As always, begin with good force diagrams!

First, draw all the forces on the hanging mass:

From that, we can apply Newton's Second Law. Since this mass is accelerating downward, let's take down as positive.

F_net = m a

F_net = w - T = m a

1.47 N - T = (0.150 kg) a

This provides only one equation but there are two unknowns. So we need additional information. Look at the forces -- and torques -- on the rotating frame.

T is the tension in the string. This force is exerted perpendicular to the radius from the axis of rotation so it provides a torque (or "rotational force") of

= r F

= r T

= (0.0075 m) T

The rotational equivalent of Newton's Second Law is

= I

We know the value of I, the moment of inertia (or the "rotational mass");

I = 0.0080 kg m²

= I

(0.0075 m) T = (0.0080 kg m²)

Now we have two equations -- but we now have three unknowns! However, the angular acceleration of the frame is easily and directly related to a the linear acceleration of the hanging mass. The string attached to the rotating frame is attached to the hanging mass so these two are related by

a = r

or

= a / r

That means we can write

(0.0075 m) T = (0.0080 kg m²) (a/r)

(0.0075 m) T = (0.0080 kg m²) (a/0.0075 m)

(0.0075 m) T = [(0.0080 kg m²) / (0.0075 m)] a

(0.0075 m) T = (1.067 kg m) a

T = [(1.067 kg m) / (0.0075 m)] a

T = (142 kg) a

And -- finally -- we have two equations with two unknowns!

1.47 N - T = (0.150 kg) a

1.47 N - [ (142 kg) a ] = (0.150 kg) a

1.47 N = (0.150 kg) a + (142 kg) a

1.47 N = (142.15 kg) a

(142.15 kg) a = 1.47 N

a = 1.47 N / 142.15 kg

a = 0.01 m/s²

(c) 2000, Doug Davis; all rights reserved.