/*! 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 56 Two balls of mass \(M=9 \mathrm{... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 56

Two balls of mass \(M=9 \mathrm{~g}\) and \(m=3 \mathrm{~g}\) are attached by massless threads \(A O\) and \(O B\). The length \(A B\) is \(1 \mathrm{~m}\). They are set in rotational motion in a horizontal plane about a vertical axis at \(O\) with constant angular velocity \(\omega\). The ratio of length \(A O\) and \(O B\left(\frac{O B}{A O}\right)\) for which the tension in threads are same will be.

Short Answer

Expert verified
The required ratio for which the tension in the strings is the same is \(\frac{OB}{AO} = 3\).

Step by step solution

01

Understanding the given information

We are given: 1. Mass of ball A: \(M = 9 \mathrm{~g}\) 2. Mass of ball B: \(m = 3 \mathrm{~g}\) 3. Length AB: \(AB = 1 \mathrm{~m}\) 4. Constant angular velocity: \(\omega\) We need to find the ratio \(\frac{OB}{AO}\) for which the tension in both strings is the same.
02

Analyzing forces on balls A and B

As both balls are moving in a horizontal circle, they will experience a centripetal force acting towards the center of the circle. Let's call the tension in strings AO and OB as \(T_A\) and \(T_B\) respectively. Using Newton's second law for both balls: 1. For ball A: \(T_A = M \times a_A = M \times R_A \times \omega^2\) 2. For ball B: \(T_B = m \times a_B = m \times R_B \times \omega^2\) Here, \(a_A\) and \(a_B\) are the centripetal accelerations of balls A and B, while \(R_A\) and \(R_B\) are the radii (lengths) of the strings AO and OB respectively.
03

Finding the required ratio

Since the tension in both strings is the same, we can equate the expressions for \(T_A\) and \(T_B\): \(M \times R_A \times \omega^2 = m \times R_B \times \omega^2\) Now, cancelling the common terms from both sides, we get: \(\frac{M}{m} = \frac{R_B}{R_A}\) Given that the ratio \(\frac{OB}{AO} = \frac{R_B}{R_A}\), we can replace it in the equation: \(\frac{M}{m} = \frac{OB}{AO}\) Finally, plugging in the given values of M and m: \(\frac{9 \mathrm{~g}}{3 \mathrm{~g}} = \frac{OB}{AO}\)
04

Calculate the ratio

To find the ratio, simplify the fraction: \(\frac{OB}{AO} = \frac{9}{3} = 3\) So, the ratio of the lengths is \(\frac{OB}{AO} = 3\).

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Most popular questions from this chapter

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