A puck of mass m = 47.0 g is attached to a taut cord passing through a small hol

Question

A puck of mass m = 47.0 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 = 1.70 m/s in a circle of radius ri = 0.310 m. The cord is then slowly pulled from below, decreasing the radius of the circle to r = 0.150 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 0.310 m to 0.150 m? J

Explanation / Answer

Apply conservation of angular momentum to the puck initial and final position
m*vi*ri=m*vf*rf

vf=vi*ri/rf

= 1.7 ( 0.310/ 0.150 )

=3.513 m/s
b)

tension of the cord is
T=mv^2/r=m*(vi*ri/rf)^2/r=m(vi*ri)^2/rf^3 = 0.047 kg ( 1.7 ( 0.310)^2/ (0.150)^3 = 2.27 N
c)

from the wor energy thoerem the change in kinetic energy is equal to work done

so work done by the hand.
W = dKE =0.5 * m*(v2² - v1²) =0.5 * 0.047 ( 3.513^2 - 1.70 ^2 ) = 0.222 J

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