The triceps muscle in the back of the upper arm is primarily used to extend the forearm. Suppose this muscle in a professional boxer exerts a force of 1.75 Ă— 103 N with an effective perpendicular lever arm of 3.05 cm, producing an angular acceleration of the forearm of 115 rad/s2. What is the moment of inertia of the forearm, in kilogram meters squared?

Respuesta :

To solve this problem it is necessary to apply the concepts related to Torque as a function of the Force and the distance radius where it is applied.

By definition the Torque can be expressed as

[tex]\tau = F\times r[/tex]

Where

F = Force exerted

r = Radius

Substituting we have to

[tex]\tau = (1.75*10^3)(0.0305)[/tex]

[tex]\tau = 53.375N\cdot m[/tex]

Through the second definition of the rotational Torque we can then find the moment of inertia for which we have to

[tex]\tau = I\alpha[/tex]

Where

I = Moment of inertia

[tex]\alpha =[/tex] Angular acceleration

Replacing

[tex]53.373 = I*115[/tex]

[tex]I = 0.4641Kg \cdot m^2[/tex]

Therefore the moment of inertia is

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