Assuming the mass at the end has negligible size , what is the I of a particle of mass m about a pivot point , and then note the I of the rod-mass system is the sum of the I's of its parts.xd14 said:Homework Statement
a 3 kg mass is on the end of a metal rod which is pivoted at one end. the mass of the rod is 2kg its length is 4 meters
Homework Equations
I=(ML^2)/3
The Attempt at a Solution
the rotational inertia of the rod itself is 10.6 but i don't know how the 3kg mass at the end would effect things
Rotational inertia, also known as moment of inertia, is a measure of an object's resistance to change in rotational motion. It depends on the object's mass, distribution of mass, and axis of rotation.
Rotational inertia is similar to linear inertia in that it describes an object's resistance to change in motion. However, while linear inertia depends on an object's mass and velocity, rotational inertia also takes into account the object's distribution of mass and axis of rotation.
The position of mass on a rod can significantly affect its rotational inertia. The farther away the mass is from the axis of rotation, the higher the rotational inertia will be. This means that a rod with mass concentrated at one end will have a higher rotational inertia than a rod with evenly distributed mass.
Yes, the shape of a rod can affect its rotational inertia. For example, a thin and long rod will have a lower rotational inertia than a thick and short rod with the same mass. This is because the thin rod has more of its mass concentrated farther away from the axis of rotation, resulting in a higher rotational inertia.
The rotational inertia of a rod can be calculated using the formula I = MR^2, where I is the rotational inertia, M is the mass of the rod, and R is the distance from the axis of rotation to the center of mass. This formula assumes that the mass is evenly distributed along the rod.