A solid aluminum sphere of radius R has moment of inertia I about an axis through its center. What is the moment of inertia abou
t a central axis of a solid aluminum sphere of radius 2R
1 answer:
Answer:
I1 = 2/5 M1 R^2 for a sphere about its center
I2 = 2/5 M2 (2 R)^2 = 2/5 M2 R^ * 4 = 8/5 M2 R^2
Remember that M2 is greater than M1 by a factor 0f 2^3 = 8
Then I2 exceeds I1 by a factor of 32
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Good morning.
We have:
Where
j is the unitary vector in the direction of the
y-axis.
We have that
We add the vector
-a to both sides:
Therefore, the magnitude of
b is
47 units.
Self productive and it depends on how whom is behaving.
B: Extension Lines! You could have just searched this up on google
Answer:
I would say the 3rd one
Explanation:
