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21 September, 22:52

A certain 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?

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  1. 22 September, 02:06
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    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
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