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I've also attached the picture of this problem at the end of the post.
A uniform rod of mass 4:5 kg is 10mlong. The
rod is pivoted about a horizontal, frictionless
pin at the end of a thin extension (of negligible
mass) a distance 10 m from the center of
mass of the rod. Initially the rod makes an
angle of 60± with the horizontal. The rod is
released from rest at an angle of 60± with the
horizontal, as shown in the ¯gure below
The acceleration of gravity is 9:8 m=s2 :
Hint: The moment of inertia of the rod
about its centerofmass is Icm = (1/12)m*l^2
What is the angular speed of the rod at
the instant the rod is in a horizontal position?
Answer in units of rad=s.
I tried doing mgh = 1/2 I * omega^2, but it doesn't feel right at all. Especially since I was unsure of whether I could use the height of the center of mass of the rod for the h in mgh. Can I? There's gotta be a better way to solve this problem...
A uniform rod of mass 4:5 kg is 10mlong. The
rod is pivoted about a horizontal, frictionless
pin at the end of a thin extension (of negligible
mass) a distance 10 m from the center of
mass of the rod. Initially the rod makes an
angle of 60± with the horizontal. The rod is
released from rest at an angle of 60± with the
horizontal, as shown in the ¯gure below
The acceleration of gravity is 9:8 m=s2 :
Hint: The moment of inertia of the rod
about its centerofmass is Icm = (1/12)m*l^2
What is the angular speed of the rod at
the instant the rod is in a horizontal position?
Answer in units of rad=s.
I tried doing mgh = 1/2 I * omega^2, but it doesn't feel right at all. Especially since I was unsure of whether I could use the height of the center of mass of the rod for the h in mgh. Can I? There's gotta be a better way to solve this problem...
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