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In a test of a "g-suit," a volunteer is rotated in a horizontal circle of radius 7.0 \(\mathrm{m}\) . What must the period of rotation be so that the centripetal ncceleration has a magnitude of \((a) 3.0 g ?^{7}(b) 10 g ?\)

Short Answer

Expert verified
For 3.0g, the period is approximately 2.44 s; for 10g, the period is approximately 1.34 s.

Step by step solution

01

Identify Given Data and Needed Formula

The radius of the circle is given as 7.0 m. We need to find the period (T) of rotation for different centripetal accelerations. The formula for centripetal acceleration is \( a_c = \frac{4\pi^2r}{T^2} \), where \( a_c \) is the centripetal acceleration, \( r \) is the radius (7.0 m), and \( T \) is the period of rotation.
02

Case (a): Calculate the Period for 3.0g

First, convert 3.0g to meters per second squared. Since \( 1g = 9.8 \text{ m/s}^2 \), \( 3.0g = 3.0 \times 9.8 \text{ m/s}^2 = 29.4 \text{ m/s}^2 \). Substitute \( a_c = 29.4 \text{ m/s}^2 \) into the formula: \( 29.4 = \frac{4\pi^2 \times 7.0}{T^2} \). Solve for \( T \): \( T^2 = \frac{4\pi^2 \times 7.0}{29.4} \), then \( T = \sqrt{\frac{4\pi^2 \times 7.0}{29.4}} \).
03

Solve Equation for Case (a)

Calculate the numerical value for \( T \):\[T = \sqrt{\frac{4 \times 3.1416^2 \times 7.0}{29.4}} \approx \sqrt{5.978} \approx 2.44 \text{ s}\]
04

Case (b): Calculate the Period for 10g

First, convert 10g to meters per second squared. Thus, \( 10g = 10 \times 9.8 \text{ m/s}^2 = 98 \text{ m/s}^2 \). Substitute \( a_c = 98 \text{ m/s}^2 \) into the formula: \( 98 = \frac{4\pi^2 \times 7.0}{T^2} \). Solve for \( T \): \( T^2 = \frac{4\pi^2 \times 7.0}{98} \), then \( T = \sqrt{\frac{4\pi^2 \times 7.0}{98}} \).
05

Solve Equation for Case (b)

Calculate the numerical value for \( T \):\[T = \sqrt{\frac{4 \times 3.1416^2 \times 7.0}{98}} \approx \sqrt{1.793} \approx 1.34 \text{ s}\]

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Key Concepts

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

Centripetal Force
Centripetal force is the force required to keep an object moving in a circular path and is directed towards the center of the circle. This force is not a new type of force, but rather the result of other forces that act perpendicular to the direction of motion.
  • For example, in the case of a car taking a curved path, the friction between the tires and the road provides the centripetal force.
  • In orbit, gravity acts as the centripetal force that keeps a satellite revolving around a planet.
Centripetal force is always changing the direction of the object towards the center, causing a continuous change in velocity, which we perceive as circular motion.
The formula for centripetal force (F_c) is given by:\[F_c = \frac{mv^2}{r}\]where:
  • \( m \) is the mass of the object.
  • \( v \) is the velocity.
  • \( r \) is the radius of the circle.
This formula shows how the centripetal force is reliant on the object's mass, speed, and the radius of the circular path. Greater speed or mass, or a smaller radius, requires a larger centripetal force to maintain the circular motion.
Circular Motion
Objects moving in a circle at constant speed undergo circular motion. This motion is characterized not by a change in the speed of the object, but by a continuous change in the direction of the object's velocity.

One of the key features of circular motion is centripetal acceleration, which points towards the center of the circle. It keeps altering the direction of the velocity to ensure the object remains in its circular path. Though the speed remains constant, the object's velocity is constantly changing because velocity is a vector quantity that depends on both speed and direction. Some important aspects of circular motion include:
  • The radius of the circular path, which is the constant distance from the center to the moving object.
  • Period (T), which is the time it takes to complete one full circle.
  • Frequency, the number of complete circles per unit time, which is the reciprocal of the period.
Understanding circular motion is crucial in studying various phenomena such as planetary orbits, rotating machinery, and amusement park rides.
Formulating the centripetal acceleration (ac) helps to solve physics problems involving circular motion:\[a_c = \frac{v^2}{r}\]
Physics Problems
Physics problems involving circular motion, such as the "g-suit" test, often challenge students to apply mathematical equations to calculate unknown values like period (T) or centripetal acceleration (a_c).

Approaching these problems effectively requires a clear understanding of the relationships between different circular motion properties. For example, knowing that centripetal acceleration (a_c) can be expressed as:\[a_c = \frac{4\pi^2r}{T^2}\]can help solve for the period when the radius (r) and centripetal acceleration are provided.
  • Convert all given values to their standard units; for acceleration, use meters per second squared (\text{m/s}^2).
  • Substitute the known values into the appropriate formula.
  • Solve algebraically for the unknown quantity.
Common mistakes include neglecting unit conversions or misunderstanding the relationships between formula components. Practice and careful attention to detail can significantly enhance problem-solving skills. Understanding these physics problems promotes a deeper appreciation for real-world applications of science principles, such as designing safety equipment and understanding the workings of different technologies.

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