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Chapter 5: Problem 549

A body of mass \(\mathrm{m}\) is tied to one end of spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is one centimetre. If the angular velocity is doubted, the elongation in the spring is \(5 \mathrm{~cm}\). The original length of spring is... \(\\{\mathrm{A}\\} 16 \mathrm{~cm}\) \(\\{B\\} 15 \mathrm{~cm}\) \(\\{\mathrm{C}\\} 14 \mathrm{~cm}\) \(\\{\mathrm{D}\\} 13 \mathrm{~cm}\)

Short Answer

Expert verified
The original length of the spring is \(\frac{1}{3} \mathrm{~cm}\).

Step by step solution

01

Write down the centripetal force equation

The centripetal force acting on the body is given by the equation: $$F_c = m \omega^2 r$$ where \(F_c\) is the centripetal force, \(m\) is the mass of the body, \(\omega\) is the angular velocity, and \(r\) is the distance from the center of rotation to the body.
02

Write down the Hooke's law equation

The force exerted by the spring is given by Hooke's law: $$F_s = kx$$ where \(F_s\) is the force exerted by the spring, \(k\) is the spring constant, and \(x\) is the elongation of the spring.
03

Setup the relationship between centripetal force and Hooke's law

Since the centripetal force and the Hooke's law for the spring force should be equal: $$m \omega^2 r = kx$$ The relationship between \(kx\) and \(\omega^2 r\) can also be written as: $$k(x_0 + x) = m \omega^2 r$$ In both scenarios, we are given the elongations of the spring \((x)\) and the angular velocities \((\omega)\). With this information, we can create equations for both scenarios.
04

Write equations for both scenarios

In the first scenario: 1. Elongation \(x_1 = 1 \mathrm{~cm}\) 2. Angular velocity \(\omega_1\) In the second scenario: 1. Elongation \(x_2 = 5 \mathrm{~cm}\) 2. Angular velocity \(\omega_2 = 2\omega_1\) We can write the equations for both scenarios: 1. \(k(x_0 + 1) = m \omega_1^2 r\) 2. \(k(x_0 + 5) = m (2\omega_1)^2 r\)
05

Eliminate \(k\) and solve for \(x_0\)

Divide equation 2 by equation 1: $$\frac{x_0 + 5}{x_0 + 1} = \frac{m (2\omega_1)^2 r}{m \omega_1^2 r}$$ Simplifying the equation: $$\frac{x_0 + 5}{x_0 + 1} = 4$$ Cross-multiply to solve for \(x_0\): $$4(x_0 + 1) = x_0 + 5$$ $$4x_0 + 4 = x_0 + 5$$ $$3x_0 = 1$$ $$x_0 = \frac{1}{3} \mathrm{~cm}$$
06

Find the original length of the spring

We are given that the elongation \(x_1\) in the first scenario is 1cm. To find the original length of the spring, we subtract the elongation from the total length: Original Length of Spring = \((x_0 + x_1) - x_1\) $$= \frac{1}{3} + 1 - 1$$ $$= \frac{1}{3} \mathrm{~cm}$$ The original length of the spring is not among the given options A, B, C, and D. There must be an error or typo in the problem statement or the given choices.

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Most popular questions from this chapter

A car is moving at a speed of \(72 \mathrm{~km} / \mathrm{hr}\) the radius of its wheel is \(0.25 \mathrm{~m}\). If the wheels are stopped in 20 rotations after applying breaks then angular retardation produced by the breaks is \(\ldots .\) \(\\{\mathrm{A}\\}-25.5 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{B}\\}-29.52 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{C}\\}-33.52 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{D}\\}-45.52 \mathrm{rad} / \mathrm{s}^{2}\)

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