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
a)  = 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 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?

1)
2)
3)
4)

Ans: Table 10-1 deals
with constant angular
acceleration case hence
calculate acceleration
for each equation:
 = d2 /dt2=0
 = d2 /dt2=-30t+8
 = d2 /dt2 = 12/t4-8/t2
 = 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 merrygo-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
a)radial acceleration
b) tangential acceleration
at=  r
aR= 2 r
Then
a) 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
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
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
Rotation
 Work-Kinetic Energy theorem
K=Kf-Ki=I(f2-i2)/2=W
Work-Kinetic Energy theorem
 Work in rotation about fixed
K=Kf-Ki=m(vf2-vi2)/2=W
axis : W=.d
 Work in one dimension
 Work in rotation about fixed
motion: W=F.dx
axis under constant torque  :
 Work in one dimension
W=d= 
motion under constant force
Power:(rotation about fixed axis )
W=Fdx = F X
x
Power: (one dimension motion)
P= dW/dt= F.v
P= dW/dt= .