Acertain merry-go-round is accelerated uniformly from rest and attains an angular speed of 1.2 rad/s in the first 18 seconds. if the net applied torque is 1200 n · m, what is the moment of inertia of the merry-go-round?

Moment of inertia is the term specified to rotational inertia which is the rotational equivalent of mass for linear motion. It gives the idea in the relationship for changing aspects of rotational motion the moment of inertia must be stated with respect to a preferred axis of rotation.

Given the following

Net applied torque         = 1200 Nm

Angular acceleration      = 1.2 rad/s / 18 seconds (angular speed over time)

Derive the formula: net applied torque = moment of inertia x angular acceleration into moment of inertia = net applied torque over angular acceleration.

Therefore:

I             = 1200 Nm / (1.2 rad/s / 18 seconds)

I             = 18000 kg.m² moment of inertia

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west