I'm getting stuck on friction (heh). Here's a simplified problem from my textbook illustrating my confusion.

We have a solid cylinder of mass m and radius r, lying on its side on a table, with a lightweight wire wrapped around it. We pull on the wire with a known force T, causing the cylinder to roll without slipping. What is the acceleration of the cylinder?

My attempt: let $T$ be the force from the wire and $f$ be the force of friction. Treat friction as acting on the bottom of the cylinder, opposing the applied force. The net force produces a linear acceleration:

$$F = T - f = ma$$

The torque produces a rotational acceleration. Here friction agrees with the applied rotation:

$$\tau = Tr + fr = I \alpha$$

To roll without slipping, we must have $a = r \alpha$. Plugging in:

$$\frac{T - f}m = r \times \frac{Tr + fr}{I}$$

A solid cylinder has $I=\frac{1}{2}mr^2$, so this simplifies to:

$$T - f = 2(T + f)\\ f = -\frac{1}{3}T $$

The negation implies that the force of friction is not acting backwards, but instead forwards, causing the cylinder to accelerate more than it would in the absence of friction:

$$ma = T-f = T + \frac{1}{3}T = \frac{4}{3}T $$

This result seems absurd! How can friction make something go faster? Does it actually? This sort of problem is a textbook mainstay, so I am confused.

My best guess is: torque acting on the rim of a solid cylinder can never produce rolling without slipping: its moment of inertia is too small, it will always over-rotate. Problems which assume this is possible are taking certain liberties.

Thank you for any help.


1 Answer 1


SEnse of rotation

The sense (direction) of the fiction force $F_f$ (in my diagram) is determined by the relative motion of the cylinder's surface and the ground.

The moment you start applying force on the string the cylinder 'wants' to start rotating clockwise and its surface wants to move left wrt the floor. Friction of course always opposes motion, that's just what it does!

So here, friction tries to prevent that relative motion by providing a friction force pointing to the right.

Your observation is counter-intuitive but correct.

Now take this into account in your FBD computations.

  • $\begingroup$ I got it! When the cylinder over-rotates, its contact point moves left against the ground. This means that friction is directed right. I imagine a wheel spinning in the mud: even if the wheel is accelerating, once it catches it accelerates even faster, because friction pushes it. Friction can accelerate when rotational velocity exceeds linear velocity. Energy is conserved because you have to pull the wire twice as far: once to move the cylinder and again to unwind it. It's so cool. $\endgroup$ Nov 12, 2021 at 1:51
  • 1
    $\begingroup$ Cool, right? :-) Thanks for the acceptance! $\endgroup$
    – Gert
    Nov 12, 2021 at 1:53

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