Question

with what angular speed, in rad/s, does the earth rotate about its axis?
rad/s
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1. -/125 points

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 2.00 cm, producing an angular acceleration of the forearm of 105 rad/s². what is the moment of inertia (in kg m²) of the boxers forearm?
kg m²

Explanation:

Step1: Calculate the angular speed of the Earth

The Earth makes one full rotation (\(\theta = 2\pi\) radians) in \(T=24\times3600\ s\).
The formula for angular speed \(\omega=\frac{\theta}{T}\).
Substitute \(\theta = 2\pi\) and \(T = 24\times3600\):
\(\omega=\frac{2\pi}{24\times3600}\)
\(\omega=\frac{2\times3.14159}{24\times3600}\)
\(\omega=\frac{6.28318}{86400}\approx7.27\times 10^{-5}\ rad/s\)

Step2: Calculate the moment of inertia of the boxer's forearm

The torque \(\tau = rF\), where \(r = 2.00\ cm=0.02\ m\) and \(F = 2.00\times 10^{3}\ N\). So \(\tau=(0.02)(2.00\times 10^{3}) = 40\ N\cdot m\)
From the rotational form of Newton's second law \(\tau = I\alpha\), where \(\alpha=105\ rad/s^{2}\)
We can solve for \(I\): \(I=\frac{\tau}{\alpha}\)
Substitute \(\tau = 40\ N\cdot m\) and \(\alpha = 105\ rad/s^{2}\): \(I=\frac{40}{105}\approx0.381\ kg\cdot m^{2}\)

Answer:

The angular speed of the Earth is approximately \(7.27\times 10^{-5}\ rad/s\). The moment of inertia of the boxer's forearm is approximately \(0.381\ kg\cdot m^{2}\)