A force applied to a cylinder on a flat surface is usually expected to make the cylinder roll in the direction of the force. Contrary to expectation, an example shows that the frictional force can exceed the applied force, causing the cylinder to roll in a direction opposite to the force (and expectation).
Similar situations are found in the literature. Some references include: JewettJ. W., ‘Energy and the confused student 1: Work’, Physics Teacher, 46 (2008), 39–40; HibbelerR. C., Engineering Mechanics – Dynamics, 7th edition (Prentice-Hall, New York, 1995), p. 442; LindeburghM. R., Mechanical Engineering Reference Manual for the PE Exam, 12th edition (Professional Publications, Belmont, CA, 2006), p. 56–10.
2.
This question was asked about 15 years ago during the author's junior year in college.
3.
This situation is different from others considered in the literature, such as the case of a cylinder that rolls without slipping on a block that is lying on a horizontal frictionless surface. A pull on the cylinder can cause the block to move in the opposite direction (due to the rotation of the cylinder): BergR. E., ‘Traction force on accelerated rolling bodies’, Physics Teacher, 28 (1990), 600–601. Here we consider the case of a push on the cylinder that causes the cylinder itself to move in the opposite direction. [4] Note that the value for the sign of the friction force would be negative if it was actually pointed in the other direction – for a clockwise cylinder rotation. We will see shortly that the cylinder does in fact rotate counterclockwise.
4.
The moment of inertia, Ic, a positive number, is calculated using the parallel axis theorem: IC = IO + mrb2, where IO = 1/2 mra2. Note that we can assume that the lip at radius rf has negligible mass and does not contribute to the moment of inertia of the cylinder.
