Moment of Inertia in Planet System

[lc99]

Homework Statement

Two planets are masses stuck to end of long thin rods. These model systems (Rod + mass) will be rotated by a motor at their other end of the rod, as shown in the figure. the length of the rod is proportional to the orbital distance of the planet, and the mass at the end is proportional to the mass of the planet. For this problem we focus on the moments of inertia of the Earth and Mars model systems, about the point ofattachment to the motor.

Given:
* Rod is made of material with mass density of 1.0 kg per meter (does this mean that the rod is uniform?)
*The Earth is presented by a rock of mass 1kg, at the end of a 1 meter rod
* The mass of Mars is 0.1 x the mass of earth
*the orbital radius of Mars is 1.5 the radius of earth

1) inerta Earth is 1/4 (approx) of inertia mars
2) inertia Earth is 1/2 mars
3)inertia Earth is 4 times mars
4)inertia Earth is 2 times mars
5)inertia Earth is equal to mars

Homework Equations

Inertia of point masses around orbit = MR^2

The Attempt at a Solution

My answer is 3) Earth's moment of inertia is 4 times the inertia of mars

I ignored the rod because it's light weight
Earth's Moment of Inertia = MR^2 = 1*1^2 =1
Mar's moment of Inertia = MR^2 = 0.1*1*1.5^2=0.225

1/0.225 = approx 4.44
So, Earth is about 4 x mar's moment of inertia

 

[Gene Naden]

Seems reasonable

 

[lc99]

Seems reasonable

Do you think that it is reasonable to ignore that thin rods? I wasn't sure when i was doing the problem

 

[haruspex]

I ignored the rod because it's light weight

It is? How does the mass of the rod compare with the mass of the planet in each case?

 

[Gene Naden]

Well actually, since they specify the density of the rod per unit length, it seems we are supposed to take it into account. So the moment of inertia of the planet plus the rod would include a contribution from the integral of linear density*r^2 for r from 0 to the radius of the planet. Does your physics course include calculus?

 

[haruspex]

Well actually, since they specify the density of the rod per unit length, it seems we are supposed to take it into account. So the moment of inertia of the planet plus the rod would include a contribution from the integral of linear density*r^2 for r from 0 to the radius of the planet. Does your physics course include calculus?

It surely includes formulas for moments of inertia of simple shapes, like thin rods.

 

[lc99]

It surely includes formulas for moments of inertia of simple shapes, like thin rods.

Yeah, we have the formula: thin rod about the center is 1/12ML^2. Does this mean my answer is incorrect?

i think with the rod, the answer is B) half

Inertia of Earth = 1/12ML^2 +md^2
inertia of Mars = 1/12ML^2 + md^2

d of Earth = .5
d of Mars = 1.5/2

m of Earth = 1 + 1 =2
m of Mars = 0.1 +1.5

M of Earth = 1
M of Mars = 0.1

L of Earth rod = 1
L of Mars rod = 1.5

Plugging in the numbers, i got 0.49 (M of earth/ M of mars)

 

[haruspex]

about the center

Right, but the rotation is not about its centre.

 

[lc99]

Right, but the rotation is not about its centre.

Oh, it's 1/3 because the axis is at the end

 

[haruspex]

Oh, it's 1/3 because the axis is at the end

Yes.