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At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats facing the axis, their backs against the outer wall. At one instant the outer wall moves at a speed of 3.2 m/s, and an 83-kg person feels a 560-N force pressing against his back. What is the radius of the chamber?

Short Answer

Expert verified
The radius of the chamber is approximately 1.52 meters.

Step by step solution

01

Identify the Given Variables and Forces

We are given two key pieces of information: the speed of the outer wall, which is 3.2 m/s, and the force felt by the person, which is 560 N. The mass of the person is 83 kg. The force pressing against the person is the centripetal force, which maintains the circular motion.
02

Apply the Centripetal Force Formula

The formula for centripetal force is given by \( F_c = \frac{mv^2}{r} \), where \( F_c \) is the centripetal force, \( m \) is the mass, \( v \) is the speed, and \( r \) is the radius of the circular path. Here, \( F_c = 560 \) N, \( m = 83 \) kg, and \( v = 3.2 \) m/s.
03

Rearrange the Formula to Solve for Radius

We rearrange the centripetal force formula to solve for \( r \):\[ r = \frac{mv^2}{F_c} \]Substitute the given values into the formula: \( m = 83 \), \( v = 3.2 \), and \( F_c = 560 \).
04

Calculate the Radius

By substituting the values in, we find:\[ r = \frac{83 \times (3.2)^2}{560} \]Calculate \( 3.2^2 = 10.24 \), then \( 83 \times 10.24 = 849.92 \), and finally, divide by \( 560 \): \[ r = \frac{849.92}{560} \approx 1.518 \] Therefore, the radius of the chamber is approximately 1.52 meters.

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Key Concepts

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

Circular Motion
Circular motion occurs when an object moves in a circular path at a constant speed. In our amusement park ride scenario, the passengers move in a circle around a central axis. This continuous circular movement requires a constant inward force known as centripetal force.

It's important to understand that while speed remains constant in uniform circular motion, velocity does not. The velocity is a vector, meaning it has both magnitude and direction. In a circle, the direction continuously changes even if the speed is constant.

To maintain this circular path, the centripetal force must constantly act perpendicular to the velocity of the object. This force acts inward, towards the center of the circle, ensuring the path remains circular. Without this force, the object would simply move off in a straight line due to inertia.
Centripetal Acceleration
Centripetal acceleration is directly related to circular motion. When an object undergoes circular motion, it experiences acceleration towards the center of the circle, this is centripetal acceleration.

Mathematically, centripetal acceleration is determined using the formula:
  • \[ a_c = \frac{v^2}{r} \]
where \( a_c \) is the centripetal acceleration, \( v \) is the velocity, and \( r \) is the radius of the circular path. Even though the speed of the object is constant, centripetal acceleration changes velocity direction, not its magnitude.

This acceleration is crucial for keeping the object in a circular path. It ensures the force we calculate as the centripetal force exactly matches the requirements to keep this exact radius and speed. Think of it as the necessary tweak to maintain perfect circular motion.
Physics Problem Solving
Solving physics problems can be much easier when certain systematic approaches are applied. Let's dive deeper into solving circular motion problems like our chamber ride example.

Firstly, identify given variables and unknowns. Knowing the speed, force, and mass gives us a clear path. Next, apply a relevant formula—in this case, the centripetal force formula:\[ F_c = \frac{mv^2}{r} \]\( F_c \) is the centripetal force, \( m \) is the mass, \( v \) is the velocity, and \( r \) is the radius we want.

Rearranging the formula can help find unknowns. Rearrange to solve for \( r \) and substitute your known values. Calculations require care to avoid common mistakes. Step-by-step checking can reveal errors. Finally, review solutions in the context of the problem, ensuring they fit realistic expectations like a positive radius.

Developing these skills allows you to tackle physics problems of varying complexities with confidence.

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Most popular questions from this chapter

A computer is reading data from a rotating CD-ROM. At a point that is 0.030 m from the center of the disc, the centripetal acceleration is 120 \(\mathrm{m} / \mathrm{s}^{2}\) What is the centripetal acceleration at a point that is 0.050 m from the center of the disc?

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