Math Problem Statement

A rope breaks when the tension reaches 305 N. What is the largest mass it can swing at 14.2 m/s in a circle of radius 0.425 m?

Solution

To solve this problem, we need to calculate the maximum mass (mm) that the rope can swing without breaking. We are given:

  • Maximum tension the rope can handle, T=305NT = 305 \, \text{N}.
  • Speed of the mass in circular motion, v=14.2m/sv = 14.2 \, \text{m/s}.
  • Radius of the circle, r=0.425mr = 0.425 \, \text{m}.

Since the mass is moving in a circular path, the tension in the rope provides the centripetal force required to keep the mass moving in a circle. The centripetal force FcF_c needed for circular motion is given by:

Fc=mv2rF_c = \frac{mv^2}{r}

Here, the tension TT in the rope is equal to the centripetal force FcF_c, so:

T=mv2rT = \frac{mv^2}{r}

Rearranging to solve for mm:

m=Trv2m = \frac{T \cdot r}{v^2}

Substituting the values:

m=305×0.42514.22m = \frac{305 \times 0.425}{14.2^2}

Calculating the denominator first:

14.22=201.6414.2^2 = 201.64

Then, the mass mm:

m=305×0.425201.64m = \frac{305 \times 0.425}{201.64}

m129.125201.640.64kgm \approx \frac{129.125}{201.64} \approx 0.64 \, \text{kg}

Answer:

The largest mass the rope can swing at 14.2 m/s in a circle of radius 0.425 m without breaking is approximately 0.64 kg.

Would you like a deeper explanation, or do you have any questions?


Here are some additional related questions you might find helpful:

  1. What happens to the maximum mass if the speed increases to 20 m/s?
  2. How does the radius of the circle affect the maximum allowable mass?
  3. If the rope's breaking tension was increased to 500 N, what would be the new maximum mass?
  4. How would you calculate the maximum tension for a different rope length?
  5. How does the concept of centripetal force apply to other circular motion scenarios, like cars on curved roads?

Tip: In circular motion problems, remember that increasing speed significantly increases the required centripetal force because it’s proportional to the square of the velocity (v2v^2).

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Math Problem Analysis

Mathematical Concepts

Centripetal Force
Tension in Circular Motion
Newton's Second Law

Formulas

Centripetal force formula: F_c = mv^2 / r
Rearranged formula for mass: m = T * r / v^2

Theorems

Newton's Second Law of Motion

Suitable Grade Level

Grades 10-12