8 ROTATIONAL MOTION 8.11 Combined translational and rotational motion

Given that the tire starts from rest, its angular velocity after t seconds takes
the form

ω = α t = 12.59 × 2.3 = 28.96 rad./s.

Worked example 8.2: Accelerating a wheel

Question: The net work done in accelerating a wheel from rest to an angular

speed of 30 rev./min. is W = 5500 J. What is the moment of inertia of the wheel?

Answer: The final angular speed of the wheel is

ω = 30 × 2 π/60 = 3.142 rad./s.

Assuming that all of the work W performed on the wheel goes to increase its
rotational kinetic energy, we have

W =

1

I ω^2 ,

2

where I is the wheel’s moment of inertia. It follows that
2 W
I =
ω^2

=^

2 × 5500

3.142^2

= 1114.6 kg m^2.

Worked example 8.3: Moment of inertia of a rod

Question: A rod of mass M = 3 kg and length L = 1.2 m pivots about an axis,
perpendicular to its length, which passes through one of its ends. What is the
moment of inertia of the rod? Given that the rod’s instantaneous angular velocity

is 60 deg./s, what is its rotational kinetic energy?

Answer: The moment of inertia of a rod of mass M and length L about an axis,

perpendicular to its length, which passes through its midpoint is I = (1/12) M L^2.
This is a standard result. Using the parallel axis theorem, the moment of inertia

about a parallel axis passing through one of the ends of the rod is

IJ = I + M

L

! 2

1

M L^2 ,

(^2 3)