CHAPTER-10 Rotation

Ch 10-2 Rotation Rotation of a rigid body about a fixed axis Every point of the body moves in a circle, whose center lies on the axis of rotation Every point of the body moves through the same angle during a particular interval of time Angular position  : Angle of reference line (fixed on rigid body and  rotational axis) relative to Zero angular position;  (rad)= s/r; 1 rad = 57.3

Ch 10-2 Rotational Variable Linear displacement x = xf - xi Average linear velocity vavg=x/t=(xf - xi)/t Instantaneous linear velocity v= lim v/t = dv/dt Average linear acceleration aavg = v /t=(vf - vi )/(tf – ti) Instantaneous linear acceleration a= lim v/t = dv/dt= d2/dt2 Angular displacement  = f - i Average angular velocity avg=/t=(f - i)/t Instantaneous angular velocity = lim /t = d/t Average angular acceleration avg =  /t=(f - i )/(tf – ti) Instantaneous angular acceleration = lim /t = d/dt= d2/dt2

Ch 10-3 Are Angular quantities Vectors? Yes, they are Right hand curl rule

Ch 10-Check Point 1 A disk can rotate about its central axis like the one . Which of the following pairs of values for its initial and final angular position , respectively, give a negative angular displacement? A) -3 rad, +5 rad B) -3 rad, -7 rad C) 7 rad, -3 rad  = f-i = 5-(-3)=8 rad  = f -i= -7-(-3)=-4 rad  = f - i = -3-7= -10 rad Ans: b and c

Ch 10-4 Rotation with Constant Angular Acceleration- Equations of Motion Linear Motion x=tvavg= t(vf+vi)/2 vf = vi+at vf2 = vi2+2ax x =vit+at2/2 Rotational Motion = tavg=t(f+i)/2 f= i+t f2= i2+2 = it+ t2/2

Ch 10 Check Point 2 In four situations, a rotating body has an angular position (t) given by a)  =3t-4 2)  =-5t3+4t2+6 3)  =2/t2-4/t 4)  =5t2-3 To which of these situations do the equations of Table 2-1 apply? Ans: Table 10-1 deals with constant angular acceleration case hence calculate acceleration for each equation: 1)  = d2 /dt2=0 2)  = d2 /dt2=-30t+8 3)  = d2 /dt2 = 12/t4-8/t2 4)  = d2  /dt2 = 10 Ans: 1 and 4 ( constant angular acceleration case)

Ch 10-5: Relating the Linear and Angular Variables Position: s=r Speed: ds/dt=r d/dt v= r Period T= 2r/v= 2/ Acceleration: Tangential acceleration at=dv/dt=r d/dt = r Radial acceleration aR=v2/r = r 2

Ch 10 Check Point 3 A cockroach rides the rim of a rotating merry-go-round . If the angular speed of the sytem ( merry-o-round + cockroach) is constant , does the cockroach have a) radial acceleration b) tangential acceleration If  is decreasing , does the cockroach have radial acceleration at=  r aR= 2 r Then Yes aR ; b) No at If  is decreasing then a) yes ; b) yes

Ch 10-6: Kinetic Energy of Rotation Kinetic energy of a rapidly rotating body: sum of particles kinetic energies (vcom=0) K =Kparticle= ½(miv2i) but vi=riI Then K=Ki=½  mi(rii)2 = ½ (mri)2 2where i2= 2 I= (mri)2 ; I is rotational inertia or moment of inertia Then rotational kinetic energy K =½ I2 Rotational analogue of m is I A rod can be rotated easily about an axis through its central axis (longitudinal) [ case a] than an axis  to its length [case b]

Ch 10-7: Calculating Rotational Inertia I= (mri)2 =mdr2 Parallel-Axis Theorem I=Icom+Mh2 Example (a): For rod Icom=ML2/12 And for two masses m , each has moment of inertia Im=mL2/4 and then Itot=Irod+2Im Itot= ML2/12 +2(mL2/4) =L2(M/12 +m/2) For case (b) Then Irod=Icom+Mh2= ML2/12+M(L/2)2 = ML2/3 Itot= Irod + Im= ML2/3 +mL2 = L2(M/3 +m)

Ch 10 Check Point 4 The figure shows three small spheres that rotates about a vertical axis. The perpendicular distance between the axis and the center of each sphere is given. Rank the three spheres according to their rotational inertia about that axis, greatest first. I=mr2 1) I=36 x 12=36 kg.m2 2) I=9 x 22=36 kg.m2 3) I=4 x 32=36 kg.m2 Answer: All tie

Ch 10 Check Point 5 Parallel Axis Theorem I=Icom+Mh2 The figure shows a book-like object (one side is longer than the other) and four choices of rotation axis, all perpendicular to the face of the object. Rank the choices according to the rotational inertia of the object about the axis, greatest first. Parallel Axis Theorem I=Icom+Mh2 Moment of inertia in decreasing order I1; I2; I4 and I3

Ch 10-8: Torque Torque is turning or twisting action of a body due to a force F : If a force F acts at a point having relative position r from axis of rotation , then Torque  = r F sin=rFt= rF, where ( is angle between r and F) Ft is component of F  to r, while r is  distance between the rotation axis and extended line running through F. ris called moment arm of F. Unit of torque: (N.m) Sign of  : Positive torque for counterclockwise rotation : Negative torque for clockwise rotation

Ch 10-9: Newton’s Second Law for Rotation Newton’s Second Law for linear motion : Fnet= ma Newton’s Second Law for Rotational motion: net = I Proof: net=Ftr=matr=m(r)r=mr2  where Ft=mat; at=r net=Ftr=matr=mr2=I  expressed in radian/s2

Ch 10 Check Point 6  = rt x F F2= 0= F5 F3= F1 = maximum F4= next to maximum Ans: F1 and F3 (tie), F4, then F1 and F5( Zero, tie) The figure show an overhead view of a meter stick that can pivot about the dot at the position marked 20 (20 cm). All five forces on the stick are horizontal and have the same magnitude. Rank the forces according to magnitude of the torque they produce, greatest first

Ch 10-10 Work and Rotational Kinetic Energy Linear Motion Work-Kinetic Energy theorem K=Kf-Ki=m(vf2-vi2)/2=W Work in one dimension motion: W=F.dx Work in one dimension motion under constant force W=Fdx = Fx X Power: (one dimension motion) P= dW/dt= F.v Rotation Work-Kinetic Energy theorem K=Kf-Ki=I(f2-i2)/2=W Work in rotation about fixed axis : W=.d Work in rotation about fixed axis under constant torque  : W=d=  Power:(rotation about fixed axis ) P= dW/dt= .