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Chapter 4: Problem 60

A conical pendulum consists of a mass \(M\) suspended from a strong sling of length \(l\). The mass executes a circle of radius \(R\) in a horizontal plane with speed \(v\). At time \(t\), the mass is at position \(R_{i}^{\prime}\) and has \(v \hat{j}\) velocity. At time \(t\), the angular momentum vector of mass \(M\) about the point from which the string passes on the ceiling is (1) \(M v R \hat{k}\) (2) \(M v l \hat{k}\) (3) \(M v l\left[\frac{\sqrt{R-R^{2}}}{l} \hat{i}+\frac{R}{l} \hat{k}\right]\) (4) \(-M v l\left[\frac{\sqrt{p-R^{2}}}{l} \hat{i}+\frac{R}{l} \hat{k}\right]\)

Short Answer

Expert verified
The correct answer is option (4).

Step by step solution

01

Determine the Components of the Pendulum

First, understand that the mass is undergoing circular motion. The pendulum has a length \(l\) and moves in a circle with radius \(R\) in a horizontal plane. We need to identify how this relates to its angular momentum.
02

Define the Angular Momentum

Angular momentum \(L\) with respect to a point is given by \(L = r imes p\), where \(r\) is the position vector of the mass from the point, and \(p\) is its linear momentum. In this case, \(p = Mv \hat{j}\).
03

Calculate the Position Vector r

The position vector \(r\) from the ceiling to the mass has a horizontal component \(R\) in the \(x\)-direction and vertical components due to the pendulum length \(l\): \(r = R \hat{i} + \sqrt{l^2 - R^2} \hat{k}\).
04

Cross Product for Angular Momentum

Using the cross product \(L = r imes p = (R \hat{i} + \sqrt{l^2 - R^2} \hat{k}) imes (Mv \hat{j})\), calculate each component. The cross product is \(-MvR \hat{k} + Mv \sqrt{l^2 - R^2} \hat{i}\).
05

Identify the Correct Expression

This solution matches option (4), which is \(-M v l\left[\frac{\sqrt{l^2-R^{2}}}{l} \hat{i}+\frac{R}{l} \hat{k}\right]\). This form corresponds to our calculated components when expanded.

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Key Concepts

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

Circular Motion
In the context of a conical pendulum, circular motion refers to the movement of the mass as it describes a circular path around a fixed point. This is a crucial aspect of circular motion, where the speed of the mass remains constant while the direction of its velocity changes continuously.
When considering circular motion, it is important to focus on:
  • Radius of Motion: The mass is attached to a string of length \(l\) and moves in a circle with a radius \(R\). The radius is perpendicular to the axis around which the mass moves, creating a conical shape.
  • Velocity: The speed of the mass remains constant, denoted as \(v\), and points tangentially to the path of the circle in the horizontal plane.
The centripetal force that keeps the mass moving in a circle is provided by the horizontal component of the tension in the string. This constant force is always directed toward the center of the circle, making it possible for the mass to maintain its circular path without changing speed.
Angular Momentum
Angular momentum is a measure of the rotational momentum of an object. For the conical pendulum, it relates to how much rotational motion the mass \(M\) exhibits with respect to a fixed point, which is the ceiling from where the pendulum is suspended.
The angular momentum \(L\) can be expressed as:
  • \(L = r \times p\)
  • where \(r\) is the position vector of the mass, and \(p\) is its linear momentum.
Here, for our system:
  • Position Vector, \(r\): The position vector is composed of a horizontal component \(R \hat{i}\) and a vertical component \(\sqrt{l^2 - R^2} \hat{k}\).
  • Linear Momentum, \(p\): This is given by the product of the mass \(M\) and its velocity \(v\), in the direction \(\hat{j}\).
By calculating the cross product of \(r\) and \(p\), we determine the angular momentum vector, which takes into account both the direction and magnitude of rotation. This value describes how the mass moves and rotates around the central axis of the pendulum's circle.
Cross Product
The cross product is a vector operation that is central to calculating angular momentum. It provides a way to determine a vector that is perpendicular to the plane formed by two other vectors; in this case, the position vector and momentum vector.
For the conical pendulum:
  • \(r \times p\) gives us the angular momentum.
Let's break it down:
  • Calculate Components: The position vector \(r\) is \(R \hat{i} + \sqrt{l^2 - R^2} \hat{k}\) and the momentum \(p\) is \(Mv \hat{j}\).
  • Apply Cross Product: Using the order of operations for cross products, \((R \hat{i} + \sqrt{l^2 - R^2} \hat{k}) \times (Mv \hat{j})\), results in components \(-MvR \hat{k} + Mv\sqrt{l^2 - R^2} \hat{i}\).
The result of the cross product gives a vector that is perpendicular to both \(r\) and \(p\), indicating the rotation's axis and the magnitude of rotation for the pendulum. This calculation is fundamental in revealing the behavior and dynamics of systems involving rotational motion, such as our conical pendulum.

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Most popular questions from this chapter

A solid sphere rolls down two different inclined planes of the same height but of different inclinations (1) the speeds will be same but time of descent will be different (2) in both cases, the speeds and time of descent will be same (3) speeds and time of descent both will be different (4) the speeds will be different but time of descent will be same

A thin uniform rod of mass \(m\) and length \(l\) is free to rotate about its upper end. When it is at rest, it receives an impulse \(J\) at its lowest point, normal to its length. Immediately after impact, (1) the angular momentum of the rod is \(J l\) (2) the angular velocity of the rod is \(3 \mathrm{~J} / \mathrm{ml}\) (3) the kinetic energy of the rod is \(3 J / 2 m\) (4) the linear velocity of the midpoint of the rod is \(3 J / 2 m\)

A solid cylinder of mass \(M\) and radius \(R\) is resting on a horizontal platform (which is parallel to the \(x-y\) plane) with its axis fixed along \(Y\)-axis and free to rotate about its axis. The platform is given a motion in the \(X\)-direction given by \(x=A \cos (\omega t)\). There is no slipping between the cylinder and the platform. The maximum torque acting on the cylinder during its motion is (1) \(\frac{M \omega^{2} A R}{3}\) (2) \(\frac{M \omega^{2} A R}{2}\) (3) \(\frac{2}{3} \times M \omega^{2} A R\) (4) the situation is not possible

A horizontal plane supports a fixed vertical cylinder of radius \(R\) and a particle is attached to the cylinder by a horizontal thread \(A B\) as shown in figure. The particle initially rest on a horizontal plane. A horizontal velocity \(v_{0}\) is imparted to the particle, normal to the threading during subsequent motion. Point out the false statements. (1) Angular momentum of particle about \(O\) remains constant. (2) Angular momentum about \(B\) remains constant. (3) Momentum and kinetic energy both remain constant. (4) Kinetic energy remains constant.

A sphere of radius \(r\) and mass \(m\) has a linear velocity \(v_{0} \mathrm{~m} / \mathrm{s}\) directed to the left and no angular velocity as it is placed on a horizontal platform moving to the right with a constant velocity \(10 \mathrm{~m} / \mathrm{s}\). If after sliding on the platform the sphere is to have no linear velocity relative to the ground as it starts rolling on the platform without sliding. The coefficient of kinetic friction between the sphere and the platform is \(\mu_{k}\) Determine the required value of \(v_{0}\) in \(\mathrm{m} / \mathrm{s}\).
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