# How to find the direction of friction in rotation?

Suppose we have a spool which is plucked constantly from top with a force F. The spool is on a rough ground. We have to find the direction of friction, nd direction of motion of spool.

           .   .----------->  F
.           .
.                 .
.                     .
.                       .
.                     .
.                 .
.           .
.   .
^^^^^^^^^^^^^^^^^^^^^^^^


How do we get to know where does the friction act? Does friction tend to stop the rotation of spool?

• Just assume the friction to be in any direction, the sign of the final quantity (if positive) will ascertain whether you assumed correctly. – Abhinav Mar 30 '16 at 17:39
• Note that it has to be along the line of motion, or likewise as the problem demands. – Abhinav Mar 30 '16 at 17:40
• Think about where the base tends to move. The friction is in the opposite direction. The tip about the assumption of the direction is great, you should follow it, but you should train yourself in order to be able to predict the direction. That way you build intuition and be able to suspect if your calculations gave you the wrong direction of friction. – TheQuantumMan Mar 30 '16 at 18:26
• Hello, and welcome to Physics SE! Look around, and take the tour. You should have at least started to learn how to sketch out a force diagram. What forces will act where? What if there is no friction anywhere, only the force F in your sketch? – Jon Custer Mar 30 '16 at 18:32

From definition of Friction:

Friction is the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other.

Now consider your problem. Let the force be $F$, mass of spool be $m$ ## If Friction is absent

If Friction were absent, then due to torque of $F$, the spool will rotate clockwise. Now, the bottom edge of spool may have a velocity in forward direction, or backward direction, given by: \begin{align}\mathbb{v} &= v_{trans} -v_{rot} \\ \\ &=\frac{Ft}{m} - r\alpha t \\ \\ &= \frac{Ft}{m} - r\left(\frac{r\times F}{I}\right)t \\ \\ &= \frac{Ft}{m} - r\left(\frac{r F}{I}\right)t \\ \\ \end{align} But we know that $\max(I) = mR^2$, so $\mathbb{v}$ above reduces to zero. \begin{align}&= \frac{Ft}{m} - r\left(\frac{r F}{mR^2}\right)t \\ \\ &= 0 \end{align}

But we also do know that for most rigid bodies, $I < mR^2$.

From this, we have : $$v_{rot} > \frac{Ft}{m}$$

Therefore we can say in general, for rigid body, that: $$\boxed{v_{rot} > v_{trans}}$$ Due to absence of friction, $\mathbb{v} \neq 0$ and also that the bottom point of wheel will have a net velocity in backward direction, and spool will slip.

## If Friction is present

If Friction is present, it would make $\mathbb{v} = 0$, since it tends to remove relative motion between the bottom of spool and ground.

Now, the direction of friction will be decided by the velocity which was greater in the earlier case.

As we saw earlier, $v_{rot} > v_{trans}$ for a rigid body, $\mathbb{v}$ came out to be negative, hence friction will try to make $\mathbb{v}$ zero, or will be exerted in positive $x$ direction on spool.

Friction always opposes relative motion. So on ground, friction will act in -x direction (by third law of Newton).