/*! 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 16 A speed skater goes around a tur... [FREE SOLUTION] | 91Ó°ÊÓ

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A speed skater goes around a turn that has a radius of 31 m. The skater has a speed of \(14 \mathrm{m} / \mathrm{s}\) and experiences a centripetal force of \(460 \mathrm{N}\). What is the mass of the skater?

Short Answer

Expert verified
The mass of the skater is approximately 72.6 kg.

Step by step solution

01

Understand the Problem

We need to find the mass of a speed skater who is taking a turn with a given radius while experiencing a certain centripetal force.
02

Recall the Formula for Centripetal Force

The formula for the centripetal force is given by \( F = \frac{mv^2}{r} \), where \( F \) is the centripetal force, \( m \) is the mass, \( v \) is the speed, and \( r \) is the radius of the turn.
03

Rearrange the Formula to Solve for Mass

We need to solve for the mass \( m \). Rearranging the formula gives \( m = \frac{Fr}{v^2} \).
04

Substitute Known Values into the Formula

Substitute the given values into the rearranged formula: \( F = 460 \text{ N} \), \( r = 31 \text{ m} \), and \( v = 14 \text{ m/s} \). This becomes \( m = \frac{460 \times 31}{14^2} \).
05

Calculate the Mass

Calculating the expression, we find that \( m = \frac{460 \times 31}{196} \approx 72.6 \text{ kg} \).

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Key Concepts

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

Mass Calculation
Understanding how to calculate mass is an essential skill when solving physics problems involving motion. In the context of centripetal force, mass calculation becomes necessary if you know other parameters like force, velocity, and radius.
In our specific problem, we have a speed skater experiencing a centripetal force while moving in a circle. To find the skater's mass, we use the formula for centripetal force, \( F = \frac{mv^2}{r} \).
By rearranging it, we isolate the mass \( m \), obtaining \( m = \frac{Fr}{v^2} \).
Here:
  • \( F \) is the centripetal force acting on the body.
  • \( v \) is the velocity or speed.
  • \( r \) is the radius of the circle.
This formula helps us directly calculate mass from the given values, making it straightforward to understand how changes in speed, force, or the circular path radius can influence the mass determination.
Circular Motion
Circular motion is a fascinating concept seen in everyday life, from cars turning around a corner to planets orbiting stars. Understanding it involves various aspects, including speed, radius, and centripetal force.
When an object moves in a circle, it constantly changes direction, even if its speed remains constant. This change in direction is due to centripetal force, which acts towards the center of the circle, keeping the object in its path. For instance, our speed skater maintains circular motion due to the centripetal force applied towards the center of their turn.
Important aspects when considering circular motion:
  • Radius \( r \): The distance from the center to the path of motion.
  • Velocity \( v \): The speed of the object along the path.
  • Centripetal Force \( F \): The net force causing the inward acceleration.
Understanding these factors and how they interrelate supports broader comprehension of any object following a curved path.
Physics Problem
Solving physics problems can sometimes feel tricky but becomes manageable with a structured approach. Breaking down a problem into digestible parts, like identifying what is known and what you need to find, is essential.
In our exercise, we dissect the situation by:
  • Identifying the given values: radius \( r = 31 \) meters, speed \( v = 14 \) meters per second, and centripetal force \( F = 460 \) Newtons.
  • Recognizing the wanted value: the mass of the skater.
  • Using the appropriate formula: for centripetal force to express mass in terms of accessible variables.
  • Executing the calculation: substituting the known values into the rearranged formula \( m = \frac{Fr}{v^2} \).
This structured method enhances accuracy and confidence in tackling physics problems. As we solve the particular problem, we can see how fundamental each step is for reaching the correct answer of approximately 72.6 kg for the skater's mass.

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Most popular questions from this chapter

A stone is tied to a string (length \(=1.10 \mathrm{m}\) ) and whirled in a circle at the same constant speed in two different ways. First, the circle is horizontal and the string is nearly parallel to the ground. Next, the circle is vertical. In the vertical case the maximum tension in the string is \(15.0 \%\) larger than the tension that exists when the circle is horizontal. Determine the speed of the stone.

In a skating stunt known as crack-the-whip, a number of skaters hold hands and form a straight line. They try to skate so that the line rotates about the skater at one end, who acts as the pivot. The skater farthest out has a mass of \(80.0 \mathrm{kg}\) and is \(6.10 \mathrm{m}\) from the pivot. He is skating at a speed of \(6.80 \mathrm{m} / \mathrm{s}\). Determine the magnitude of the centripetal force that acts on him.

A car is safely negotiating an unbanked circular turn at a speed of \(21 \mathrm{m} / \mathrm{s} .\) The road is dry, and the maximum static frictional force acts on the tires. Suddenly a long wet patch in the road decreases the maximum static frictional force to one-third of its dry-road value. If the car is to continue safely around the curve, to what speed must the driver slow the car?

An \(830-\mathrm{kg}\) race car can drive around an unbanked turn at a maximum speed of \(58 \mathrm{m} / \mathrm{s}\) without slipping. The turn has a radius of curvature of \(160 \mathrm{m}\). Air flowing over the car's wing exerts a downward-pointing force (called the downforce) of \(11000 \mathrm{N}\) on the car. (a) What is the coefficient of static friction between the track and the car's tires? (b) What would be the maximum speed if no downforce acted on the car?

A car travels at a constant speed around a circular track whose radius is \(2.6 \mathrm{km}\). The car goes once around the track in \(360 \mathrm{s}\). What is the magnitude of the centripetal acceleration of the car?

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