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 radB) -3 rad, -7 radC) 7 rad, -3 rad

) = f-i = 5-(-3)=8 rad

b) = f -i= -7-(-3)=-4 rad

c) = f - i = -3-7= -10 rad

d) Ans: b and c

Ch 10-4 Rotation with Constant Angular Acceleration- Equations of

MotionLinear Motionx=tvavg=

t(vf+vi)/2

vf = vi+at

vf2 = vi

2+2ax

x =vit+at2/2

Rotational Motion= tavg=t(f+i)/2

f= i+t

f2= i

2+2

= it+ t2/2

In four situations, a rotating body has an angular position (t) given by

a) =3t-42) =-5t3+4t2+63) =2/t2-4/t4) =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=02) = d2 /dt2=-30t+83) = d2 /dt2 = 12/t4-8/t2

4) = d2 /dt2 = 10

Ans: 1 and 4 ( constant angular acceleration case)

Ch 10 Check Point 2

Ch 10-5: Relating the Linear and Angular Variables

Position: s=r Speed: ds/dt=r d/dt

v= r Period T= 2r/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 accelerationIf is decreasing , does the cockroach have a)radial acceleration b) tangential acceleration

at= r

aR= 2 rThen a) Yes aR ; b) No atIf 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=riI

Then K=Ki=½ mi(rii)2

= ½ (mri)2 2where i2= 2

I= (mri)2 ; I is rotational inertia or moment of inertia

Then rotational kinetic energy K =½ I2

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.m22) I=9 x 22=36 kg.m2

3) I=4 x 32=36 kg.m2

Answer: All tie

Ch 10 Check Point 5

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 TheoremI=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= rF, 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.

ris 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

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

= 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)

Ch 10-10 Work and Rotational Kinetic Energy

Linear Motion

Work-Kinetic Energy theorem

K=Kf-Ki=m(vf2-vi

2)/2=W Work in one dimension

motion: W=F.dx Work in one dimension

motion under constant force

W=Fdx = Fx X

Power: (one dimension motion)

P= dW/dt= F.v

Rotation Work-Kinetic Energy

theorem K=Kf-Ki=I(f

2-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= .