Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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# Problem P. 4298. (December 2010)

P. 4298. A carpet runner is placed onto a long straight inclined plane parallel to the steepest line in the plane, from the bottom of the slope till its top. The carpet is thin and flexible, and it cannot slide down because of the friction, but it can easily be lifted, it does not stick to the plane. The top end of the carpet is rolled, a bit, and this hoop of carpet is released. The hoop rolls down the slope with greater and greater speed, while its diameter increases, and finally it reaches the bottom of the slope during a time of t1. If a solid cylinder is released at the top of the slope, it reaches the bottom in a time of t2. a) Which motion lasts longer the motion of the carpet or the motion of the cylinder? b) Calculate the ratio of t1/t2.

(6 pont)

Deadline expired on January 10, 2011.

Sorry, the solution is available only in Hungarian. Google translation

Megoldás. A szőnyeg mozgása tart hosszabb ideig, az időtartamok kérdéses aránya $\displaystyle \sqrt{2}$.

### Statistics:

 30 students sent a solution. 6 points: Batki Bálint, Bolgár Dániel, Jenei Márk, Kovács 444 Áron, Laczkó Zoltán Balázs, Maknics András, Mázik László, Nagy 111 Miklós, Nagy Lajos, Szabó 928 Attila, Szélig Áron, Timkó Réka, Tóth Balázs, Várnai Péter, Vuchetich Bálint. 5 points: Alcer Dávid, Barta Szilveszter Marcell, Jéhn Zoltán. 4 points: 3 students. 3 points: 6 students. 2 points: 1 student. 0 point: 1 student. Unfair, not evaluated: 1 solutions.

Problems in Physics of KöMaL, December 2010