College Physics- Inertia?
Find the moment of inertia about each of the following axes for a rod that is 0.300 in diameter and 1.60 long, with a mass of 5.00×10−2 .
How do you find this ? I know how to find the perpendicular and end .
but this?
About an axis along the length of the rod.
If the rod has a measurable radius, it is a cylinder.
The moment of inertia of cylinder about the axis passing through its center
is
I=mR²/2=mD²/8 = 5•10⁻⁴•0.3²/8 =…
To find the moment of inertia about an axis along the length of the rod, you can use the formula for the moment of inertia of a thin rod rotating about an axis passing through its center and perpendicular to its length:
I = (1/12) * m * (L^2)
Where:
- I is the moment of inertia
- m is the mass of the rod
- L is the length of the rod
Given that the rod has a diameter of 0.300 (which means a radius of 0.150), and a length of 1.60, and a mass of 5.00×10^(-2), you can substitute these values into the formula to calculate the moment of inertia.
I = (1/12) * (5.00×10^(-2)) * (1.60^2)
Simplifying this equation will give you the moment of inertia about an axis along the length of the rod.