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The direction of infinitesimal change in angular momentum of the pendulum at any instant about point \(O\) is (A) parallel to instantaneous velocity. (B) perpendicular to instantaneous velocity. (C) perpendicular to length of the string. (D) None of these as angular momentum is constant.

Short Answer

Expert verified
(B) perpendicular to instantaneous velocity

Step by step solution

01

Understand the angular momentum

The angular momentum \(L\) of a point mass rotating around a point is given by \(L= \vec{r} \times \vec{p}\), where \(\vec{r}\) is the position vector (from the origin to the point mass) and \(\vec{p}\) is the linear momentum of the point mass. The '×' denotes a cross product, which results in a vector that is orthogonal (perpendicular) to the plane, formed by the two input vectors.
02

Define the vectors

In the case of the pendulum, the position vector \(\vec{r}\) is the length of the string. The linear momentum \(\vec{p}\) is mass times velocity, where velocity is the instantaneous tangent direction of the swinging pendulum.
03

Angular momentum for a pendulum

Because of how cross product works, it can be concluded that the change in angular momentum is perpendicular to both the position vector (\( \vec{r}\), or in other words, the string of the pendulum) and the linear momentum (\( \vec{p}\), or in other words, the instantaneous velocity of the pendulum).
04

Option Selection

Thus, based on the analysis, the correct answer would be 'perpendicular to instantaneous velocity' and 'perpendicular to length of the string.' However, since there's no such option, the closest correct choice is (B) 'parallel to instantaneous velocity.'

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Key Concepts

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

Cross Product
The cross product is a fundamental operation in vector mathematics, particularly important in physics. It allows us to determine a vector that is perpendicular to two other vectors. This is especially useful when dealing with rotational motion or angular momentum.

To calculate the cross product of two vectors \(\vec{A} = (A_x, A_y, A_z)\) and \(\vec{B} = (B_x, B_y, B_z)\), we use the formula:
  • \(\vec{A} \times \vec{B} = (A_yB_z - A_zB_y, A_zB_x - A_xB_z, A_xB_y - A_yB_x)\)
The result of this operation is a new vector that is orthogonal to both \(\vec{A}\) and \(\vec{B}\).

In the context of the pendulum, the cross product helps determine the direction of angular momentum. By using the position vector of the pendulum and its linear momentum, we can find the direction in which the angular change occurs. The vector resulting from the cross product lies perpendicular to the plane formed by these two vectors.
Linear Momentum
Linear momentum is a key concept in physics, representing the quantity of motion an object has. It is defined for a mass \(m\) moving with a velocity \(\vec{v}\) as \(\vec{p} = m\vec{v}\). This simple product is fundamental, especially in systems involving motion and force analysis.

For the pendulum, the linear momentum is crucial to understanding its motion. As the pendulum swings, its mass and velocity dictate its momentum at any instant. This momentum contributes to the pendulum's overall motion and helps calculate other quantities such as angular momentum.
  • Linear momentum changes as the pendulum swings due to gravitational forces acting upon it.
  • Even though the speed might remain constant at certain points, the direction of velocity changes, affecting momentum.
Understanding linear momentum allows us to appreciate how forces impact the motion of bodies, providing a foundation for further studies in mechanics.
Pendulum Motion
Pendulum motion is a classic example of simple harmonic motion in physics. A pendulum swings back and forth, demonstrating the continuous exchange between kinetic and potential energy.

The pendulum consists of a mass (known as the bob) attached to a string or rod, with its motion affected by gravity. The forces acting on it create a predictable pattern:
  • At the highest points, the pendulum has maximum potential energy and minimal kinetic energy.
  • At the lowest point, the scenario reverses with maximum kinetic energy and minimal potential energy.

During its motion, a pendulum's angular momentum provides insights into its rotational characteristics. By understanding how pendulum motion relates to other physical quantities, students can develop a deeper comprehension of dynamics and energy conservation principles.

In laboratory and theoretical studies, pendulum motion is a fundamental concept that leads to more complex ideas, like resonance and damping, influencing various applications in engineering and science.

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Most popular questions from this chapter

Column-I (A) Particle moving with speed \(v_{0}\) strikes to a rod placed on a smooth table and sticks to it. (B) A thin rod of mass \(m\) and length \(\ell\) inclined at an angle \(\theta\) with horizontal is dropped on a smooth horizontal plane without any angular velocity. Its tip does not rebound after impact. (C) A solid sphere of mass \(m\) and radius \(R\) is rolling with velocity \(v_{0}\) along a horizontal plane. It suddenly encounters an obstacle. (D) Two cylinders of radii \(r_{1}\) and \(r_{2}\) rotating about their axis with angular speed \(\omega_{1}\) and \(\omega_{2}\) moved closer to touch each other keeping their axis parallel. Cylinders first slip over each other at the contact point but slipping ceases after some time due to friction. Column-II (1) Kinetic energy will change. (2) Momentum is conserved. (3) Angular momentum is conserved about any point. (4) Angular momentum is conserved just before and after impact only about point of impact.

A hollow sphere, ring, disc and solid sphere each of mass \(1 \mathrm{~kg}\) and radius \(1 \mathrm{~m}\) is released from rest on an identical inclined plane of inclination \(37^{\circ} . \tan 37^{\circ}=3 / 4\) and $g=10 \mathrm{~m} / \mathrm{s}^{2}$ ). The co-efficient of friction between body and surface is \(\mu\). Then match the column.A hollow sphere, ring, disc and solid sphere each of mass \(1 \mathrm{~kg}\) and radius \(1 \mathrm{~m}\) is released from rest on an identical inclined plane of inclination $37^{\circ} . \tan 37^{\circ}=3 / 4\( and \)g=10 \mathrm{~m} / \mathrm{s}^{2}$ ). The co-efficient of friction between body and surface is \(\mu\). Then match the column.

A hollow cylinder, a spherical shell, a solid cylinder and a solid sphere are allowed to roll on an inclined rough surface of coefficient of friction \(\mu\) and inclination \(\theta\). The correct statements are (A) if cylindrical shell can roll on inclined plane, all other objects will also roll. (B) if all the objects are rolling and have same mass, the \(\mathrm{KE}\) of all the objects will be same at the bottom of inclined plane. (C) work done by the frictional force will be zero, if objects are rolling. (D) frictional force will be equal for all the objects, if mass is same.

Three marbles roll down on three different smooth tracks of same vertical height. Track (i) is inclined at \(75^{\circ}\) to the ground, track (ii) is inclined at \(60^{\circ}\) to the ground and track (iii) is inclined at \(30^{\circ}\) to ground. These marbles reach the ground with respective velocities \(u_{1}, u_{2}\) and \(u_{3}\) (A) \(u_{1}\) is greatest of all (B) \(u_{3}\) is least of all (C) \(u_{1}=u_{2}\) (D) \(u_{2}=u_{3}\)

A thick-walled hollow sphere has outer radius \(R .\) It rolls down on an inclined plane without slipping and its speed at bottom is \(v_{0}\). Now the incline is waxed so that the friction becomes zero. The sphere is observed to slide down without rolling and the speed now is \(\left(5 v_{0} / 4\right)\). The radius of gyration of the hollow sphere about the axis through its centre is \(\frac{n R}{4} .\) Then the value of \(n\) is.

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