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 15.0 cm in diameter? (Round your answer to at least two decimal places.)

Complete question is;

(a) A light, rigid rod of length, l = 1.00 m joins two particles, with masses m = 4.00 kg and m, = 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 6.4 m/s. (Enter the magnitude to at least two decimal places in kg. m/s.)

(b) What If? What would be the new angular momentum of the system (in kg. m/s) if each of the masses were instead a solid sphere 15.0 cm in diameter? (Round your answer to at least two decimal places.)

A) L = 22.4 kg.m²/s and it's direction will be along the positive(+ve) z-axis

B) L = 3.36 kg.m²/s

Explanation:

The image is missing, so i have attached it.

Formula for the moment of Inertia would be; I = mr²

m1 = 4 kg

m2 = 3 kg

r = 1/2 = 0.5 m

So, sum of moment of inertia for the 2 masses would be;

I = (4 × 0.5²) + (3 × 0.5²)

I = 1.75 kg.m²

Now, angular velocity is given by the formula;

ω = v/r

We are given v = 6.4 m/s

So;

ω = 6.4/0.5

ω = 12.8 rad/s

Now, let's find angular momentum.

Angular momentum; L = Iω

L = 1.75 × 12.8

L = 22.4 kg.m²/s

Now, using the right hand rule, the direction will be along the positive(+ve) z-axis.

B) Now, the new diameter is 15 cm = 0.15 m

Thus,

radius;r = 0.15/2 = 0.075 m

Similar to a above;

I = (4 × 0.075²) + (3 × 0.075²)

I = 0.039375 kg.m²

ω = v/r

We are given v = 6.4 m/s

ω = 6.4/0.075

ω = 85.33 rad/s

Angular momentum; L = Iω

L = 0.039375 × 85.33

L = 3.36 kg.m²/s

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energy can be converted from one form to another, but it cannot be created or destroyed in ordinary chemical or physical means. do not understand what you are asking but hope this