Find Center of Mass of Hemisphere Homework

  • Thread starter Thread starter JJfortherear
  • Start date Start date
  • Tags Tags
    Hemisphere
AI Thread Summary
To find the center of mass of a hemisphere with radius R, the correct approach involves using volume density instead of surface density. The mass element, dm, is defined as ρA dz, where A is the area of a circular disc with radius r = √(R² - z²). The integral for the z-component of the center of mass is set up as CM(z) = (1/M) ∫₀ᴿ π(R² - z²)ρz dz. Evaluating this integral correctly yields the center of mass at 3R/8. The discussion emphasizes the importance of using the appropriate density for solid objects in such calculations.
JJfortherear
Messages
14
Reaction score
0

Homework Statement


A hemisphere of radius R, covering +/- R in the x and y directions and 0 to R in the Z direction (only the top half of a sphere centered at the origin)
Find the center of mass.

Homework Equations


Cm=(1/m)\int(z dm)

The Attempt at a Solution


dm is sigma dA, and dA is x dz, where x = \sqrt{}R^2-z^2. The limits are 0 to R, and the answer should be 3/8 R, but when I evaluate the integral I get nothing like that.
 
Physics news on Phys.org
since you are trying to find for a hemisphere not for a hemispherical shell, your equation for 'dm' is wrong.
You should consider not the surface density, but instead the volume density ρ since it is a solid object
\rho = \frac{M}{\frac{2}{3} \pi R^3}

You then have

dm = \rho A dz where A is the area of the circular disc centered on the z-axis with radius r = \sqrt{R^2 - z^2}.

The final integral then becomes

CM (z-comp) = \frac{1}{M} \int_0^R \ dz\ \pi (R^2 - z^2) \rho\ z

which on solving gives the result \frac{3 R}{8}
 
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