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While a person is walking, his arms swing through approximately a 45\(^\circ\) angle in \(\frac{1}{2}\) s. As a reasonable approximation, assume that the arm moves with constant speed during each swing. A typical arm is 70.0 cm long, measured from the shoulder joint. (a) What is the acceleration of a 1.0-g drop of blood in the fingertips at the bottom of the swing? (b) Draw a free-body diagram of the drop of blood in part (a). (c) Find the force that the blood vessel must exert on the drop of blood in part (a). Which way does this force point? (d) What force would the blood vessel exert if the arm were not swinging?

Short Answer

Expert verified
(a) 1.73 m/sĀ², (b) Centripetal and gravitational forces, (c) 0.00173 N towards shoulder, (d) 0.0098 N upwards.

Step by step solution

01

Convert Angle to Radians

First, we need to convert the angle in degrees to radians, because angular calculations typically use radians.\(\theta = 45^\circ = \frac{45 \times \pi}{180} = \frac{\pi}{4} \text{ radians}\)
02

Calculate Angular Velocity

Using the time and angle from the previous step, calculate the angular velocity. Angular velocity \( \omega \) is calculated using the formula:\( \omega = \frac{\text{angle}}{\text{time}} = \frac{\pi/4}{0.5} = \frac{\pi}{2} \text{ rad/s} \)
03

Calculate Centripetal Acceleration

Centripetal acceleration can be calculated using the formula \( a_c = r \omega^2 \), where \( r \) is the length of the arm.\[ r = 70.0 \text{ cm} = 0.7 \text{ m} \]\( a_c = 0.7 \times \left(\frac{\pi}{2}\right)^2 = 0.7 \times \frac{\pi^2}{4} \approx 1.73 \text{ m/s}^2 \)
04

Draw Free-Body Diagram

In the free-body diagram, depict the forces acting on the drop of blood. The forces include: - The centripetal force towards the center of the swing. - The gravitational force downwards (which is minimal compared to the centripetal).
05

Calculate Force Exerted by Blood Vessel

Calculate the force that the blood vessel needs to exert using Newtonā€™s second law \( F = ma \), where \( m = 1.0 \text{ g} = 0.001 \text{ kg} \).\( F = 0.001 \times 1.73 = 0.00173 \text{ N} \)The force exerted by the blood vessel points in the direction of the centripetal acceleration, i.e., towards the shoulder.
06

Analyze Force Exertion When Arm is Stationary

When the arm is not swinging, the only force acting on the drop of blood is due to gravity. The force is calculated as:\( F_{gravity} = mg = 0.001 \times 9.8 = 0.0098 \text{ N} \).The force exerted by the blood vessel would exactly balance the gravitational force, pointing upwards.

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Key Concepts

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

Angular Velocity
In physics, angular velocity is a measure of how quickly an object rotates or revolves around a central point. Itā€™s often symbolized by the Greek letter \( \omega \). Angular velocity is expressed in radians per second, which aligns with the calculations you would typically perform in physics problems.
To calculate angular velocity, you need to know how many radians an object covers in a given amount of time. Here's the simple formula used:
  • \( \omega = \frac{\text{angle in radians}}{\text{time in seconds}} \).
For example, in our exercise, the arm swings through a 45 degree angle, which was converted into \( \frac{\pi}{4} \) radians. If this swing happens over 0.5 seconds, the angular velocity is:
  • \( \omega = \frac{\pi/4}{0.5} = \frac{\pi}{2} \) rad/s
This gives the measure of how quickly the arm swings.ā€œAngular velocityā€ is crucial for computing centripetal forces in rotational motion.
Centripetal Acceleration
Centripetal acceleration describes the inward acceleration of an object moving in a circular path. This acceleration is always directed towards the center of the circle. It's absolutely essential when considering objects that move in curved, rather than straight, paths.
Centripetal acceleration \( a_c \) is calculated using the square of the angular velocity \( \omega \) and the radius \( r \). Hereā€™s the formula:
  • \( a_c = r \omega^2 \)
In our swing problem, the fingertips travel at a constant radius of 0.7 meters from the shoulder. With an angular velocity of \( \frac{\pi}{2} \) rad/s, the centripetal acceleration becomes:
  • \( a_c = 0.7 \times \left(\frac{\pi}{2}\right)^2 = 1.73 \text{ m/s}^2 \)
This indicates how quickly the blood in the fingertips is being pulled inward during the swing, showing how centripetal forces keep the blood moving in a curved path, following the swing of the arm.
Newton's Second Law
Newton's Second Law states that the force acting on an object is equal to the mass of that object multiplied by its acceleration. This is expressed in the equation \( F = ma \). Itā€™s a central concept in understanding dynamics of motion and forces.
In our exercise, this law helps us determine the force that the blood vessel must exert to maintain the movement of the blood within the swinging arm. With a mass \( m \) of 1.0 gram (converted to 0.001 kg) and a centripetal acceleration \( a \) of 1.73 m/s\(^2\), the force is:
  • \( F = 0.001 \times 1.73 = 0.00173 \text{ N} \)
This computed force tells us how much effort the blood vessel needs to exert in order to counter this circular inward pull. When the arm is stationary, the only force it needs to exert is due to gravity, calculated as \( F = 0.0098 \text{ N} \) pulling downward. Newton's Second Law thus gives us the framework to quantify these essential forces.

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Most popular questions from this chapter

A bowling ball weighing 71.2 N 116.0 lb2 is attached to the ceiling by a 3.80-m rope. The ball is pulled to one side and released; it then swings back and forth as a pendulum. As the rope swings through the vertical, the speed of the bowling ball is 4.20 m/s. At this instant, what are (a) the acceleration of the bowling ball, in magnitude and direction, and (b) the tension in the rope?

Some sliding rocks approach the base of a hill with a speed of 12 m/s. The hill rises at 36\(^\circ\) above the horizontal and has coefficients of kinetic friction and static friction of 0.45 and 0.65, respectively, with these rocks. (a) Find the acceleration of the rocks as they slide up the hill. (b) Once a rock reaches its highest point, will it stay there or slide down the hill? If it stays, show why. If it slides, find its acceleration on the way down.

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A stockroom worker pushes a box with mass 16.8 kg on a horizontal surface with a constant speed of 3.50 m/s. The coefficient of kinetic friction between the box and the surface is 0.20. (a) What horizontal force must the worker apply to maintain the motion? (b) If the force calculated in part (a) is removed, how far does the box slide before coming to rest?

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