/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 The triceps muscle in the back o... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(2.00 \times 10^{3} \mathrm{N}\) with an effective perpendicular lever arm of \(3.00 \mathrm{cm},\) producing an angular acceleration of the forearm of \(120 \mathrm{rad} / \mathrm{s}^{2}\). What is the moment of inertia of the boxer's forearm?

Short Answer

Expert verified
The moment of inertia of the boxer's forearm is \(0.5 \, \text{kg} \cdot \text{m}^{2}\).

Step by step solution

01

Understanding Torque

Torque, represented by \( \tau \), is the rotational equivalent of linear force. It measures how much a force acting on an object causes that object to rotate. The torque can be calculated by the formula \( \tau = F \cdot r \) where \( F \) is the force applied, and \( r \) is the lever arm or the perpendicular distance from the axis of rotation to the point where the force is applied.
02

Converting Units

Make sure that all the units are consistent. In this exercise, the lever arm is given in centimeters, so it needs to be converted to meters (the standard SI unit for distance in physics). Converting \( 3.00 \, \text{cm} \) to meters, we get \( r = 3.00 \, \text{cm} = 0.0300 \, \text{m} \).
03

Calculating the Torque

Now, calculate the torque using the force and the lever arm. Use the formula \( \tau = F \cdot r \). The force \(F\) is \(2.00 \times 10^{3} \, \text{N}\) and the converted lever arm \( r \) is \(0.0300 \, \text{m} \). Plugging in the values, we get \( \tau = 2.00 \times 10^{3} \, \text{N} \cdot 0.0300 \, \text{m} \) which results in \( \tau = 60 \, \text{N} \cdot \text{m} \).
04

Understanding the Relationship Between Torque and Angular Acceleration

The relationship between torque, angular acceleration \( (\alpha) \) and the moment of inertia \( (I) \) is given by Newton's second law for rotation: \( \tau = I \cdot \alpha \). Angular acceleration \( \alpha \) is given as \(120 \, \text{rad}/\text{s}^{2}\).
05

Calculating the Moment of Inertia

Rearrange the torque equation to solve for the moment of inertia: \( I = \frac{\tau}{\alpha} \). Substituting in the values for torque \( (60 \, \text{N} \cdot \text{m}) \) and angular acceleration \( (120 \, \text{rad}/\text{s}^{2})\) we get \( I = \frac{60}{120} \) which simplifies to \( I = 0.5 \, \text{kg} \cdot \text{m}^{2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Torque
Torque, denoted as \( \tau \), plays a crucial role in rotational dynamics, similar to how force is essential in linear motion. Envision torque as a twist or a turn that can cause an object to rotate around an axis. The greater the force or the longer the lever arm, the higher the torque, and consequently, the greater the object's rotational change.

To calculate torque, we use the equation \( \tau = F \cdot r \), where \( F \) is the force applied perpendicular to the lever arm, and \( r \) represents the lever arm's length. It's essential to keep units consistent, typically converting them to the metric system. In our example, a professional boxer's triceps muscle generates a torque calculated by converting the lever arm to meters and multiplying by the force exerted, yielding \( \tau = 60 \, \text{N} \cdot \text{m} \). This torque acts as the rotational force enabling the boxer's forearm to accelerate.
Exploring Angular Acceleration
Angular acceleration, symbolized by \( \alpha \), describes how quickly an object's rotational speed changes. It's analogous to linear acceleration in straight-line motion but specifically applies to rotation. The unit for angular acceleration is \( \text{rad}/\text{s}^{2} \), indicating a change in rotational velocity over time.

When the boxer applies a torque on the forearm, it results in an angular acceleration, which is a measure of how swiftly the forearm's angular velocity increases. In our context, the boxer's forearm experiences an angular acceleration of \( 120 \, \text{rad}/\text{s}^{2} \), a direct consequence of the exerted torque. The link between torque and angular acceleration is defined by Newton's second law for rotation, \( \tau = I \cdot \alpha \), which sets the foundation for determining the moment of inertia—denoted \( I \)—of an object.
The Lever Arm Effect
The lever arm is essentially the distance from the axis of rotation to the line of action of the force. More specifically, it is the shortest perpendicular distance from the axis to the force vector. This concept is a vital aspect of calculating torque because it determines the torque's magnitude based on the position where the force is applied.

In the boxer's scenario, the lever arm is remarkably short, at just 3.00 cm, or \(0.0300\) m after conversion from centimeters to meters. A longer lever arm would create a greater torque if the same force were applied. The lever arm length is crucial; even a powerful force can produce minimal torque with a short lever arm, while a moderate force can generate substantial torque with a long lever arm. Understanding the influence of the lever arm helps us visualize how the boxer is able to extend the forearm rapidly by strategically applying force through the triceps muscle.

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Most popular questions from this chapter

Show that if two resistors \(R_{1}\) and \(R_{2}\) are combined and one is much greater than the other \(\left(R_{1}>>R_{2}\right):\) (a) Their series resistance is very nearly equal to the greater resistance \(R_{1}\). (b) Their parallel resistance is very nearly equal to smaller resistance \(R_{2}\)

While exercising in a fitness center, a man lies face down on a bench and lifts a weight with one lower leg by contacting the muscles in the back of the upper leg. (a) Find the angular acceleration produced given the mass lifted is \(10.0 \mathrm{kg}\) at a distance of \(28.0 \mathrm{cm}\) from the knee joint, the moment of inertia of the lower leg is \(0.900 \mathrm{kg} \cdot \mathrm{m}^{2}\), the muscle force is 1500 N, and its effective perpendicular lever arm is \(3.00 \mathrm{cm}\). (b) How much work is done if the leg rotates through an angle of \(20.0^{\circ}\) with a constant force exerted by the muscle?

Everyday application: Suppose a yo-yo has a center shaft that has a \(0.250 \mathrm{cm}\) radius and that its string is being pulled. (a) If the string is stationary and the yo-yo accelerates away from it at a rate of \(1.50 \mathrm{m} / \mathrm{s}^{2},\) what is the angular acceleration of the yo-yo? (b) What is the angular velocity after 0.750 s if it starts from rest? (c) The outside radius of the yo-yo is \(3.50 \mathrm{cm}\). What is the tangential acceleration of a point on its edge?

What is the final velocity of a hoop that rolls without slipping down a 5.00-m-high hill, starting from rest?

Regarding the units involved in the relationship \(\tau=R C\), verify that the units of resistance times capacitance are time, that is. \(\Omega \cdot \mathrm{F}=\mathrm{s}\)

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