Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

An astronaut in gravity-free space is twirling a mass \(m\) on the end of a string of length \(R\) in a circle, with constant angular velocity \(\omega\) Write down Newton's second law (1.48) in polar coordinates and find the tension in the string.

Short Answer

Expert verified
The tension is \( T = m \omega^2 R \).

Step by step solution

01

Understanding Newton's Second Law

Newton's Second Law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. In symbols, it's written as \( \vec{F} = m \vec{a} \). To apply this in polar coordinates to circular motion, we need to convert the acceleration into polar components.
02

Acceleration in Polar Coordinates

For an object moving in a circle at a constant angular velocity \( \omega \), the radial component of acceleration \( \vec{a}_r \) arises from the centripetal force. This is given by \( a_r = -\omega^2 R \), directed towards the center of the circle. The tangential component of acceleration \( \vec{a}_\theta \) is zero because the speed is constant.
03

Applying Newton's Second Law

In polar coordinates, the radial force \( F_r \) is equal to \( m \times a_r \). Therefore, the equation using Newton's Second Law is: \( F_r = m (-\omega^2 R) \).
04

Finding the Tension

The tension \( T \) in the string provides the centripetal force required to keep the mass revolving in a circle. Therefore, \( T = -F_r = m \omega^2 R \). Here, the negative sign is ignored in the expression for tension because it represents the magnitude of the force directed inward, opposite the direction we originally assumed.

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 Vaia!

Key Concepts

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

Centripetal Force
Centripetal force is essential for any object moving in a circular path. It's the force that directs the object inward, toward the center of the circle. Without this force, the object would continue moving in a straight line due to inertia, as described by Newton's First Law.
When something moves in a circle, even with constant speed, its velocity is changing direction. This change in velocity is an acceleration, known specifically as centripetal acceleration. A key formula to remember is that centripetal force \( F_c \) is given by \( F_c = m a_r \), where \( m \) is the mass of the object and \( a_r \) is the radial component of acceleration, which in circular motion is \( a_r = -\omega^2 R \), with \( \omega \) being the angular velocity and \( R \) being the radius of the circle.
Thus, in circular motion, the centripetal force is always directed toward the center of the path. It's interesting to note how this inward force is balanced by other forces acting on the object, such as tension in a string when twirling a mass.
Circular Motion
Circular motion involves an object moving along the circumference of a circle. This type of motion can vary in speed, but often, for theoretical purposes, it's studied as uniform circular motion, where the speed remains constant.
An important aspect is understanding why an object in circular motion remains in its path, which ties back to forces. For the motion to be sustained, a continuous force must act perpendicular to the velocity at every point. This force is what we call the centripetal force. But in a situation like an astronaut twirling a mass in space, this centripetal force is provided by the tension in the string.
Remember that while speed might be constant, the direction of velocity continuously changes, making it an example of accelerated motion. This is why studying circular motion can give us deep insights into the dynamics of forces and how they keep bodies in motion.
Tension in String
When twirling a mass around with a string, the tension acts as the centripetal force needed to maintain circular motion. Essentially, tension is a force exerted by the string on the object, pulling it towards the center.
In our specific scenario, the tension \(T\) must match the centripetal force required, hence \( T = m \omega^2 R \). The twist comes in understanding that tension is an internal force along the string that adjusts to maintain the balance necessary for the circular motion.
Additionally, it’s crucial to note that the direction of tension is always towards the center of the circular path. It provides a real-world example of how a force can operate without directly being in contact with another object (since the string might not attach to anything but wraps around the twirling mass). Understanding tension is vital as it depicts how forces transmit in connected bodies.
Angular Velocity
Angular velocity \( \omega \) is a measure of how quickly an object is rotating around a particular point or axis. Unlike linear velocity, which describes movement in a straight line, angular velocity tackles rotational motion.
It's defined as the rate of change of angular position and is typically measured in radians per second. The formula \( \omega = \frac{\Delta \theta}{\Delta t} \) shows us how the angle \( \theta \) changes over time \( t \).
In circular motion, angular velocity remains constant when an object is rotating at a steady speed. However, it's important to realize that although angular velocity remains the same, the linear velocity at any point depends on the distance from the axis of rotation. The relationship between linear velocity \( v \) and angular velocity is given by \( v = \omega R\, \) where \(R\) is the radius of the circular path.
This concept is incredibly useful when analyzing different types of rotational dynamics and is fundamental in understanding how rotational systems operate.

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

Find expressions for the unit vectors \(\hat{\boldsymbol{\rho}}, \hat{\boldsymbol{\phi}},\) and \(\hat{\mathbf{z}}\) of cylindrical polar coordinates (Problem 1.47) in terms of the Cartesian \(\hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{z}}\). Differentiate these expressions with respect to time to find \(d \hat{\boldsymbol{\rho}} / d t, d \hat{\boldsymbol{\phi}} / d t,\) and \(d \hat{\mathbf{z}} / d t\)

In case you haven't studied any differential equations before, I shall be introducing the necessary ideas as needed. Here is a simple excercise to get you started: Find the general solution of the firstorder equation \(d f / d t=f\) for an unknown function \(f(t) .\) [There are several ways to do this. One is to rewrite the equation as \(d f / f=d t\) and then integrate both sides.] How many arbitrary constants does the general solution contain? [Your answer should illustrate the important general theorem that the solution to any \(n\) th-order differential equation (in a very large class of "reasonable" equations) contains \(n\) arbitrary constants.]

A cannon shoots a ball at an angle \(\theta\) above the horizontal ground. (a) Neglecting air resistance, use Newton's second law to find the ball's position as a function of time. (Use axes with \(x\) measured horizontally and \(y\) vertically.) (b) Let \(r(t)\) denote the ball's distance from the cannon. What is the largest possible value of \(\theta\) if \(r(t)\) is to increase throughout the ball's flight? [Hint: Using your solution to part (a) you can write down \(r^{2}\) as \(x^{2}+y^{2},\) and then find the condition that \(r^{2}\) is always increasing.]

In Section 1.5 we proved that Newton's third law implies the conservation of momentum. Prove the converse, that if the law of conservation of momentum applies to every possible group of particles, then the interparticle forces must obey the third law. [Hint: However many particles your system contains, you can focus your attention on just two of them. (Call them 1 and 2.) The law of conservation of momentum says that if there are no external forces on this pair of particles, then their total momentum must be constant. Use this to prove that \(\mathbf{F}_{12}=-\mathbf{F}_{21}\).]

A particle moves in a circle (center \(O\) and radius \(R\) ) with constant angular velocity \(\omega\) counter-clockwise. The circle lies in the \(x y\) plane and the particle is on the \(x\) axis at time \(t=0 .\) Show that the particle's position is given by $$\mathbf{r}(t)=\hat{\mathbf{x}} R \cos (\omega t)+\hat{\mathbf{y}} R \sin (\omega t)$$ Find the particle's velocity and acceleration. What are the magnitude and direction of the acceleration? Relate your results to well-known properties of uniform circular motion.

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.

Sign-up for free