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A particle of mass \(m\) is moving on a frictionless horizontal table and is attached to a massless string, whose other end passes through a hole in the table, where I am holding it. Initially the particle is moving in a circle of radius \(r_{\mathrm{o}}\) with angular velocity \(\omega_{\mathrm{o}},\) but I now pull the string down through the hole until a length \(r\) remains between the hole and the particle. What is the particle's angular velocity now?

Short Answer

Expert verified
The new angular velocity \( \omega \) is \( \frac{r_o^2 \omega_o}{r^2} \).

Step by step solution

01

Identify Initial Angular Momentum

The initial angular momentum of the particle can be expressed as the product of its moment of inertia and angular velocity. Initially, the particle is moving in a circle of radius \( r_o \) with angular velocity \( \omega_o \). The moment of inertia \( I_i \) at this point is \( m r_o^2 \). Therefore, the initial angular momentum \( L_i \) is given by: \[ L_i = I_i \omega_o = m r_o^2 \omega_o. \]
02

Express Final Angular Momentum

After pulling the string down, the remaining length of the string is \( r \), and the new moment of inertia \( I_f \) is \( m r^2 \). If the new angular velocity is \( \omega \), the final angular momentum \( L_f \) is: \[ L_f = I_f \omega = m r^2 \omega. \]
03

Apply Conservation of Angular Momentum

Since no external torques are acting on the system, the angular momentum is conserved. This gives: \[ L_i = L_f. \] Substituting the expressions from Steps 1 and 2: \[ m r_o^2 \omega_o = m r^2 \omega. \]
04

Solve for Final Angular Velocity

To find the particle's angular velocity \( \omega \) after pulling the string, we solve the equation: \[ m r_o^2 \omega_o = m r^2 \omega \rightarrow \omega = \frac{r_o^2 \omega_o}{r^2}. \]

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Key Concepts

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

Moment of Inertia
Moment of inertia is a fundamental concept when it comes to understanding rotational motion. It is the rotational equivalent of mass in linear motion. In simple terms, it measures how much an object resists changes in its rotational state. Think of it as the rotational "mass," determining how easily an object can spin.

The formula for the moment of inertia depends on the mass distribution of an object. For a particle moving in a circle, like in this exercise, it can be calculated as the product of the mass and the square of the radius of motion:
  • Moment of Inertia, \( I = m imes r^2 \) where \( m \) is the mass and \( r \) is the radius.
When you decrease the radius by pulling the string, the moment of inertia decreases since it depends on \( r^2 \). This change affects how the particle rotates, as seen in the principle of angular momentum conservation.
Angular Velocity
Angular velocity is an essential part of understanding how quickly an object is rotating. It tells us how fast something spins around a central point, measured in radians per second.

In the given exercise, the initial angular velocity \( \omega_o \) describes the rate at which the particle moves in its original circular path. With the string's length pulled to a smaller radius, the angular velocity changes to maintain the balance of angular momentum. As the radius decreases, the angular velocity must increase to conserve the system's rotational motion due to the reduced moment of inertia.

Here's how you can determine the new angular velocity:
  • Initial angular velocity: \( \omega_o \)
  • New angular velocity: \( \omega = \frac{r_o^2 \omega_o}{r^2} \)
This equation shows the inverse relationship between radius and angular velocity, driven by the conservation of angular momentum.
Circular Motion
Circular motion occurs when an object moves continuously along a circular path. It's characterized by the motion's radius, speed (or velocity), and the force keeping it on its circular path.

In this exercise, the particle is initially moving in a uniform circular path. The key forces at play here are centripetal, meaning they act towards the center of the circle, maintaining the particle's path.

When the string is pulled, not only does the path radius change, but so do other parameters:
  • The centripetal force adjusts, but its nature stays the same - always towards the circle's center.
  • The path becomes tighter or broader according to how the string length varies.
  • The particle's speed and angular velocity adapt to maintain its motion without any external torques acting.
Understanding circular motion helps predict how objects will behave when their paths or conditions change, like in the exercise where pulling the string modifies the circle's size and hence the motion dynamics.

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Most popular questions from this chapter

A juggler is juggling a uniform rod one end of which is coated in tar and burning. He is holding the rod by the opposite end and throws it up so that, at the moment of release, it is horizontal, its \(\mathrm{CM}\) is traveling vertically up at speed \(v_{\mathrm{o}}\) and it is rotating with angular velocity \(\omega_{\mathrm{o}} .\) To catch it, he wants to arrange that when it returns to his hand it will have made an integer number of complete rotations. What should \(v_{\mathrm{o}}\) be, if the rod is to have made exactly \(n\) rotations when it returns to his hand?

A particle moves under the influence of a central force directed toward a fixed origin \(O\). (a) Explain why the particle's angular momentum about \(O\) is constant. (b) Give in detail the argument that the particle's orbit must lie in a single plane containing \(O\).

Show that the moment of inertia of a uniform solid sphere rotating about a diameter is \(\frac{2}{5} M R^{2}\). The sum (3.31) must be replaced by an integral, which is easiest in spherical polar coordinates, with the axis of rotation taken to be the \(z\) axis. The element of volume is \(d V=r^{2} d r \sin \theta d \theta d \phi\). (Spherical polar coordinates are defined in Section 4.8. If you are not already familiar with these coordinates, you should probably not try this problem yet.)

Consider a gun of mass \(M\) (when unloaded) that fires a shell of mass \(m\) with muzzle speed \(v\). (That is, the shell's speed relative to the gun is \(v\).) Assuming that the gun is completely free to recoil (no external forces on gun or shell), use conservation of momentum to show that the shell's speed relative to the ground is \(v /(1+m / M).\)

Use spherical polar coordinates \(r, \theta, \phi\) to find the CM of a uniform solid hemisphere of radius R, whose flat face lies in the \(x y\) plane with its center at the origin. Before you do this, you will need to convince yourself that the element of volume in spherical polars is \(d V=r^{2} d r \sin \theta d \theta d \phi\). (Spherical polar coordinates are defined in Section 4.8. If you are not already familiar with these coordinates, you should probably not try this problem yet.)

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