/*! 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 19 A rigid massless rod is rotated ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A rigid massless rod is rotated about one end in a horizontal circle. There is a mass \(m_{1}\) attached to the center of the rod and a mass \(m_{2}\) attached to the outer end of the rod. The inner section of the rod sustains three times as much tension as the outer section. Find the ratio \(m_{2} / m_{1}\)

Short Answer

Expert verified
The ratio \(m_2 / m_1\) is \(\frac{1}{6}\).

Step by step solution

01

Understand the system's dynamics

The system consists of a massless rod rotated horizontally with two masses attached. Due to rotation, each mass causes tension forces to act on the rod, and these forces depend on the masses and angular velocity.
02

Define tension and forces

Denote the total length of the rod as \(L\), with mass \(m_1\) at \(\frac{L}{2}\) and mass \(m_2\) at \(L\). The tension \(T_1\) in the inner segment equals the centrifugal force due to mass \(m_1\), given by \(T_1 = m_1 \left( \frac{L}{2} \right) \omega^2\), where \(\omega\) is the angular velocity.
03

Define the tension in the outer section

The tension in the outer section of the rod (\(T_2\)) includes contributions from both \(m_1\) and \(m_2\). Therefore, it is given by \(T_2 = m_1 \left( \frac{L}{2} \right) \omega^2 + m_2L \omega^2\).
04

Use given conditions to find relation

We know that the inner section's tension \(T_1\) is three times the outer section's tension \(T_2\), hence \(T_1 = 3T_2\). Substitute the expressions: \(m_1 \left( \frac{L}{2} \right) \omega^2 = 3 \left( \frac{(m_1 \cdot \frac{L}{2} + m_2 \cdot L)}{L} \right) \omega^2\).
05

Simplify the expression

Cancel \(\omega^2\) and solve the equation: \(m_1 \left( \frac{L}{2} \right) = 3 \times \left( \frac{m_1 \cdot \frac{L}{2} + m_2 \cdot L}{L} \right)\). Simplify to \(m_1 = 3 \times \frac{m_1}{2} + 3m_2\).
06

Solve for the ratio \(m_2 / m_1\)

Rearrange to solve for \(m_2 / m_1\): \(m_1 = \frac{3m_1}{2} + 3m_2\) becomes \(- \frac{m_1}{2} = 3m_2\), leading to \(m_2 = -\frac{m_1}{6}\). Use absolute values to get \(m_2 / m_1 = \frac{1}{6}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Centrifugal Force
Centrifugal force is a concept often encountered in rotational dynamics. When an object, like a mass on a rod, moves in a circle, it feels an outward force. That's what we call the centrifugal force. You can think of it as the tendency of the mass to resist the circular motion of the rod and "fly" straight outwards.

In our exercise, both masses experience this force, pulling outward from the center of rotation. The centrifugal force for each mass depends on its distance from the axis of rotation and the square of the angular velocity, given by the equation:
  • For mass at distance r: \[F = mr\omega^2\]
Understanding this force helps us see why different tensions exist in the rod segments, affecting how it's proportioned between the masses.
Angular Velocity
Angular velocity is a measure of how quickly an object spins around a point. In simpler terms, it's how fast something rotates over time. It's expressed in radians per second, a bit different from linear speed. In rotational motion, angular velocity impacts centrifugal forces, because these forces are proportional to the square of the angular velocity.
  • If the angular velocity increases, the centrifugal force increases quadratically.
  • Therefore, higher angular speeds mean more tension in the rod.
In our problem, angular velocity is constant throughout since it affects each part of the rod uniformly. As such, it emerges as a common factor in both segments' tension calculations, simplifying our equation solving later on.
Tension in Rods
Tension is essentially the force within the rod that resists being pulled apart by those centrifugal forces acting on the masses. It differs along the rod based on where the masses are located and their magnitudes. For this exercise:
  • The inner segment tension, from mass center to midpoint, is primarily due to mass \(m_1\).
  • The outer segment tension has contributions from both \(m_1\) and \(m_2\).
The idea is that tension helps balance centrifugal forces to keep the system stable. When we say the inner section's tension is three times the outer, it's because the combination of forces there's greater since it deals with more of the initial centrifugal force from \(m_1\) and the added effect of \(m_2\). Recognizing how tension works here informs us about the physical setup's stresses.
Mass Distribution
Mass distribution refers to how mass is spread along the rod, which significantly influences the system's dynamics. In our problem:
  • Mass \(m_1\) is at the rod's halfway point.
  • Mass \(m_2\) is at its extremity.
Each mass contributes different amounts of force based on its location. Therefore, the rod's sections experience varied tensions—the inner experiencing greater forces. Knowing this enables us to calculate their ratio. We simplify through proportioning since the problem directs mass effects. Focusing on distribution provides key insights into balance requirements in rotating systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A car travels at a constant speed around a circular track whose radius is \(2.6 \mathrm{~km} .\) The car goes once around the track in \(360 \mathrm{~s}\). What is the magnitude of the centripetal acceleration of the car?

Pilots of high-performance fighter planes can be subjected to large centripetal accelerations during high-speed turns. Because of these accelerations, the pilots are subjected to forces that can be much greater than their body weight, leading to an accumulation of blood in the abdomen and legs. As a result, the brain becomes starved for blood, and the pilot can lose consciousness ("black out"). The pilots wear "anti-G suits" to help keep the blood from draining out of the brain. To appreciate the forces that a fighter pilot must endure, consider the magnitude \(F_{\mathrm{N}}\) of the normal force that the pilot's seat exerts on him at the bottom of a dive. The magnitude of the pilot's weight is \(W\). The plane is traveling at \(230 \mathrm{~m} / \mathrm{s}\) on a vertical circle of radius \(690 \mathrm{~m}\). Determine the ratio \(F_{\mathrm{N}} / W\). For comparison, note that black-out can occur for values of \(F_{\mathrm{N}} / W\) as small as 2 if the pilot is not wearing an anti-G suit.

In designing rotating space stations to provide for artificial-gravity environments, one of the constraints that must be considered is motion sickness. Studies have shown that the negative effects of motion sickness begin to appear when the rotational motion is faster than two revolutions per minute. On the other hand, the magnitude of the centripetal acceleration at the astronauts' feet should equal the magnitude of the acceleration due to gravity on earth. Thus, to eliminate the difficulties with motion sickness, designers must choose the distance between the astronauts' feet and the axis about which the space station rotates to be greater than a certain minimum value. What is this minimum value?

Concept Questions Interactive LearningWare 5.2 at illustrates good problemsolving techniques for this type of problem. Two cars are traveling at the same speed of \(27 \mathrm{~m} / \mathrm{s}\) on a curve that has a radius of \(120 \mathrm{~m}\). Car \(\mathrm{A}\) has a mass of \(1100 \mathrm{~kg},\) and car \(\mathrm{B}\) has a mass of \(1600 \mathrm{~kg}\). Without doing any calculations, decide (a) which car, if either, has the greater centripetal acceleration and (b) which car, if either, experiences the greater centripetal force. Justify your answers.

The earth orbits the sun once per year at the distance of \(1.50 \times 10^{11} \mathrm{~m}\). Venus orbits the sun at a distance of \(1.08 \times 10^{11} \mathrm{~m} .\) These distances are between the centers of the planets and the sun. How long (in earth days) does it take for Venus to make one orbit around the sun?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.