Peoples Physics Concepts

http://www.ck12.org Chapter 9. Rotational Motion

Solution

First we’ll draw a free body diagram for the situation. In the diagram below the force of gravity and the tension

from the string have been labeled and are drawn from they points they are acting.

Let’s start by applying Newton’s second law to this object and we’ll let the positive direction be downward.

ΣF=ma

mg−T=ma

As you can see, we have two unknowns in this equation(Tanda), so we’re going to use Newton’s second law for

rotation as our second equation and solve forT. The rotation axis is at the cm of the cylinder, thus there is no

torque from gravity.

Στ=Iα

Tr=Iα

T=

Iα

r

Now we will substitute in values forIandα. In order to calculate the moment of inertia of the wooden disk, we use
the known formula for a disk being rotated about it’s center.