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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 motorcycle has a constant speed of 25.0 m/s as it passes over the top of a hill whose radius of curvature is 126 m. The mass of the motorcycle and driver is 342 kg. Find the magnitudes of (a) the centripetal force and (b) the normal force that acts on the cycle.

There is a clever kitchen gadget for drying lettuce leaves after you wash them. It consists of a cylindrical container mounted so that it can be rotated about its axis by turning a hand crank. The outer wall of the cylinder is perforated with small holes. You put the wet leaves in the container and turn the crank to spin off the water. The radius of the container is 12 cm. When the cylinder is rotating at 2.0 revolutions per second, what is the magnitude of the centripetal acceleration at the outer wall?

A rigid massless rod is rotated about one end in a horizontal circle. There is a particle of mass \(m_{1}\) attached to the center of the rod and a particle of mass \(m_{2}\) attached to the outer end of the rod. The inner section of the rod sustains a tension that is three times as great as the tension that the outer section sustains. Find the ratio \(m_{1} / m_{2}\) .

The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the plane of the equator. Assuming the earth is a sphere with a radius of \(6.38 \times 10^{6}\) m, determine the speed and centripetal acceleration of a person situated (a) at the equator and (b) at a latitude of 30.0 north of the equator.

Multiple-Concept Example 7 reviews the concepts that play a role in this problem. Car A uses tires for which the coefficient of static friction is 1.1 on a particular unbanked curve. The maximum speed at which the car can negotiate this curve is 25 m/s. Car B uses tires for which the coefficient of static friction is 0.85 on the same curve. What is the maximum speed at which car B can negotiate the curve?

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