The figure below shows a hollow cylinder that has a length equal to 1.80 m, a ma
ID: 1325571 • Letter: T
Question
The figure below shows a hollow cylinder that has a length equal to 1.80 m, a mass M equal to 0.77 kg, and radius equal to 0.20 m. The cylinder is free to rotate about a vertical axis that passes through its center and is perpendicular to the cylinder. Two objects are inside the cylinder. Each object has a mass m equal to 0.22 kg and is attached to a spring that has a force constant k and an unstressed length equal to 0.40 m. The inside walls of the cylinder are frictionless. (Take the objects of mass m to be thin disks. Assume the hollow cylinder is open-ended.)
(a) Determine the value of the force constant if the objects are located 0.80 m from the center of the cylinder when the cylinder rotates at 21 rad/s.
N/m
(b) How much work is required to bring the system from rest to an angular speed of 21 rad/s?
J
Explanation / Answer
(a) angular speed = w= 21 rad/sec
radius of circle for two masses= 0.8 m
elongation= x= 0.8-0.4 = 0.4 m
so centripertal force= spring force
m r w2 = kx
0.22 (0.8)(21)2 = K (0.4)
77.616 = k (0.4)
k= 77.616/0.4
k= 194.04 N/m
(b) momenen of inertia of two masses= 2 mr2
Im= 2(0.22)(0.8)2 =0.2816 kgm2
moment of inertia of cylinder = 1/2 Mr2 + 1/3 ML2
Ic=1/2(0.77)(0.2)2 +1/3 (0.77)(1.8)2
Ic=0.847 kgm2
total moment of inertia= I=0.2816 +0.847 = 1.1286 kgm2
final angular speed= wf= 0
work done = change in knietic energy
W= 1/2 Iwf2 - 1/2 Iwo2
W= 1/2 (1.1286) 02 - 1/2 (1.1286)(21)2
W= 248.86 joules
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