Use conservation of energy to determine the angular speed of the spool shown in the figure below after the 3.00 kg bucket has fallen 4.05 m, starting from rest. The light string attached to the bucket is wrapped around the spool and does not slip as it unwinds.

Question

the spool, weights 5.00kg and the bucket weights 3.00kg the radius of the spool is 0.600m.

to solve this iused this equation to solve for wf^2 ....
mgyi-1/2mr^2-1/2(1/2Mr^2)=2wf^2

when i plugged everything in i got a wrong answer is is not 7.67 rad/s.

Answer 1

fell distance 4.05
so lost potential energy m g h
= 3 * 9.81 * 4.05

That is kinetic energy of bucket plus spool
v = omega R = .6 omega
call omega = w

Ke = (1/2) m v^2 + (1/2)I w^2
= (1/2)v^2 (m + I/.36)
I = 5 * .5 * .36 =
so I/.36 = 2.5
so
3 * 9.81 * 4.05 = .5 * v^2 *(5.5)

remember omega = v/.6

Answer 2

v = 6.58
so
omega = 11 radians/second

Answer 3

Be sure to check my arithmetic carefully !