Problem P. 4438. (April 2012)

P. 4438. A uniform solid disc of mass m=5 kg, and of radius R=10 cm is rotated at an angular speed of ₀=120 1/s about its symmetry axis, which is horizontal, and then it is carefully placed to the horizontal ground, such that it has no initial speed. The coefficient of friction between the disc and the ground is =0.4.

Determine the work done by the frictional force on the disc, while the disc covers a distance of s=5 m. Show that the result is the same if the work-kinetic energy theorem is used or if the product of the force and the displacement is calculated.

(5 pont)

Deadline expired on May 10, 2012.

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

Megoldásvázlat. A korong mozgási energiája 120 J-lal csökken. A súrlódási erő munkájára is ugyanekkora érték adódik, ha az erőt a relatív elmozdulással szorozzuk.

Statistics:

51 students sent a solution.
5 points:	Agócs Fruzsina, Antalicz Balázs, Barta Szilveszter Marcell, Bolgár Dániel, Bősze Zsófia, Büki Máté, Csathó Botond, Csóka József, Czipó Bence, Dinev Georgi, Fehér Zsombor, Filep Gábor, Garami Anna, Győrfi 946 Mónika, Hegyfalvi Csaba, Homonnay Bálint, Janzer Barnabás, Janzer Olivér, Koncz Gabriella, Laczkó Zoltán Balázs, Olosz Balázs, Papp Roland, Sárvári Péter, Szász Norbert Csaba, Szigeti Bertalan György, Tossenberger Tamás, Varju Ákos.
4 points:	Cseuz Áron, Epinger Bálint.
3 points:	2 students.
2 points:	1 student.
1 point:	13 students.
0 point:	4 students.
Unfair, not evaluated:	2 solutionss.

Problems in Physics of KöMaL, April 2012