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.)
Answer:
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
The 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.) Image check below.
A) L is = 22.4 kg.m²/s and it's direction will be along the positive(+ve) z-axis
B) L is = 3.36 kg.m²/s
We are applying the Formula for the moment of Inertia would be; I = mr²
Then, m1 = 4 kg
After that, m2 is = 3 kg
Now, r = 1/2 = 0.5 m
So, When the sum of the moment of inertia for the 2 masses would be;
Then, I = (4 × 0.5²) + (3 × 0.5²)
After that, I = 1.75 kg.m²
Now, when the angular velocity is given by the formula;
Then, ω = v/r
We are given v is = 6.4 m/s
Then, ω = 6.4/0.5
After that, ω = 12.8 rad/s
Now, let's find the angular momentum.
Then, the Angular momentum; L = Iω
L is = 1.75 × 12.8
Theregore, L = 22.4 kg.m²/s
Now, we are using the right-hand rule, the direction will be along the positive(+ve) z-axis.
B) Now, wehn the new diameter is 15 cm = 0.15 m
Thus, radius;r = 0.15/2 = 0.075 m
Then, Similar to a above;
After that, I = (4 × 0.075²) + (3 × 0.075²)
Now, I = 0.039375 kg.m²
Then, ω = v/r
We are given v = 6.4 m/s
ω is = 6.4/0.075
ω is = 85.33 rad/s
When the Angular momentum; L = Iω
Then, L = 0.039375 × 85.33
Therefore, L = 3.36 kg.m²/s
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