Use conservation of energy to determine the angular speed of the spool shown in
ID: 1288353 • 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 fallen 4.75 m, 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
moment of inertia I = 0.5*M*R^2 =0.5*5*0.6^2 = 3.6 kgm^2...
K.E of spool is (1/2)*I*w^2 = 1.8*w^2...
from law of conservation of energy...
1.8*w^2 = mgh+(1/2)*m*v^2.......
v = sqrt(2*g*h) = sqrt(2*9.8*4.75) = 9.65 m/sec...
1.8*w^2 = 3*9.8*4.75 + 0.5*3*9.65^2 = 279.3...
moment of inertia I = 0.5*M*R^2 =0.5*5*0.6^2 = 3.6 kgm^2...
K.E of spool is (1/2)*I*w^2 = 1.8*w^2...
from law of conservation of energy...
1.8*w^2 = mgh+(1/2)*m*v^2.......
v = sqrt(2*g*h) = sqrt(2*9.8*4.75) = 9.65 m/sec...
1.8*w^2 = 3*9.8*4.75 + 0.5*3*9.65^2 = 279.3...
w = sqrt(279.3/1.8) = 12.45 rad/sec
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