I have never dealt with moment of inertia before, this is a physics lab i need to do some pre planning for which involves topics we have never covered and are expected to learn. I've been busy working two jobs and am struggling to get time to pick up this before my lab tomorrow so some help would be extremely helpful.
We have a rod with length 2d, spinning at its centre. With 2 masses on either side of mass m2, the masses are spherical with radius r. What is the moment of inertia?
The answers is:
[itex] I=2 m_2 (d^2 + \frac{2}{5} r2) [/itex]
By my logic the moment of inertia is the sum of each component, we have two masses m2 and their centre of gravity is d+r away from the centre hence each sphere will contribute, based on the inertia of a sphere:
[itex] I_(spheres) =2 \frac{2}{5} m_2 (r + d)^2 [/itex]
And then add the component of the rod, which I believe is ignored as we don't have any value for the mass of the rod. Obviously my assumption is wrong as it isn't the same as the answer. Any tips would be choice. Cheers.
We have a rod with length 2d, spinning at its centre. With 2 masses on either side of mass m2, the masses are spherical with radius r. What is the moment of inertia?
The answers is:
[itex] I=2 m_2 (d^2 + \frac{2}{5} r2) [/itex]
By my logic the moment of inertia is the sum of each component, we have two masses m2 and their centre of gravity is d+r away from the centre hence each sphere will contribute, based on the inertia of a sphere:
[itex] I_(spheres) =2 \frac{2}{5} m_2 (r + d)^2 [/itex]
And then add the component of the rod, which I believe is ignored as we don't have any value for the mass of the rod. Obviously my assumption is wrong as it isn't the same as the answer. Any tips would be choice. Cheers.
