# Axis of rotation of a spool unwinding and falling under gravity

There is a problem that asks for forces that apply torque around the axis of rotation on a spool of string that falls and unwinds under gravity. The spool falls straight down with a vertical string. My query is identifying the axis of rotation, as I understand, it should be passing through the center like in the figure since force due to gravity also passes through the center, and so tension along the string must be the only force that applies torque around the axis of rotation.

"The spool is rotating around the point where it loses contact with the string. Since the tension force is applied at this point, it creates no torque. The gravitational force is applied at the center of the spool, which is a distance $$r$$ from the rotation axis, so it applies a torque."

I don't understand how they have chosen the axis of rotation. Can someone explain?

• This isn't asking to give the answer to a homework problem. It is asking about a concept. That is permitted even for homework problems. Sep 16, 2023 at 15:17

If the spool was on a fixed axis and you pulled upward on the string, the spool would rotate around the fixed axis. The center of the spool would not move.

Here, the spool is simultaneously rotating and falling. The straight part of the string does not move. So the point where the edge of the spool touches the straight part of the string does not move.

At some point in time, the center of the spool is moving downward at velocity $$v$$. The edge of the spool also rotates with speed $$v$$. The velocities add. On the left edge, the downward $$v$$ from falling cancels the upward $$v$$ from rotating. On the right edge, the the downward $$v$$ from falling cancels the downward $$v$$ from rotating, so the total velocity is $$2v$$. Other points are more complicated. But if you figure them out, they are the same as the velocity of a spool on a fixed axis through the edge.

This is a little confusing because the motion of the falling spool only matches that of a fixed axis spool for an instant. At a later time the falling spool matches a fixed axis spool at a lower point.

But for any instant, it is true that the falling spool rotates just like a spool with a fixed axis through the edge. And that is why we put the axis there.

This may help - Rotation + Translation = Rotation. Animated proof

I found this honorable mention video in Summer of Math Exposition 3. See 25 Math explainers you may enjoy | SoME3 results for a summary.

This example can be looked at in a number of ways.

Because you are dealing with a rigid body the angular velocity and angular acceleration are independent of position.
If the angular velocity about the centre of mass, $$C$$, is $$5\,\rm rad/s$$ then the angular velocity about $$A$$ is also $$5\,\rm rad/s$$.
Explained here, Friction and Rolling.

Here is the free body diagram for the spool acted on by two forces, the tension in the string, $$T$$, and the weight of the spool, $$mg$$.

Linear acceleration, $$a$$ of centre of mass $$F=ma\Rightarrow mg-T=m\ddot z$$.
There are two "convenient" axes to use $$\tau = I\,\dot\omega$$.
One through $$A$$ and the other through $$C$$.
$$A$$ is not accelerating and so there are no problems regarding pseudo-forces, $$\Rightarrow mgr = (I_{\rm c}+mr^2)\,\dot \omega$$.
$$C$$ is accelerating but as any pseudo-forces must act through $$C$$ those forces cannot contribute to the torque about $$C, \Rightarrow Tr=I_{\rm c}\,\dot\omega$$.

Finally, there is the constraint, $$\dot z =r\,\omega$$ and $$\ddot z = r \dot \omega$$.