W=KE₂-KE₁=mv₂²/2 -mv₁²/2=
=0.5m(v₂² - v₁²)
Law of conservation of angular momentum
I₁ω₁=I₂ω₂
(m₁ r₁²)(v₁/r₁)=(m₂ r₂²)(v₂/r₂)
v₂ =(r₁/r₂)v₁
W= 0.5mv₁²{(r₁/r₂)²-1}
The puck in the figure below has a mass of 0.160 kg. Its original distance from the center of rotation is 40.0 cm, and it moves with a speed of 60.0 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.)
1 answer
1 answer