A puck of mass m = 1.50 kg slides in a circle of radius r = 20.0 cm on a frictionless table while attached to a hanging cylinder of mass M = 2.50 kg by a cord through a hole in the table. What speed keeps the cylinder at rest?
The answer is 1.81 m/s but I don't know how to get this answer.
2006-10-06 07:10:56 · 2 answers · asked by afchica101 1 in Education & Reference ➔ Homework Help
OK...this was hard to envision, but I think I understand.
Essentially, the hole through which the cord passes through is at the center of the puck's circular motion, and the cord goes from the puck, through the hole in the table, to hold up the cylinder.
The cylinder is at rest when the weight of the cylinder exerts the centripetal force necessary to keep the puck going in a circle.
The cylinder has a weight of 2.5 * 9.8 = 24.5N.
The centripetal force for the puck is F = mv^2/r
24.5N = 1.50 kg * v^2 / .2m
3.266... = v^2
v = 1.81 m/s
2006-10-10 07:17:41 · answer #1 · answered by ³√carthagebrujah 6 · 1⤊ 0⤋
I don't either, so I'll learn along with you as soon as someone explains it.
2006-10-06 07:36:19 · answer #2 · answered by old lady 7 · 0⤊ 1⤋