Question

The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2.00 103 N with an effective perpendicular lever arm of 3.40 cm, producing an angular acceleration of the forearm of 165 rad/s2. What is the moment of inertia of the boxer's forearm

Answer 1

Answer: The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2.00 103 N with an effective perpendicular lever arm of 3.40 cm, producing an angular acceleration of the forearm of 165 rad/s2. Then, the moment of inertia of the boxer's forearm will be 0.412Nm/rad/sec2.

Explanation: To find the correct answer, we have to know more about the moment of force or torque.

What is Torque?

• Torque is the measure of turning effect of a force.
• If the object rotates about an axis, then the perpendicular distance from the axis to the line of action of the force is called the lever arm.
• Torque is measured by the product of force and the lever arm.
• If r is the position vector of the point of application of force, then torque T is,

                        $T=rFsin\alpha$, where, $\alpha$ will be the angle between r and F.

• Torque in terms of moment of inertia $I$ and the angular acceleration $\beta$ will be,

                                   $T=I\beta$

                     Where,  $I=r$×m

How to solve the problem?

• Given that,  

                  $F=2*10^3N\\r=3.40*10^-2m\\\beta =165 rad/sec^2$

• From the above equation of T, we can produce the equation of moment of inertia as,

                          $I=\frac{T}{\beta } =\frac{rFsin\alpha }{\beta } \\where, sin\alpha =1.\\Thus,\\T=\frac{rF}{\beta } =\frac{68}{165} =0.412Nm/rad/s^2.$

Thus, we can conclude that, moment of inertia of the boxer's forearm will be, 0.412Nm/rad/sec².

Learn more about the Torque here:

brainly.com/question/28044611

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