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Chapter 10: Problem 1387

A spring is attached to the center of a frictionless horizontal turn table and at the other end a body of mass \(2 \mathrm{~kg}\) is attached. The length of the spring is \(35 \mathrm{~cm}\). Now when the turn table is rotated with an angular speed of \(10 \mathrm{rad} \mathrm{s}^{-1}\), the length of the spring becomes \(40 \mathrm{~cm}\) then the force constant of the spring is $\ldots \ldots \mathrm{N} / \mathrm{m}$. (A) \(1.2 \times 10^{3}\) (B) \(1.6 \times 10^{3}\) (C) \(2.2 \times 10^{3}\) (D) \(2.6 \times 10^{3}\)

Short Answer

Expert verified
The force constant of the spring is \(1.6 \times 10^3 N/m\), which corresponds to option (B).

Step by step solution

01

Calculate the centripetal force of the rotating body

To calculate the centripetal force, \(F_c\), we use the formula: \[F_c = m \cdot \omega^2 \cdot r\] where - \(m = 2 kg\) = mass of the body attached to the spring end - \(\omega = 10 rad/s\) = angular speed of the turntable - \(r = 0.4 m\) = stretched length of the spring (converted from 40 cm to meters by multiplying by 0.01) Plugging these values into the formula, we get: \(F_c = 2 \times (10)^2 \times 0.4\)
02

Apply Hooke's Law to find spring force

Hooke's Law states that the force exerted by a spring, \(F_s\), is proportional to its elongation/extension, \(\Delta x\): \[F_s = k \cdot \Delta x\] where - \(k\) = force constant of the spring (which we want to find) - \(\Delta x = 0.4 - 0.35 = 0.05 m\) = extension of the spring when the turntable is rotating (converted lengths to meters)
03

Relate the centripetal force to spring force

Since there's no other external force in the horizontal plane, the centripetal force equals the spring force: \[F_c = F_s\] So we can write the equation as: \[m \cdot \omega^2 \cdot r = k \cdot \Delta x\]
04

Solve for the force constant of the spring

Now we can plug in all the known values and solve for \(k\): \(2 \times (10)^2 \times 0.4 = k \times 0.05\) Solving for \(k\), we get: \[k = \frac{2 \times (10)^2 \times 0.4}{0.05} = 1.6 \times 10^3 N/m\] So the force constant of the spring is \(1.6 \times 10^3 N/m\), which corresponds to option (B).

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