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.
