Rotation · AP C/F=ma · 7 min
Rolling links translation and rotation; it is not just rotation plus friction.
01 The trap
Wrong path
Students either ignore rotational kinetic energy or assume friction always dissipates energy.
Why it feels right
Friction often means energy loss in sliding problems, so seeing friction in rolling feels like a loss term.
02 Correct model
For ideal rolling on a fixed surface, v = ωR and the contact point is instantaneously at rest. Static friction can provide torque without doing work.
03 Mini-example
Prompt
A solid sphere rolls down an incline without slipping. Does static friction do work?
Common wrong answer
Yes, because friction always removes energy.
Correct reasoning
No in the ideal model. The contact point is instantaneously at rest.
Diagnostic cue
The phrase 'without slipping' is a constraint equation and an energy clue.
04 Guided practice
Warm-up isolates the principle. Bridge changes the context. Contest-style requires a complete setup on less familiar geometry.
A solid disk rolls without slipping down through a vertical drop h. Write its speed at the bottom.
Use K_trans + K_rot and I = ½MR².
Mgh = ½Mv² + ½(½MR²)(v/R)² = 3/4 Mv², so v = √(4gh/3).
Trap: Using v = √(2gh), which ignores rotational kinetic energy.
A solid sphere rolls without slipping down a fixed rough ramp. In the ideal model, does static friction do work on the sphere?
Ask about the contact point's instantaneous motion.
No. The contact point is instantaneously at rest relative to the ramp, so static friction does zero work. It still provides torque.
Trap: Treating all friction as energy loss.
A hoop and a solid disk roll without slipping from the same height. Which reaches the bottom faster, and why?
Compare how much energy goes into rotation.
The disk reaches the bottom faster. The hoop has larger I/(MR²), so more energy is tied up in rotation and less in translation.
Trap: Assuming equal drop height means equal final translational speed for all rolling objects.
A solid disk of mass M and radius R rests on a horizontal surface. A light string is wrapped around its rim and pulled horizontally at the top with tension T. Assume rolling without slipping. Find the center-of-mass acceleration and the friction direction.
Take rightward and clockwise as positive. The top pull creates clockwise torque; friction direction must be solved.
Let rightward be positive and clockwise angular acceleration be positive, so rolling gives a = αR. Translation: T + f = Ma. Torque about the center: TR - fR = Iα = (1/2)MR²(a/R). Thus T - f = (1/2)Ma. Solving with T + f = Ma gives a = 4T/(3M) and f = T/3 to the right. Friction points right here; it helps the translation while opposing the contact-point slipping tendency.
Trap: Assuming friction must point left because the disk moves right.
A uniform solid cylinder rolls without slipping from rest through a vertical drop of 1.20 m. What is its center-of-mass speed at the bottom?
Use Mgh = ½Mv² + ½I(v/R)².
Mgh = ¾Mv², so v = √(4gh/3) = √[4(9.8)(1.20)/3] = 3.96 m/s.
Trap: Using √(2gh), which treats the cylinder as a sliding particle.
A horizontal force F is applied to the top of a uniform solid cylinder of mass M and radius R on a rough horizontal floor. The cylinder rolls without slipping. (a) Derive its center-of-mass acceleration. (b) Find the magnitude and direction of static friction. (c) Determine the minimum coefficient of static friction required.
Use F + f = Ma and torques about the center with a = |α|R.
For I = ½MR², the translation and rotation equations give a = 4F/(3M) and f = F/3 to the right. Since N = Mg, the no-slip condition requires μs ≥ f/N = F/(3Mg).
Trap: Pointing friction left merely because the cylinder moves right.