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The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2.00\times 10^32.00×10 ^3 N with an effective perpendicular lever arm of 3.00 cm, producing an angular acceleration of the forearm of 120 rad/s^2 .
What is the moment of inertia of the boxer's forearm?

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Answer:

Explanation:

From the given information:

The torque produced due to the force can be expressed as:

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

where;

[tex]\tau[/tex] = torque

F = force exerted

r = lever's arm radius

[tex]\tau[/tex] = [tex]2.00 \times 10^3 \times 0.03 m[/tex]

[tex]\tau[/tex] = 60 N.m

However, equating the torque with the moment of inertia & angular acceleration, we use the equation:

[tex]\tau[/tex] = I∝

60 Nm = I × 120 rad/s²

I = 60 Nm/120 rad/s²

I = 0.5 kg.m²

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