Finding mass of a nonuniform sphere, given a function for density

AI Thread Summary
To find the mass of a nonuniform sphere with a density that decreases with distance from the center, the mass can be calculated using the formula for a spherical shell, where the mass of each shell is given by dm=4πr²ρ(r)dr. The integral of this expression over the radius will yield the total mass of the sphere. For the moment of inertia about a diameter, a similar approach involving integration of the density function is required. The discussion highlights the importance of understanding spherical coordinates and the use of integrals in solving the problem. The solution emphasizes that the problem can be approached with a single integral rather than a triple integral.
Tubefox
Messages
9
Reaction score
0

Homework Statement



A sphere with radius .175 cm has a density ρ that decreases with distance r from the center of the sphere according to

ρ= ((2.75*10^3) kg/m^3) - ((9.25*10^3) kg/m^4)r

A) Calculate the total mass of the sphere.

B) Calculate the moment of inertia for an axis along the diameter.


Homework Equations



2.75*10^3 kg/m^3 - (9.25*10^3 kg/m^4)r

Probably some sort of integral.

The Attempt at a Solution




I'm sorry, but I have absolutely no idea how to even begin to do this. I tried googling, and the only result I found contained a triple integral in spherical coordinates. Given that this is on my Calc-based physics I homework and the only prerequisite is calculus I, it seems implausible that this is actually the best solution. I've never used spherical coordinates in my life, and only vaguely know how to solve a triple integral (I've taught myself a bit online, but none of my classes have ever covered them before).
 
Last edited:
Physics news on Phys.org
Tubefox said:

Homework Statement



A sphere with radius .175 cm has a density ρ that decreases with distance r from the center of the sphere according to

ρ= ((2.75*10^3) kg/m^3) - ((9.25*10^3) kg/m^4)r

Homework Equations



2.75*10^3 kg/m^3 - (9.25*10^3 kg/m^4)r

Probably some sort of integral.

The Attempt at a Solution

I'm sorry, but I have absolutely no idea how to even begin to do this. I tried googling, and the only result I found contained a triple integral in spherical coordinates. Given that this is on my Calc-based physics I homework and the only prerequisite is calculus I, it seems implausible that this is actually the best solution. I've never used spherical coordinates in my life, and only vaguely know how to solve a triple integral (I've taught myself a bit online, but none of my classes have ever covered them before).

This has not been covered in the lecture at any point, and there isn't a lecture before this homework is due.

The mass of a spherical shell is ##dm=4 \pi r^2 \rho(r) dr##, surface area times density times thickness, so you can also do it as a single integral since it's spherically symmetric. Perhaps that's what they are expecting you to use.
 
Oh, yeah, you're absolutely correct. Thank you so much - I didn't realize that you could use the surface area, and needless to say the fact that the only similar problem I could find involved a triple integral did not exactly help me maintain confidence that I could solve it.

Thanks.
 
mass = volume integral of density.

Divvy up your sphere into shells of thickness dr, then integrate over r.
Careful with your R given in cm, not m.
 
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