paodealho
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In the figure, the cylinder and pulley turn without friction about stationary horizontal axies. A rope of negligible mass is wrapped around the cylinder, passes over the pulley, and has a 3.00-kg box suspended from its free end. There is no slipping between the rope and the pulley surface. The uniform cylinder has mass 5.00 kg and radius 40.0 cm. The pulley is a uniform disk with mass 2.00 kg and radius 20.0 cm. The box is released from rest and descends as the rope unwraps from the cylinder.
Find the speed of the box after it has descended 1.5m as well as the angular speeds of the cylinder and disk
tnet = Iw
fnet = ma
t = mgr
alpha = a/r
conservation of energy: mgd = 1/2mv^2 + 1/2*IWdisk^2 + 1/2*IWcylinder^2
3*9.8*1.5 = 1/2*3*v^2 + 1/2*.04*Wdisk^2 + 1/2*.4*Wcylinder^2
88.2 = 3*v^2 + .04 *Wdisk^2 + .4*Wcylinder^2
wdisk = v/r wcylinder = v/r wdisk = v/.2 wcylinder = v/.4
Substituting in: 88.2 = 3*v^2 + v^2 + 2.5v^2 >> velocity box = 3.68 m/s
Speed disk = 3.68/.2 = 18.4 rad/s speed cylinder = 3.68/.4 = 9.2 rad/s
Thanks!
Find the speed of the box after it has descended 1.5m as well as the angular speeds of the cylinder and disk
Homework Equations
tnet = Iw
fnet = ma
t = mgr
alpha = a/r
The Attempt at a Solution
conservation of energy: mgd = 1/2mv^2 + 1/2*IWdisk^2 + 1/2*IWcylinder^2
3*9.8*1.5 = 1/2*3*v^2 + 1/2*.04*Wdisk^2 + 1/2*.4*Wcylinder^2
88.2 = 3*v^2 + .04 *Wdisk^2 + .4*Wcylinder^2
wdisk = v/r wcylinder = v/r wdisk = v/.2 wcylinder = v/.4
Substituting in: 88.2 = 3*v^2 + v^2 + 2.5v^2 >> velocity box = 3.68 m/s
Speed disk = 3.68/.2 = 18.4 rad/s speed cylinder = 3.68/.4 = 9.2 rad/s
Thanks!