Find the center of mass of a semi-circular plate

AI Thread Summary
To find the center of mass of a semi-circular plate with radius r, the x-coordinate is intuitively zero, while the y-coordinate can be calculated using the formula ycentroid = ∫∫_R y dA / A, assuming uniform mass distribution. The volume of the solid formed when the plate is rotated around its straight side can be determined using the theorem of Pappus, which states that volume equals the area of the region multiplied by the distance traveled by the centroid, specifically 2πycentroid. The expected volume from this rotation should match that of a sphere, calculated as 4/3πr^3, providing a check for accuracy. This method effectively combines centroid calculations with rotational volume principles.
Debelius
Messages
1
Reaction score
0

Homework Statement


Find the center of mass of a semi-circular plate of radius r

Find the volume when the plate (above) is rotated around a line along its straight side

Homework Equations



2(pi) integral of r dr

The Attempt at a Solution



I honestly don't know how to do centroids. :-( I'd like to know how to actually solve this problem.
 
Physics news on Phys.org


The x coordinate of the center of mass should be zero (intuitively).
For the y coordinate, assuming that the mass is uniformly distributed, use this equation (which can generalize for x):
ycentroid = \frac{\int\int_R y dA}{A}

The volume is thus computed from the theorem of Pappus which states that
volume = (area of R) * (distance traveled by the centroid)
where the distance traveled by the centroid is 2*pi*ycentroid

Hope that helps, let me know if you need further clarification.
(Also, the volume when rotated about the straight side should come out to be that of a sphere, which is 4/3*pi*r^3, so you can use that to make sure you did it right.)
 
Last edited:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top