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 X 10^3 N with an effective perpendicular lever arm of 3.10 cm, producing an angular acceleration of the forearm of 125 rad/s2. What is the moment of inertia of the boxer's forearm? kg · m2

Answer 1

Answer:

$0.496 kg m^2$

Explanation:

The torque exerted is given by

$\tau = Fd$

where

$F=2.00 \cdot 10^3 N$ is the force applied

d = 3.10 cm = 0.031 m is the length of the lever arm

Substituting,

$\tau=(2.00\cdot 10^3 N)(0.031 m)=62 Nm$

The equivalent of Newton's second law for rotational motion is:

$\tau = I \alpha$

where

$\tau = 62 Nm$ is the net torque

I is the moment of inertia

$\alpha = 125 rad/s^2$ is the angular acceleration

Solving the equation for I, we find

$I=\frac{\tau}{\alpha}=\frac{62 Nm}{125 rad/s^2}=0.496 kg m^2$