Use conservation of energy to determine the angular speed of the spool shown in
ID: 2059800 • Letter: U
Question
Use conservation of energy to determine the angular speed of the spool shown in the figure below after the 3.00 kg bucket has fallen3.80m, starting from rest. The light string attached to the bucket is wrapped around the spool and does not slip as it unwinds.
rad/s
Explanation / Answer
(I am extremely sorry as i didnt see that angular speed was needed and not velocity)
SOLUTION
Now by energy conservation
P.E. of bucket =K.E. of bucket + Inertia energy of spool
P.E. of bucket = m.g.h = 3 x 9.81 x 3.80 = 111.834 J
also if V= velocity of bucket then V = .R as no slipping and = angular velocity of spool
K.E. of bucket = 1/2.m.V2
Inertia energy of spool = 1/2.I.2 = 1/2x (m.R2/2).2 {as moment of inertia of spool = m.R^2 /2}
111.834 = 1/2.m1.V2 + 1/2x (m2.R2/2).2
here m1 = mass of bucket = 3 kg and m2 = mass of spool = 5 kg
111.834 = 11/4 x V2 {putting V = .R }
V = 111.834 x 4 / 11 = 6.377 m/s
angular speed of spool = V/r = 6.377/0.6 = 10.628 rad/s
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