A Classical Approach of Newtonian Mechanics

8 ROTATIONAL MOTION 8.11 Combined translational and rotational motion

The angular equation of motion of the pulley is written

I ω ̇ = τ,

where I is its moment of inertia, and τ is the torque acting on the pulley. Now, the

only force acting on the pulley (whose line of action does not pass through the

pulley’s axis of rotation) is the tension in the cable. The torque associated with

this force is the product of the tension, T, and the perpendicular distance from
the line of action of this force to the rotation axis, which is equal to the radius, b,

of the pulley. Hence,

τ = T b.

If the cable does not slip with respect to the pulley, then its downward velocity, v,

must match the tangential velocity of the outer surface of the pulley, b ω. Thus,

v = b ω.

It follows that

̇v = b ω ̇.

The above equations can be combined to give

̇v =

g

,

1 + I/m b^2

m g

T =

1 + m b^2 /I

.

Now, the moment of inertia of the pulley is I = (1/2) M b^2. Hence, the above
expressions reduce to

̇v =

g

1 + M/2 m

m g

T =

1 + 2 m/M

9.81

=

1 + 6.4/2 × 2.6

=

2.6 × 9.81

1 + 2 × 2.6/6.4

= 4.40 m/s^2 ,

= 14.07 N.

Worked example 8.5: Hinged rod

Question: A uniform rod of mass m = 5.3 kg and length l = 1.3 m rotates about a
fixed frictionless pivot located at one of its ends. The rod is released from rest at