# Derviation of Moment of inertia

our lecturer defined MI as inertia of body in circular motion(i don't know anything about inertia)
except that it is something to oppose
my question is that
when M.I in circular motion depends on both mass and radius(distance from axis) of the body.
why MI is represented as mr^2 why is it not mr
how can some quantity depend on the square of another term when they are also directly prop. to that quantity without square....

Why do you think it should be mr? Momentum of inertia *depends* on the radius from the axis of rotation, but is not *proportional* to the radius. In fact it is proportional to the square of the radius. It is defined this way so that (momentum of inertia) times (angular velocity) = (angular momentum).

my question is why is the square
force=ma
not
force=ma^2

Newton's law is a completely separate equation. Perhaps you are confused by the terminology "moment of inertia"?

Newton's second law is (force) = (mass) * (acceleration). It tells you the rate of change of an object's velocity if you apply a given force to it.

Moment of inertia is defined by (moment of inertia) = (mass) * (distance from axis of rotation)^2. In conjunction with the equation (torque) = (moment of inertia) * (angular acceleration) it lets you find the rate of change of angular velocity if you apply a torque. Moment of inertia is kind of an analog to mass: if something has a large mass, you have to apply a large force to accelerate it much. If something has a large moment of inertia, you have to apply a large torque to get it to rotate very quickly.

i am just confused or i mean trying to figure out why some quantities depends on square of a given quantity and some degree of 1 of it.....

tiny-tim
Homework Helper
i am just confused or i mean trying to figure out why some quantities depends on square of a given quantity and some degree of 1 of it.....

hi anmolnanda! "moment of …" means "r cross …", so anything with moment has an extra r (and you can take any formula like F = ∫ a dm, do "r cross" of the whole thing, and get another formula …

in this case, r x F = ∫ r x a dm = ∫ r x (r x α) dm, or τ = Iα)

Suppose a particle of mass m is attached to a pivot by a thin rod of length r . As the particle travels around the circle, we know that the distance it travels is equal to the angle the rod sweeps out measured in radians multiplied by the radius r . Differentiating twice shows that
a = r A

where A is the angular acceleration (i.e. the rate at which the angular velocity of the rod is changing) and a is the instantaneous linear acceleration the particle experiences out on the circle.

By Newton's second law for linear motion, if we apply a force F to the particle, then F = m a . On the other hand, since we have a rotating system, we would like to work with torque , instead of force, so we multiply both sides of the equation by r . Then

T = F r = m r a .

Finally, we use the equation derived about, to convert from linear acceleration to angular acceleration:

T = m r a = m r (A r ).

Rearranging terms gives the desired formula T = (m r 2) A.