Question

A 3.28-kg bucket is attached to a disk-shaped pulley of radius 0.108 m and mass 0.727 kg.

(a) If the bucket is allowed to fall, what is its linear acceleration?

(b) What is the angular acceleration of the pulley?

(c) How far does the bucket drop in 1.65 s?

Answer 1

T = tension in line

free body of bucket (Positive is down):
mg - T = m a

free body of pulley:
acceleration a = angular acceleration alpha * r

alpha = moment/I
I = (1/2) m r^2 =
moment = T r
so
alpha = 2 T r/ m r^2 = 2 T/m r
a = alpha r = 2 T/m
therefore
mg - T = m a but a = 2 T/m
so
m g - T = 2T
T = m g/3
now go back and get a and alpha etc

Answer 2

this is wrong. periodt.