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Understanding Circumference and Rotational Motion in Disk Dynamics

January 06, 2025Anime1235
Introduction to Disk Rotation and Circumference Understanding the rela

Introduction to Disk Rotation and Circumference

Understanding the relationship between the radius of a disk, its rotation rate, and the distance traveled by a point on its rim is essential in various scientific and engineering applications. In this article, we will explore how to calculate the distance traveled by a point on the outside rim of a disk during a specified period of time.

The Problem Statement

A disk of radius 50 cm rotates at a constant rate of 100 RPM (revolutions per minute). We need to determine the distance in meters that a point on the outside rim of the disk will travel during a 30-second rotation period.

Calculating the Circumference

The circumference of a circle is given by the formula ( C 2pi R ), where ( R ) is the radius of the circle. Given that the radius ( R 50 ) cm, the circumference of the disk is:

[ C 2pi times 50 approx 314 text{ cm} ]

Calculating Rotations and Distance Traveled

In 30 seconds, the disk rotates half a minute, which means it completes:

[ text{Rotations} frac{100 text{ RPM}}{2} 50 text{ rotations} ]

Therefore, the distance traveled by a point on the rim during this time can be calculated by multiplying the number of rotations by the circumference:

[ text{Distance} 50 times 314 approx 15700 text{ cm} ]

Converting this distance to meters:

[ text{Distance} frac{15700 text{ cm}}{100} 157 text{ meters} ]

Confirming the Calculation

The calculation can be verified as follows:

[ text{Distance} 50 times 2pi times 50 approx 5000pi text{ cm} ]

Converting cm to meters:

[ text{Distance} approx frac{5000pi}{100} 157.08 text{ meters} ]

This confirms that the point on the rim of the disk will travel approximately 157.08 meters.

Summary

Through the detailed calculation, we have determined that the distance traveled by a point on the outside rim of a disk with a radius of 50 cm, rotating at 100 RPM, during a 30-second period is approximately 157 meters. This problem showcases the practical application of both geometry and basic physics principles.

Note: The calculations are rounded to two decimal places for simplicity.