Question

A smooth circular hoop with a radius of 0.400 m is placed flat on the floor. A 0.450-kg particle slides around the inside edge of the hoop. The particle is given an initial speed of 8.50 m/s. After one revolution, its speed has dropped to 4.00 m/s because of friction with the floor.

a) Find the energy transformed from mechanical to internal in the particle—hoop—floor system as a result of friction in one revolution.
b) What is the total number of revolutions the particle makes before stopping? Assume the friction force remains constant during the entire motion.

Answer 1

a) Relevant equation: Emec = DU + DK
If there are no nonconservative forces, this equation equals zero. Or in this case, some K is transformed into Eint (heat). So the equation becomes: Emec = Ff.d. (F in Newton, distance in meters and N.m = J).
-> DU + DK = Ff.d

Then, 0,5 m vi² = 0,5 m vf² + Ff.d
What do you need is: Ff.d = 0.5 m vi² - 05. m vf². (/!\ all the expression Ff.d is the answer, do not separate it).

b) You know what is the initial speed,
so you have the Ki = 0.5x0.4x64
Or you know that each turn you loose a certain quantity of energy (answer A).
-> Ki/ansA = ...