A frictionless pulley has the shape of a uniform sold disk of mass 2.5 kg and radius .2m. A 1.5kg stone is attached to a very light wire that is wrapped around the rim of the pulley, and the system is released from rest. (a) How far must the stone fall so that the pulley has 4.5J of kinetic energy? (b) What percent of the total kinetic energy does the pulley have?
I for solid disk: I = (1/2)(MR^2)
The Attempt at a Solution
first i plugged in this equation:
K = (1/2)IW^2 + (1/2)MV^2
i replaced W with V/r
K = (1/2)I(V/r)^2 + (1/2)MV^2
first i had to find I:
using formula above: I = (1/2)(2.5)(.2^2) = .05 kgm^2
then i plugged in everything to find final velocity :
4.5J = (1/2)(.05kgm^2)(V/.2m)^2 + (1/2)(1.5kg)V^2
then i solved for V and got V = 1.8 m/s
ok now im not entirely sure what to do... i need the distance that the rock drops. but to do that i need to find the acceleration.
radial acceleration is V^2/r = 16.2 rad/s^2
then i tried V^2 = Vo^2 + 2(a)(X-Xo)
Vo=0 and Xo=0 so it becomes:
V^2 = 2(a)(x)
i have value for V. i know im supposed to be able to figure out A so i can solve for x. but im not sure if im converting from radial acceleration to regular acceleration correctly because i keep getting the wrong answer.
i thought this was the formula for that:
a^2 = Arad^2 + Atan^2
and Atan = Arad(r) right? because im doin something wrong... idk what it is
the answer is supposed to be 67.3 cm