200. newton·meters of work is put into a machine over a distance of 20. meters. The machine does 150. newton·meters of work as i
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From the case we know that:
- The moment of inertia Icm of the uniform flat disk witout the point mass is Icm = MR².
- The moment of inerta with respect to point P on the disk without the point mass is Ip = 3MR².
- The total moment of inertia (of the disk with the point mass with respect to point P) is I total = 5MR².
Please refer to the image below.
We know from the case, that:
m = 2M
r = R
m2 = 1/2M
distance between the center of mass to point P = p = R
Distance of the point mass to point P = d = 2R
We know that the moment of inertia for an uniform flat disk is 1/2mr². Then the moment of inertia for the uniform flat disk is:
Icm = 1/2mr²
Icm = 1/2(2M)(R²)
Icm = MR² ... (i)
Next, we will find the moment of inertia of the disk with respect to point P. We know that point P is positioned at the arc of the disk. Hence:
Ip = Icm + mp²
Ip = MR² + (2M)R²
Ip = 3MR² ... (ii)
Then, the total moment of inertia of the disk with the point mass is:
I total = Ip + I mass
I total = 3MR² + (1/2M)(2R)²
I total = 3MR² + 2MR²
I total = 5MR² ... (iii)
Learn more about Uniform Flat Disk here: brainly.com/question/14595971
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