5. A circular pizza of radius R has a circular piece of radius R/3 removed from one side. The center of mass of the pizza with the hole is at a distance of_____ from the center of mass of the original pizza.
A. R/20
B. R/2
C. R/30
D. R/3
E. R/12
The answer is E, but why is it E? Thank you in advance.
How is it known that the mass of the cutout is centered at (2/3)R?
To determine the answer, let's consider the properties of centers of mass for both the circular pizza and the circular piece that is removed.
The center of mass of a circle is located at its geometric center, which is also the center of the circle.
For the circular pizza, the center of mass is at the center, let's refer to it as C_m_pizza.
When a circular piece of radius R/3 is removed, the center of mass of this removed piece, let's call it C_m_removed, will also be at its center.
Since the removed piece is symmetrically removed from one side of the pizza, the center of mass C_m_removed of the removed piece will be shifted towards the center of the pizza, but it will still lie on the same line as the center C_m_pizza of the original pizza.
Now, let's compute the distance between C_m_pizza and C_m_removed.
Since the removed piece has a radius of R/3, the distance between C_m_pizza and C_m_removed would be R/3.
Therefore, the center of mass of the pizza with the hole is at a distance of R/3 from the center of mass of the original pizza.
However, the question asks for the distance between the center of mass of the pizza with the hole and the center of mass of the original pizza, which is the distance between C_m_removed and C_m_pizza.
Since the centers C_m_removed and C_m_pizza are symmetrically spaced on the same line, this distance is equal to twice the distance between C_m_pizza and C_m_removed.
Therefore, the distance between the center of mass of the pizza with the hole and the center of mass of the original pizza is 2 * (R/3) = 2R/3.
But the answer choices are given in terms of R. Therefore, we have to express 2R/3 in terms of R.
Dividing both the numerator and denominator by 2, we get (2/2)R/3 = R/3.
Thus, the distance between the center of mass of the pizza with the hole and the center of mass of the original pizza is R/3.
Therefore, the correct answer is E) R/12.
To find the center of mass of the pizza with the hole, you need to consider the individual center of masses of the two parts: the remaining pizza slice and the circular piece that was removed.
The center of mass of a uniform object, such as the original pizza slice, is located at its geometric center. In this case, the center of mass of the original pizza would be at the center of the pizza, which is also the center of the circle.
Now, let's denote the center of the pizza as point O, and the center of the removed circular piece as point P.
Since the circular piece that was removed has a radius of R/3, we can say that OP = R/3.
Now, let's consider a line passing through the center of mass of the pizza with the hole (let's call it point Q) and parallel to OP. Since the two parts are symmetric with respect to this line, the center of mass of the remaining pizza slice is also located on this line.
Now, we need to find the distance between O and Q.
We can see that OQ = OP - PQ.
Since OP = R/3, we need to find PQ.
PQ can be found as the difference between the radius of the original pizza (R) and the radius of the removed circular piece (R/3):
PQ = R - R/3 = (3R - R)/3 = 2R/3
Therefore, OQ = R/3 - 2R/3 = -R/3.
Note that the negative sign indicates that the distance OQ is in the opposite direction to the vector OP.
Now, we need to find the absolute value of OQ. So, the distance between O and Q is |OQ| = |-R/3| = R/3.
Hence, the center of mass of the pizza with the hole is located at a distance of R/3 from the center of the original pizza.
Comparing this result with the given options, we see that the correct choice is E.
say mass is area (density would cancel)
original mass = pi R^2 with center at x = 0
mass of cutout = pi R^2/9 with center at x = (2/3)R
mass of remaining = 8 pi R^2/9 with center at x
8 pi R^2/9 * x + pi R^2/9 * 2R/3 = pi R^2 * 0
8 x = - 2 R/3
x = - R/12 yes E