OPhO 2026, Единствен кръг — Задача 5. Belt friction
2026 Online Physics Olympiad · 21 август 2026 г.
Last updated
Last updated: 22nd August 03:35 UTC
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Corrections (all times are in UTC)
- Q8: Cleared up wording (21/08 04:40 and 04:52)
- Q3: improved wording (21/08 08:49)
- Q19: Improved wording (21/08 09:42)
- Q16: Attempts reset (21/08 22:00), Error bound increased (22/8 02:30)
- Q1: Improved wording (22/08 03:35)
Указания
Format. The OPhO is a team-based competition (3 members at most per team). Unlike previous years, the 2026 OPhO consists of a single round held from August 21, 12 am (UTC), to August 23, 11:59 pm (UTC). The contest is made up of three parts, all running concurrently over the full contest window:
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List of Constants
- Proton mass,
- Neutron mass,
- Electron mass,
- Avogadro’s constant,
- Universal gas constant,
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- Magnetic constant,
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- Wien’s displacement constant,
- Stefan-Boltzmann constant,
Условие
Belts are sometimes used to "connect" the rotation of two wheels. This problem will explore the mechanics of such systems. There are two uniform cylinders, Cylinders 1 and 2. Cylinders 1 and 2 have masses and , radii and respectively. The central axes of the cylinders are attached to separate horizontal frictionless axles a distance apart such that the cylinders can freely rotate. A taut inelastic belt with a mass much less than the mass of the cylinders is wrapped around both cylinders. There is significant friction between the belt and the surface of the cylinders.
Cylinder 1 is quickly spun up to an angular speed of (clockwise), fast enough that Cylinder 2 does not rotate while Cylinder 1 is being spun up. Then, the cylinders are left to rotate freely. After a time , both cylinders reach a constant angular velocity. Find the magnitude of the average vertical force exerted by the axle of Cylinder 1 on the cylinder (in Newtons), across the 0.4 seconds. Gravity is pointing downwards in the diagram provided.

Отговори
- 11.012 N
Решение
Let us ignore gravity first. The key to solving this problem is noticing that the tension in the belt varies with position due to friction with the surface of the cylinders. Specifically, one should notice that the total tangential force the belt exerts on one cylinder is equal to the increase in tension over the arc of the belt. Since the total change in tension over one "loop" is 0, the tangential force exerted by the belt on one cylinder is equal and opposite in "direction" to the other cylinder.
This allows us to get a conservation law.
Substituting
After a long time, the belt stops slipping, so
Angular momentum is not conserved as the axles exert a force on the cylinders, creating a net torque. The change in angular momentum is
Interestingly the sign depends on whether ! Here, we have taken positive change as into the page. Since , the impulse on Cylinder 1 is upwards. The magnitude of the average force is then
Adding gravity in,
You may have noticed that if you assume constant coefficient of friction , the Capstan equation seems to imply that the tension in the belt does not return to its original value after one “loop". This is resolved by the the belt undergoing stick-slip on one or both of the cylinders, thus modifying the effective coefficient of friction so that tension is conservative.
Оригинал в Архива: problems.pdf · официални решения: solutions.pdf