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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

A \(58-\) kg astronaut is in space, far from any objects that would exert a significant gravitational force on him. He would like to move toward his spaceship, but his jet pack is not functioning. He throws a 720 -g socket wrench with a velocity of \(5.0 \mathrm{m} / \mathrm{s}\) in a direction away from the ship. After \(0.50 \mathrm{s}\), he throws a 800 -g spanner in the same direction with a speed of \(8.0 \mathrm{m} / \mathrm{s} .\) After another $9.90 \mathrm{s}\(, he throws a mallet with a speed of \)6.0 \mathrm{m} / \mathrm{s}$ in the same direction. The mallet has a mass of \(1200 \mathrm{g}\) How fast is the astronaut moving after he throws the mallet?
Prove Eq. $(7-13) \Sigma \overrightarrow{\mathbf{F}}_{\mathrm{ext}}=M \overrightarrow{\mathbf{a}}_{\mathrm{CM}} .\( [Hint: Start with \)\Sigma \overrightarrow{\mathbf{F}}_{\mathrm{cxt}}=\lim _{\Delta t \rightarrow 0}(\Delta \overrightarrow{\mathbf{p}} / \Delta t),\( where \)\Sigma \overrightarrow{\mathbf{F}}_{\mathrm{ext}}$ is the net external force acting on a system and \(\overrightarrow{\mathbf{p}}\) is the total momentum of the system.]
A cannon on a railroad car is facing in a direction parallel to the tracks. It fires a \(98-\mathrm{kg}\) shell at a speed of \(105 \mathrm{m} / \mathrm{s}\) (relative to the ground) at an angle of \(60.0^{\circ}\) above the horizontal. If the cannon plus car have a mass of \(5.0 \times 10^{4} \mathrm{kg},\) what is the recoil speed of the car if it was at rest before the cannon was fired? [Hint: A component of a system's momentum along an axis is conserved if the net external force acting on the system has no component along that axis.]
Use the result of Problem 54 to show that in any elastic collision between two objects, the relative speed of the two is the same before and after the collision. [Hints: Look at the collision in its \(\mathrm{CM}\) frame - the reference frame in which the \(\mathrm{CM}\) is at rest. The relative speed of two objects is the same in any inertial reference frame.]
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