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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