What Is the Speed of a Disk's Center of Mass After Rotational Descent?

AI Thread Summary
The discussion centers on calculating the speed of a uniform solid disk's center of mass after it descends from a pivot point. The problem involves using the conservation of energy equation, specifically 1/2mv^2 = mgh, where the height (h) is critical for determining the speed. One participant suggests considering the motion as pure rotation about the pivot point rather than using linear height. The challenge lies in accurately determining the height and applying the correct rotational dynamics. The conversation highlights the importance of understanding rotational motion in solving the problem effectively.
skinnyabbey
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A uniform solid disk of radius 4.18 m and
mass 193 kg is free to rotate on a frictionless
pivot through a point on its rim.
The acceleration of gravity is 9.8 m/s2 :

If the disk is released from rest in the po-
sition shown by the solid circle, what is the
speed of its center of mass when the disk
reaches the position indicated by the dashed
circle? Answer in units of m/s.


i tried using the conservation of energy equation to solve this problem.

1/2mv^2=mgh

i used the diameter of the circle for the h, but i still couldn't find the speed of the center of mass. can anyone help?
 
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skinnyabbey said:
A uniform solid disk of radius 4.18 m and
mass 193 kg is free to rotate on a frictionless
pivot through a point on its rim.
The acceleration of gravity is 9.8 m/s2 :
If the disk is released from rest in the po-
sition shown by the solid circle, what is the
speed of its center of mass when the disk
reaches the position indicated by the dashed
circle? Answer in units of m/s.
i tried using the conservation of energy equation to solve this problem.
1/2mv^2=mgh
i used the diameter of the circle for the h, but i still couldn't find the speed of the center of mass. can anyone help?

What is "v"? Won't it be easier if you look at the motion as pure rotation about the pivoted point on the rim?
 
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