A Classical Approach of Newtonian Mechanics

8 ROTATIONAL MOTION 8.11 Combined translational and rotational motion

π

2

so

IJ =

3 × 1.2^2

3

= 1.44 kg m^2.

The instantaneous angular velocity of the rod is

ω = 60 ×

180

= 1.047 rad./s.

Hence, the rod’s rotational kinetic energy is written

K =

1

IJ ω^2 = 0.5 × 1.44 × 1.047^2 = 0.789 J.

Worked example 8.4: Weight and pulley

Question: A weight of mass m = 2.6 kg is suspended via a light inextensible

cable which is wound around a pulley of mass M = 6.4 kg and radius b = 0.4 m.

Treating the pulley as a uniform disk, find the downward acceleration of the

weight and the tension in the cable. Assume that the cable does not slip with
respect to the pulley.

lley

mg

Answer: Let v be the instantaneous downward velocity of the weight, ω the in-

stantaneous angular velocity of the pulley, and T the tension in the cable. Apply-
ing Newton’s second law to the vertical motion of the weight, we obtain

m ̇v = m g − T.

b

pu

T



weight