A thin uniform rod (length = 1.77 m, mass = 3.13 kg) is pivoted about a...
A thin uniform rod (length = 1.77 m, mass = 3.13 kg) is pivoted about a horizontal frictionless pin through one of its ends. The moment of inertia of the rod through this axis is (1/3)mL2. The rod is released when it is 58.5° below the horizontal. What is the angular acceleration of the rod at the instant it is released? (in rad/s^2)
| A: 2.712 | B: 3.173 | C: 3.713 | D: 4.344 | E: 5.082 | F: 5.946 | G: 6.957 | H: 8.140 |
Homework Answers
We know that the Torque,T = MI . ang accel
M=mass
I=Moment of inertia
So, 3.13*9.8*(1.77/2) cos 58.5 = 1/3 * 3.13* 1.772 . angular acceleration
By calculating we get, Angular acceleration,a =4.344 rad/s2
Option D
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