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Chapter 7: Problem 90

A flat, circular metal disk of uniform thickness has a radius of $3.0 \mathrm{cm} .\( A hole is drilled in the disk that is \)1.5 \mathrm{cm}$ in radius. The hole is tangent to one side of the disk. Where is the \(\mathrm{cm}\) of the disk now that the hole has been drilled? [Hint: The original disk (before the hole is drilled) can be thought of as having two pieces-the disk with the hole plus the smaller disk of metal drilled out. Write an equation that expresses \(x_{\mathrm{CM}}\) of the original disk in terms of the \(x_{\mathrm{CM}}\) 's of the two pieces. since the thickness is uniform, the mass of any piece is proportional to its area.]

Short Answer

Expert verified
Answer: The center of mass of the disk after the hole has been drilled is at approximately \(\frac{45}{8}\) cm from the edge.

Step by step solution

01

Calculate the area of the original disk

Using the formula for the area of a circle \(A = \pi r^2\), we can find the area of the original disk: \(A_\text{original} = \pi(3^2) = 3\pi\) sq.cm.
02

Calculate the area of the hole

Applying the same formula for the hole, we get: \(A_\text{hole} = \pi(1.5^2) = \frac{9}{4}\pi \) sq.cm.
03

Calculate the area of the disk with the hole (ring)

The area of the disk with the hole is the difference between the area of the original disk and the area of the hole: \(A_\text{ring} = A_\text{original} - A_\text{hole} = (3\pi - \frac{9}{4}\pi) = \frac{3}{4} \pi\) sq.cm.
04

Determine location x-coordinates of the centers of mass

Let's consider two x-coordinates for the centers of mass: \(x_\text{ring}\) and \(x_\text{hole}\). The center of mass of the hole is located at the midpoint of the hole's diameter, which is at the 1.5cm distance from the edge. Therefore, \(x_\text{hole} = 1.5\) cm. The center of mass of the disk with the hole (ring) is located at the center of the disk, which is at the 3cm distance from the edge. Therefore, \(x_\text{ring} = 3\) cm.
05

Set up the equation and find the x-coordinate of the center of mass of the resulting object

Let \(x_\text{CM}\) be the center of mass of the original disk. We can write the equation as: \(x_\text{CM} = \frac{A_\text{ring} \times x_\text{ring} + A_\text{hole} \times x_\text{hole}}{A_\text{original}}\) Now substitute values we found earlier: \(x_\text{CM} = \frac{(\frac{3}{4}\pi)(3) + (\frac{9}{4}\pi)(1.5)}{(3\pi)}\) Simplify the equation: \(x_\text{CM} = \frac{ 9\pi + \frac{27}{2}\pi}{6\pi}\) Finally, divide the numerator by the denominator and simplify: \(x_\text{CM} = \frac{ 9\pi + \frac{27}{2}\pi}{6\pi} = \frac{45}{8}\)
06

Determine the center of mass of the resulting object

The center of mass of the resulting object is \((x_\text{CM}, 0)\): \((\frac{45}{8}, 0)\) The center of mass of the disk after the hole has been drilled is at \(\frac{45}{8}\) cm.

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Most popular questions from this chapter

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