source : yahoo.com

## Angular momentum question about the earth and sun. Quite confusing. I appreciate any help.?

You start with the equation for the magnitude of angular momentum.

There’s actually two equations, but the one you need is Iw, where I is the moment of inertia and w is the angular speed.

For question 1), the moment of inertia if mr^2 if you consider the Earth as a point entity, since compared to its orbital radius, the size of the Earth is very small.

Then the angular speed is 2pi/T where T is the orbital period.

Then angular momentum is mr^2(2pi)/T

2) For this one either you can use a table of moments of inertias, but I think they want you to actually develop it. To do that you will need to use integration.

The moment of inertia for a non-point entity is ∫r^2dm where r is the distance of a differential mass from the axis of rotation.

There can be many ways to do this, but the ultimate goal is to express dm in terms as r and then integrate over a single variable.

One way to do it is to cut up the sphere vertically into many slices. The cross sectional area of each slice is going to be pi(y)^2 where y = sqrt(R^2 – x^2), and R is the radius of the Earth. Then the thickness of each slice will be dx, and r becomes x. There’s also a factor of density, let’s call it p. So you get:

∫r^2dm = ∫ (from -R to R) px^2(pi)(R^2 – x^2)dx

= pi∫ (from -R to R) p(R^2x^2 – x^4)dx

= p(pi)(R^2x^3/3 – x^5/5) | from -R to R

= p(pi)[(R^5/3 – R^5/5) – (-R^5/3 + R^5/5)]

= p(pi)(5R^5 – 3R^5 + 5R^5 – 3R^3)/15 = 4p(pi)R^5/15

The volume of a sphere is 4/3(pi)R^3 so the density p times the volume gives the mass M.

(4/3)piR^3 * p * R^2/5 = MR^2/5

hmm I’m missing a factor of 2 in there somewhere, cause the correct answer should be 2MR^2/5

Let me try to do it another way (cylindrical shell method).

y = sqrt(R^2 – x^2)

Surface area = 2pi(x)(2y)

thickness is dx

dm = 2p(pi)(x)2sqrt(R^2 – x^2)dx

= 4p(pi)xsqrt(R^2 – x^2)dx

I = ∫r^2dm, r = x

= 4p(pi)∫(from 0 to R) x^3sqrt(R^2 – x^2)dx

u = R^2 – x^2, x^2 = R^2 – u

du = -2xdx, du/(-2) = xdx

0->R^2, R^2->0

I = 4p(pi)∫(from R^2 to 0) (R^2 – u)sqrt(u)/(-2)du

= -2p(pi) ∫(from R^2 to 0) R^2sqrt(u) – u^(3/2) du

= -2p(pi) * [2R^2u^(3/2)/3 – 2u^(5/2)/5] | from R^2 to 0

= -4p(pi) * [R^2u^(3/2)/3 – u^(5/2)/5] | from R^2 to 0

= -4p(pi) * [5R^2u^(3/2) – 3u^(5/2)]/15 | from R^2 to 0

= 0 – {-4p(pi) * [5R^2*R^3 – 3R^5]/15

= 4p(pi)(2)R^5/15

= (4/3)p(pi)R^3 * 2R^2/5

= M * 2R^2/5 = 2MR^2/5

There you go. Nailed it.

Not sure why the first method didn’t work, whether there’s a mathematical mistake or a Physics conceptual mistake, but I’m sure the math is correct.

If you look at some sites, they have a factor of (1/2) because that’s the moment of inertia for disk over a parallel axis, so that’s like instead of using ∫r^2dm, they added up the moments of inertia for a whole stack of disks that formed a sphere instead.

Then angular speed is just 2pi/T where T is 24 hours.

Solved: Calculate The Magnitude Of The Angular Momentum Of – Calculate The Magnitude Of The Angular Momentum Of The Earth In A Circular Orbit Around The Question: Calculate The Magnitude Of The Angular Momentum Of The Earth In A Circular Orbit Around The Sun. The Earth Has Mass 5.97×1024 Kg , Radius 6.38×106 M , And Orbit Radius 1.50×1011 M . The Planet Completes One Rotation On Its Axis In 24 Hours And(a) calculate the magnitude of the angular momentum of the earth in a circular orbit around the sun. is it reasonable to model it as a particle? (b) calculate the magnitude of the angular momentum of the earth due to its rotation around an axis through the north and south poles, modeling it as a uniform sphere. consult appendix e and the astronomical data in appendix f.Favorite Answer. Angualar moment L = I ω. ————————————————————. For spinning earth. I = [2/5]mr^2 and ω = 2π/T = [ 2π/86400] rad/s. m = 5.978e24kg, r =…

(a) calculate the magnitude of the angular momentum of the – A hemisphere of mass 3 m and radius R is free to slide with its base on asmooth horizontal table. A particle of mass m is placed on the top of thehemisphere. If particle is displaced with a negligible velocity, then find the angular velocity of the particle relative to the centre of the hemisphere at an angular displacement θ, when velocity of hemisphere is v:And that would be m r squared. I would be m r squared, Megan. Now, since the Earth is moving in a circle around the sun, we know that Omega can be calculated as two pi r over its period. So, using our data, we know we king simplify this two pi mass radius cube over the period.Calculate from the average distance of the Earth from the sun and the time it takes to complete one orbit as about 3xx10^4 ms^(-1) The average distance of the Earth from the Sun is spproximately 150 million km or 1.5 xx 10^11m The orbit of the Earth around the sun is roughly circular (more eliptical really, but let's keep it simple), with a period of roughly 365 days.

Calculate the angular momentum of Earth due to its – The angular velocity w is measured in radians per second (the distance around the circumference of a circle is 2*pi radians), and can be calculated from w = 2*pi / T rot, where T is the period of rotation of the sphere.Thus, the angular momentum of the Earth in a circular orbit around the Sun will be: L= I ω = (1.34×1047kg.m2)(1.99×10−7rad/s) = 2.67×1040kg.m2/s L = I ω = ( 1.34 × 10 47 k g. m 2) ( 1.99Mean distance from Earth to Sun is 149.6×106 km (8.31 minutes at the speed of light) Mass of the Earth is 5.9742×1024 kg Earth takes 365 days to go one complete circle around Sun. Average angular momentum Earth angular momentum is 2.663×10 40 kgm 2 s-1 There is no special name for this combination of units. The direction of the vector is perpendicular to the orbit.