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

[skinnyabbey]

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?

 

[siddharth]

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?

 

[quicknote]

Yes, I can help with this problem. It seems like you are on the right track by using the conservation of energy equation. However, there are a few things you need to consider in order to find the speed of the center of mass.

First, let's define some variables:
- m = mass of the disk = 193 kg
- R = radius of the disk = 4.18 m
- g = acceleration due to gravity = 9.8 m/s^2
- v = speed of the center of mass (what we are trying to find)
- h = height of the center of mass from the pivot point (this will change as the disk rotates)

Now, let's consider the initial and final positions of the disk. In the initial position (solid circle), the center of mass is at a height of R from the pivot point. In the final position (dashed circle), the center of mass is at a height of 2R from the pivot point.

Using the conservation of energy equation, we can write:
1/2mv^2 = mgh

In the initial position:
h = R
Therefore, the initial potential energy is:
mgh = m(9.8 m/s^2)(R) = 193(9.8)(4.18) = 8001.2 J

In the final position:
h = 2R
Therefore, the final potential energy is:
mgh = m(9.8 m/s^2)(2R) = 193(9.8)(2)(4.18) = 16002.4 J

Since energy is conserved, the initial potential energy must equal the final potential energy:
8001.2 J = 16002.4 J

Now, we can solve for the speed of the center of mass:
1/2mv^2 = 8001.2 J
v^2 = (2)(8001.2)/m
v^2 = 32907.2/m
v = √(32907.2/m)
v = √(32907.2/193)
v = 10.8 m/s

Therefore, the speed of the center of mass when the disk reaches the dashed circle is 10.8 m/s.