What If? What would be the new angular momentum of the system (in kg · m2/s) if each of the masses were instead a solid sphere 14.5 cm in diameter? (Round your answer to at least two decimal places.)

Question

The first part of the question is :

A light, rigid rod of length ℓ = 1.00 m

 joins two particles, with masses m₁ = 4.00 kg

 and m₂ = 3.00 kg,

 at its ends. The combination rotates in the xy-plane about a pivot through the center of the rod (see figure below). Determine the angular momentum of the system about the origin when the speed of each particle is 5.80 m/s. (Enter the magnitude to at least two decimal places in kg · m²/s.)

I got the answer magnitude 20.3 kg m^2/s and in +z direction.

However part b asks :

What If? What would be the new angular momentum of the system (in kg · m²/s) if each of the masses were instead a solid sphere 14.5 cm in diameter? (Round your answer to at least two decimal places.)

I didn't know how to solve this part.

Answer 1

This is an application of Koenig's theorem

L = L0 + L', where

L is the angular momentum of a body rotating around a point O,

L0 is the angular momentum of the center of gravity, with all mass concentrated there,

L' is the angular momentum of the body relative to the center of gravity.

Before solving your problem with two specific spheres, lets solve

the problem for a general sphere where the matter is uniformly distributed.

Let

r be the radius of the sphere,

R be the distance of the center of gravity of the sphere from the center of rotation,

V = 4πr³/3 be the volume of the sphere,

s be the specific density of the sphere,

M = sV be the mass of the sphere,

ω be the angular velocity.

Notice that in your situation,

ω is the angular velocity of the center of gravity around the point O,

as well as the angular velocity of the sphere around its center of gravity.

As you have demonstrated

L0 = MωR²

To calculate L' think of the whole solid sphere as composed of concentric hollow sub-spheres,

each with radius x and thickness dx,

where x ranges from 0 to r.

The area of each sub-sphere is 4πx², its volume is 4πx²dx, its mass is s4πx²dx,

and its angular momentum is ωs4πx⁴dx.

The angular momentum of the whole sphere is the integral over all the sub-spheres.

The indefinite

∫ωs4πx⁴dx = ωs4πx⁵/5

Therefore for the definite integral from 0 to r we get

L' = ωs4πr⁵/5

In order to express it in terms of the mass M instead of density s we rewrite it as

L' = 3ωr²/5 × s4πr³/3 = 3Mωr²/5

Thus

L = MωR² + 3Mωr²/5 = Mω(R² + 3r²/5)

Now applying it to your two spheres, and expressing ω in terms of your given velocity v:

L₁ + L₂ =

m₁ω(R² + 3r²/5) + m₂ω(R² + 3r²/5) =

(m₁+m₂)ω(R² + 3r²/5) =

(m₁+m₂)(v/R)(R² + 3r²/5)

Substituting actual numbers

L1 + L2 = 7 × (5.8/0.5) × (0.5² + 3×0.145²/5) = 21.32

As a side note. You said that the direction of the angular momentum is in +z. From the statement of the problem I cannot tell whether it is +z or -z, because it does not mention the direction of rotation. But I assume that your picture had that.

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This is an application of Koenig's theorem

L = L0 + L', where

L is the angular momentum of a body rotating around a point O,

L0 is the angular momentum of the center of gravity, with all mass concentrated there,

L' is the angular momentum of the body relative to the center of gravity.

Before solving your problem with two specific spheres, lets solve

the problem for a general sphere where the matter is uniformly distributed.

Let

r be the radius of the sphere,

R be the distance of the center of gravity of the sphere from the center of rotation,

V = 4πr³/3 be the volume of the sphere,

s be the specific density of the sphere,

M = sV be the mass of the sphere,

ω be the angular velocity.

Notice that in your situation,

ω is the angular velocity of the center of gravity around the point O,

as well as the angular velocity of the sphere around its center of gravity.

As you have demonstrated

L0 = MωR²

To calculate L' think of the whole solid sphere as composed of concentric hollow cylinders,

each with radius x and thickness dx,

where x ranges from 0 to r.

The cylinders are centered around the axis of rotation parallel to the z-axis

and going though the center of the sphere.

The perimeter is of each cylinder is 2πx, and by Pythagorean theorem,

the height of each cylinder is 2√(r²-x²).

The area of each cylinder is then 4πx√(r²-x²),

its volume is 4πx√(r²-x²)dx,

its mass is s4πx√(r²-x²)dx,

and its angular momentum is ωs4πx³√(r²-x²)dx.

The angular momentum of the whole sphere is then the integral over all the cylinders.

Call the indefinite integral F(x), where

F(x) = ∫ωs4πx³√(r²-x²)dx = ωs4π∫x³√(r²-x²)dx

We solve the integral by parts using the formula

∫uv' = uv - ∫u'v

where

u = x², u' = 2x

v' = x√(r²-x²), v = (-1/3)(r²-x²)^3/2

Thus

F(x)

= ωs4π ( (-1/3)x²(r²-x²)^3/2 - ∫2x(-1/3)(r²-x²)^3/2 dx )

= ωs4π ( (-1/3)x²(r²-x²)^3/2 + (2/3) ∫x(r²-x²)^3/2 dx )

= ωs4π ( (-1/3)x²(r²-x²)^3/2 + (2/3)(-1/5)(r²-x²)^5/2 )

= ωs4π ( (-1/3)x²(r²-x²)^3/2 - (2/15)(r²-x²)^5/2 )

The value of L' is the definite integral from 0 to r, i.e.,

L' = F(r) - F(0)

F(r) = ωs4π ( (-1/3)r²(r²-r²)^3/2 - (2/15)(r²-r²)^5/2 ) = 0

F(0) = ωs4π ( (-1/3)0²(r²)^3/2 - (2/15)(r²)^5/2 ) = -(8/15)ωsπr⁵

L' = (8/15)ωsπr⁵

In order to express it in terms of the mass M instead of density s we rewrite it as

L' = (2/5)ωr² × (4/3)sπr³ = (2/5)Mωr²

Thus

L = MωR² + (2/5)Mωr² = Mω(R² + 2r²/5)

Now applying it to your two spheres, and expressing ω in terms of your given velocity v:

L₁ + L₂ =

m₁ω(R² + 2r²/5) + m₂ ω(R² + 2r²/5) =

(m₁+m₂)ω(R² + 2r²/5) =

(m₁+m₂)(v/R)(R² + 2r²/5)

Substituting actual numbers

L₁ + L₂ = 7 × (5.8/0.5) × (0.5² + 2×0.145²/5) = 20.98