A circular plate of uniform thickness has a diameter 56cm. a circular ...
Figure shows the circular disc of diameter 56 cm after a circular portion of diameter 42 cm is removed.
We know that point O is centre of mass for the full disc before removing the above mentioned circular portion.
we take the point O as reference to calculate the centre of mass by considering the full disc is made of two objects, i.e., the disc showed as coloured one(white colour circular portion is removed) and the removed white colour circular portion. Let point Q which is centre of mass of coloured object, be at a distance x cm from point O.
(m1*x + m2 x 7 )/(m1+m2) = 0 ....(1)
where m1 and m2 are masses of coloured object and white circular disc respectively.
We have written 0 in right hand side of eqn.(1) because we took point O which is centre of mass of full circular disc as reference point
from eqn.(1) we get x = - (m2*7)/m1...(2)
mass of circular objects are proportional to square of diameter because thickness and density are same.
hence we write eqn.(2) as , x = -(42*42*7) / [ (56*56)-(42*42)] = -9 cm
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A circular plate of uniform thickness has a diameter 56cm. a circular ...
Problem:
A circular plate of uniform thickness has a diameter 56cm. A circular portion of diameter 42 cm is removed from one edge of the plate. Find the center of mass of the remaining portion.
Solution:
To find the center of mass of the remaining portion, we can divide the plate into two sections - the removed portion and the remaining portion. Then we can find the center of mass of each section and calculate the overall center of mass.
Step 1: Calculate the area of the removed portion:
The area of a circle is given by the formula A = πr^2, where r is the radius.
For the removed portion, the radius is half the diameter, so r = 42/2 = 21 cm.
Therefore, the area of the removed portion is A_removed = π(21^2) = 1385.44 cm^2.
Step 2: Calculate the area of the remaining portion:
The remaining portion is the difference between the area of the entire plate and the area of the removed portion.
The radius of the entire plate is half the diameter, so r = 56/2 = 28 cm.
Therefore, the area of the entire plate is A_total = π(28^2) = 2463.24 cm^2.
The area of the remaining portion is A_remaining = A_total - A_removed = 2463.24 - 1385.44 = 1077.8 cm^2.
Step 3: Find the center of mass of the removed portion:
Since the removed portion is a circle, its center of mass is at its geometric center, which is the center of the circle. Therefore, the center of mass of the removed portion is at the center of the plate, which is the point (0, 0).
Step 4: Find the center of mass of the remaining portion:
The remaining portion is a circle with an off-center removed portion. To find its center of mass, we need to consider the position of the removed portion.
Since the removed portion is a circle with diameter 42 cm, its center is at a distance of 21 cm from the center of the plate.
To find the x-coordinate of the center of mass of the remaining portion, we can use the formula:
x_com = (A1*x1 + A2*x2) / (A1 + A2), where A1 and A2 are the areas of the sections and x1 and x2 are their respective x-coordinates.
In this case, A1 is the area of the remaining portion (1077.8 cm^2) and A2 is the area of the removed portion (1385.44 cm^2).
x1 is the x-coordinate of the center of mass of the remaining portion, which is the same as the x-coordinate of the center of the plate (0 cm).
x2 is the x-coordinate of the center of mass of the removed portion, which is also 0 cm.
Therefore, the x-coordinate of the center of mass of the remaining portion is:
x_com = (A1*x1 + A2*x2) / (A1 + A2) = (1077.8*0 + 1385.44*0) / (1077.8 + 1385.44)
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