A puck of mass m = 52.0 g is attached to a taut cord passing through a small hol

Question

A puck of mass m = 52.0 g is attached to a taut cord passing through a small hole in a frictionless, horizontal surface (see figure below). The puck is initially orbiting with speed vi = 1.30 m/s in a circle of radius ri = 0.290 m. The cord is then slowly pulled from below, decreasing the radius of the circle to r = 0.120 m.

(a) What is the puck's speed at the smaller radius?
m/s

(b) Find the tension in the cord at the smaller radius.
N

(c) How much work is done by the hand in pulling the cord so that the radius of the puck's motion changes from 0.290 m to 0.120 m?
J

Explanation / Answer

part a:

angular momentum=moment of inertia*angular speed

for the puck of mass m, orbiting at a speed of v and radius r

angular momentum=(m*r^2)*(v/r)

=m*v*r

using conservation of angular momentum principle,m*v*r is conserved.

hence if speed at smaller radius is v,

then 1.3*0.29=v*0.12

==>v=3.1417 m/s

part b:

tension=centripetal force=mass*v^2/r

=0.052*3.1417^2/0.12

=4.2771 N

part c:

work done=change in kinetic energy

=0.5*mass*final speed^2-0.5*mass*initial speed^2

=0.5*0.052*(3.1417^2-1.3^2)

=0.2127 J

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