM is RI of CM about suspension point, distance R away is MR2 MR2. So total RI is 2MR2 The Story so far... q, w, a Rotational Variables relation to linear variables vector nature Rotational kinematics with const. a Analogue equations to linear motion Rotation and Kinetic Energy Rotational Inertia