A puck has a mass of 0.260 kg. Its original distance from the center of rotation is 40.0 cm, and it moves with a speed of 80.0 cm/s. The string is pulled downward 15.0 cm through the hole in the frictionless table. Determine the work done on the puck. (Hint: Consider the change of kinetic energy of the puck.)
2007-11-09 10:35:10 · 1 answers · asked by ? 1 in Science & Mathematics ➔ Physics
From conservation of angular momentum, Iω remains constant, I being moment of inertia and ω being angular rate.
I = mr^2 and ω = v/r, so Iω = mvr = constant. Thus v remains inversely proportional to r. When r decreases from 0.4 to 0.25 m, v increases from 0.8 m/s to 1.28 m/s.
KE1 = mv1^2/2 = 0.0832 J
KE2 = mv2^2/2 = 0.212992 J
The difference, 0.129792 J, is the work done on the puck.
2007-11-09 11:38:01 · answer #1 · answered by kirchwey 7 · 0⤊ 0⤋