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Chapter 9: Problem 10

One end of a massless spring of spring constant \(100 \mathrm{~N} / \mathrm{m}\) and natural length \(0.5 \mathrm{~m}\) is fixed and the other end is connected to a particle of mass \(0.5 \mathrm{~kg}\) lying on a frictionless horizontal table. The spring remains horizontal. If the mass is made to rotate at an angular velocity of \(2 \mathrm{rad} / \mathrm{s}\) find the elongation of the spring (in \(\mathrm{cm}\) ).

Short Answer

Expert verified
The spring extends by approximately 1.02 cm.

Step by step solution

01

Understand the Problem

We need to find the elongation of the spring when a mass is attached at one end and the system is rotating at an angular velocity of \(2 \mathrm{rad} / \mathrm{s}\). The spring constant is \(100 \mathrm{~N} / \mathrm{m}\), and the mass of the particle is \(0.5 \mathrm{~kg}\).
02

Write the Forces Involved

When the mass rotates, a centrifugal force \(F_c\) acts outward on the mass due to its circular motion. This force must balance the restoring force from the spring's elongation. The centrifugal force is given by \(F_c = m\omega^2 r\), where \(m\) is the mass, \(\omega\) is the angular velocity, and \(r\) is the radius (or the current length of the spring including elongation).
03

Relate Forces with Spring Elongation

The elongation \(x\) of the spring produces a restoring force \(F_s\) given by Hooke's Law: \(F_s = kx\), where \(k\) is the spring constant. For equilibrium (or a rotating steady state), \(F_c = F_s\).
04

Set Up the Equation

Using \(F_c = F_s\), we have \(m\omega^2 r = kx\). As \(r = l_0 + x\) (since \(r\) is the extended length and \(l_0\) is the natural length of the spring), substitute into the equation: \[ m\omega^2(l_0 + x) = kx \]
05

Solve for Elongation \(x\)

Insert the given values: \(m = 0.5\) kg, \(\omega = 2\) rad/s, \(l_0 = 0.5\) m, and \(k = 100\) N/m: \[ 0.5 \times 2^2 \times (0.5 + x) = 100x \] This simplifies to: \[ 2(0.5 + x) = 100x \] \[ 1 + 2x = 100x \] Solving for \(x\) gives: \[ 1 = 98x \] \[ x = \frac{1}{98} \text{ m} \]
06

Convert Units

The elongation \(x\) found is in meters. Convert it to centimeters: \[ x = \frac{1}{98} \times 100 \text{ cm} = \frac{100}{98} \text{ cm} \approx 1.02 \text{ cm} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Centrifugal Force
Centrifugal force is a concept that describes the apparent force that draws a rotating object away from the center of rotation. It arises in a rotating reference frame and acts outwardly as if pulling the object in a straight line. Imagine you're on a merry-go-round: as it spins faster, you feel like you鈥檙e being pushed outward from the center. That's centrifugal force at play. This force is essential in understanding scenarios involving rotation, like a mass attached to a spinning spring. For a particle of mass \( m \) rotating at angular velocity \( \omega \), centrifugal force \( F_c \) is given by:
  • \( F_c = m \omega^2 r \)
Where \( r \) is the radius or the distance from the center of rotation to the object. In this problem, as the mass rotates, it generates a centrifugal force which causes the spring to stretch outward.
Hooke's Law
Hooke's Law describes the behavior of springs and other elastic materials when they are stretched or compressed. When a spring is stretched by a force, it tries to return to its original length. This attempt to return to its natural state creates a restoring force, which is proportional to the elongation of the spring.The relationship defined by Hooke's Law can be described by the equation:
  • \( F_s = kx \)
Where:
  • \( F_s \) is the restoring force exerted by the spring.
  • \( k \) is the spring constant, measuring how stiff the spring is.
  • \( x \) is the elongation or compression of the spring from its natural length.
In the given problem, the spring elongates until the restoring force equals the centrifugal force, maintaining a balance during rotation.
Angular Velocity
Angular velocity is a measure of how quickly an object rotates around a point or axis. It provides the rotational speed and direction of the rotation. For instance, if an object completes one full rotation every second, it has a higher angular velocity than one that does so every two seconds.Angular velocity \( \omega \) is expressed in radians per second (rad/s), and you can think of it like a car's speedometer but for rotating objects. It's crucial in problems involving rotational motion because it helps determine the forces acting on the system, such as centrifugal force.In our scenario:
  • The mass rotates at \( \omega = 2 \text{ rad/s} \).
  • This specific angular velocity determines the magnitude of the centrifugal force based on the radius of rotation.
Understanding angular velocity allows us to calculate how much the spring will stretch, given the relationship between rotational speed and stretching forces.

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Most popular questions from this chapter

A particle is moving with a velocity of \(\vec{v}=(3 i+4 t j) \mathrm{m} / \mathrm{s}\) Find the ratio of tangential acceleration to that of normal acceleration at \(t=1 \mathrm{sec}\) (1) \(4 / 3\) (2) \(3 / 4\) (3) \(5 / 3\) (4) \(3 / 5\)

A car moving along a circular track of radius \(50.0 \mathrm{~m}\) accelerates from rest at \(3.00 \mathrm{~ms}^{-2}\). Consider a situation when the car's centripetal acceleration equals its tangential acceleration. Then (1) The angle around the track does the car travel is 1 rad (2) The magnitude of the car's total acceleration at that instant is \(3 \sqrt{2} \mathrm{~m} \mathrm{~s}^{-1}\) (3) Time elapses before this situation is \(\sqrt{\frac{50}{3}} \mathrm{~s}\). (4) The distance travelled by the car during this time is \(25 \mathrm{~m}\)

Two identical particics we string which passes through a hole at the center of a table of the particles is made to move in a circle on that string which passes througit One of the particles is made to move in a circle on the tabl with angular velocity \(\omega\), and the other is made to \(\mathrm{m}\), in a horizontal circle as a contact pendulum with angular velocity pendulum with angular velocity \(\omega_{2} .\) If \(l_{1}\) and \(l_{2}\) are the length of he string over and under the able, then in order that particle under the table neither moves own nor moves up, the ratio \(l_{1} / l_{2}\) is (1) \(\frac{\omega_{1}}{\omega_{2}}\) (2) \(\frac{\omega_{2}}{\omega_{1}}\) (3) \(\frac{\omega_{1}^{2}}{\omega_{2}^{2}}\) (4) \(\frac{\omega_{2}^{2}}{\omega_{1}^{2}}\)

Two particles describe the same circle of radius \(R\) in the same direction with the same speed \(v\), then at the given instant relative angular velocity of 2 with respect to 1 will be (1) \(\frac{2 v \sin \frac{\theta}{2}}{R}\) (2) \(\frac{v}{2 R \sin \frac{\theta}{2}}\) (3) \(\frac{v}{R}\) (4) \(\frac{v \cos \frac{\theta}{2}}{R}\)

A particle is projected from ground at an angle \(\theta .\) At a certain instant the velocity vector \(\vec{v}\) of particle makes an angle \(\alpha\) with horizontal. Choose the correct option(s). (1) \(\frac{d \vec{v}}{d t}=\vec{g}\) (2) Modulus of \(\frac{d|\vec{v}|}{d t}=g \sin \alpha\) (3) Tangential acceleration has magnitude of \(g \sin \alpha\) (4) Normal acceleration \(=g \cos \alpha\)
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