The outline of a very uniform, very circular pizza, centered at (x,y)=(0,0), is
ID: 2278044 • Letter: T
Question
The outline of a very uniform, very circular pizza, centered at (x,y)=(0,0), is shown below after a slice has been carefully removed.
Calculate the x-coordinate of the center of mass of the remaining pizza. The grid units are inches. Note that the tip and corners of the slice which was removed lie exactly on intersection points. Your answer must be accurate to within 0.01 in (inches) or 0.0254 cm.
Explanation / Answer
1)
x(c.m.) = (1/M)integral x*dm
M = mass of pie after removal of wedge.
For pie before the wedge is removed,
x(c.m.) = 0
After the wedge is removed, x(c.m.) = 0 + (1/m) *integral of x*dm from x = -6 to x = 0
and bounded by y1 = x/6 and y2 = -x/6
where m = mass of wedge
(This equation is valid because the pie is of uniform surface density).
Now, dm = ?(y1 - y2)dx = (?x/3)dx
So x(cm) = (?/m)*integral of [(x^2)/3]dx from x = -6 to x = 0
= (?/m)*24
But ?/m = 1/area of wedge
= 1/[(R^2/2)(2pi - 2?)] = 1/[(R^2)(pi - ?)]
where 2? = wedge angle = 2 arc sin (1/6) = 0.3349 rad or ? = 0.16745 rad
So x(cm) = 24/(R^2*(pi - ?) = 0.224" answer
2)
? torques about (2,4) = 0:
Let W = weight of mass = 3.30*9.8 = 32.34N
Wb = weight of rod = 1.5*9.8 = 14.7N
T = tension in wire
? = arc tan(5/10) = 26.565 deg
Then 10W + 6Wb = 10Tsin? = 4.472T
T = 92.04N
?vertical forces = 0:
-W - Wb + Tsin? + Fv = 0
where Fv = vertical force where rod meets wall
Fv = W + Wb - 0.4472T = 32.34 + 14.7 - 0.4472*92.04 = 5.88N
?horizontal forces = 0:
Fh - Tcos? or Fh = 92.04*0.894 = 82.3N
where Fh = horizontal force where rod meets wall.
Therefore, min. required static coeff. of friction = Fv/Fh = 5.88/82.3 = 0.071
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