I am confused on how to solve this problem

Question

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. The speed of each particle is 8.00 m/s. What would be the new angular momentum of the system (in kg · m²/s) if each of the masses were instead a solid sphere 13.5 cm in diameter?

Answer 1

Angular momentum has the following formula:

L = m*v*r

but there is an extra term Iω.

For point masses I = 0. so you just have

L = 4*8*0.5 + 3*8*0.5.

Note both masses require separate equations,

and the distance to the point masses is half the length of the rod.

When the masses become spheres instead you need to care about I.

I for a solid sphere is 2/5m*(r_sphere)²

ω is v/r for a spinning object. (Also known as angular velocity)

The complete formula for angular momentum is

L = m*v*r + Iω (for each individual mass)

Note they give you the diameter of the sphere instead of the radius and it is in cm not meters.

Now for the math.

ω = v/r = 8/0.5 = 16 rad/s

L = m₁*v*r_rod + I₁ω + m₂*v*r_rod + I₂ω

L = m₁*v*r_rod + 2/5*m₁*r_sphere²*ω + m₂*v*r_rod + 2/5*m₂*r_sphere²*ω

Plugging in values.

L = 4*8*0.5 + 2/5 *4*0.0675²*16 + 3*8*0.5 + 2/5 *3*0.0675²*16

L = 16 + 0.117 + 12 + 0.087

L = 28.204 kg m²/s

In the end changing the point masses to small spheres didn't change much. Only 0.204 kg m²/s

The reason why the Iω needs to be included is you need to take into account the momentum of the sphere rotating as the rod spins.