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Homework Statement
Torque = 3
Time = 3.12 s
Length of the each rod = 1 m so the radius = 0.500 m
mass of each rod is = 0.500 kg
[tex]M_1 = 4 kg[/tex]
[tex]M_2 = 2 kg[/tex]
[tex]M_3 = 4 kg[/tex]
[tex]M_4 = 2 kg[/tex]
http://imageshack.us/a/img27/5475/qu51.jpg
Homework Equations
What is the energy of the object (please refer to the picture above) that has a mass fixed to each of its four corners? If it rotates for 3.12 s? Torque and time is given.
The Attempt at a Solution
[tex]K_E = \frac{1}{2} I \omega^2[/tex]
[tex]I = \sum mr^2[/tex]
[tex]I = (4+2+4+2+0.5+0.5)(0.5)^2[/tex]
[tex]I = 3.25[/tex]
[tex]\omega = \alpha t[/tex]
Now we have "I" and Torque we can calculate alpha.
[tex]Torque = I \alpha[/tex]
[tex]3 = 2.25 \alpha[/tex]
[tex]\frac{3}{3.25}= \alpha[/tex]
so
[tex]\omega = \alpha t[/tex]
[tex]\omega = \frac{3}{3.25} 3.12 s = 2.88[/tex]
substitute back in
[tex]K_E = \frac{1}{2} I \omega^2[/tex]
[tex]K_E = \frac{1}{2} 3.25 (2.88)^2[/tex]
[tex]K_E = 13.5 J[/tex]
Can you please tell me what is wrong with my answer and if there is an easier way to solve such problems.
Thanks.
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