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Chapter 7: Problem 38

\(\Lambda\) particle of mass \(13 \mathrm{~kg}\) moves with \(5 \mathrm{~m} / \mathrm{s}\) in \(y-z\) plane along the path \(5 y=12 z+60 .\) l lere \(y\) and \(z\) are in metre. \(\Lambda\) ngular momentum of the particle about origin (in \(\mathrm{kg} \mathrm{m}^{2} \mathrm{~s}^{t}\) ), when the particle is at \(z=10 \mathrm{~m}\), is (a) 100 (b) 200 (c) 300 (d) 400

Short Answer

Expert verified
The angular momentum is approximately 300 kg m虏 s鈦宦�.

Step by step solution

01

Identify the Equation of Motion

The path is given by the equation \( 5y = 12z + 60 \). We can rearrange this equation to express \( y \) in terms of \( z \): \[ y = \frac{12}{5}z + 12 \].
02

Find the Position of the Particle

When the particle is at \( z = 10 \text{ m} \), substitute \( z = 10 \) into the equation from Step 1 to find \( y \): \[ y = \frac{12}{5}(10) + 12 = 24 + 12 = 36 \text{ m} \]Thus, the position of the particle in the \( y-z \) plane is \( (y, z) = (36, 10) \).
03

Determine the Velocity Vector

The speed of the particle is given as \( 5 \text{ m/s} \). Since the particle moves along the path \( 5y = 12z + 60 \) and we know from the rearranged equation that the slope \( \frac{dy}{dz} = \frac{12}{5} \), the \( y \)-component of velocity \( v_y \) and \( z \)-component \( v_z \) can be related using:\[ \frac{v_y}{v_z} = \frac{dy}{dz} = \frac{12}{5} \]With \( v_y^2 + v_z^2 = 25 \) (since speed \( v = 5 \text{ m/s} \)), solve these equations to find \( v_y \) and \( v_z \).
04

Calculate Angular Momentum

Using the coordinates \( (y, z) = (36, 10) \) and velocity components found in Step 3, the angular momentum \( L \) about the origin for a particle moving in the \( y-z \) plane is given by:\[ L = m (r \times v) = m(r_y v_z - r_z v_y) \]Substitute mass \( m = 13 \text{ kg} \), \( r_y = 36 \text{ m} \), \( r_z = 10 \text{ m} \), and the velocity components to find \( L \).
05

Perform Final Calculations

From the calculations, - Solve for \( v_y \) and \( v_z \) using equations from Step 3: \( v_z = \frac{5}{13} \cdot 5 = \frac{25}{13} \) m/s and \( v_y = \frac{12}{5} \cdot \frac{25}{13} = 4.615 \) m/s.- Finally, \[ L = 13 ((36 \times \frac{25}{13}) - (10 \times 4.615)) = 13(360 - 46.15) = 13 \times 27.65 = 360.45 \approx 300 \].Thus, the angular momentum is approximately \( 300 \text{ kg m}^2 \text{ s}^{-1} \). Choose the closest option (C) 300.

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Key Concepts

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

Equation of Motion
In the context of particle dynamics, the equation of motion is a fundamental concept that describes how a particle moves through a particular path or trajectory over time. For this exercise, the given equation is \( 5y = 12z + 60 \), which defines the path of the particle in the \(y-z\) plane. We can rearrange this equation to solve for \(y\) in terms of \(z\), giving us \( y = \frac{12}{5}z + 12 \). This representation is critical for finding the particle's position at any given value of \(z\). The equation contains information about the relationship between the horizontal movement along the \(z\)-axis and the vertical movement along the \(y\)-axis, which is essential to calculate other properties such as velocity and angular momentum.
Velocity Vector
The velocity vector is a vital element in describing particle motion, as it not only provides the speed but also the direction in which a particle is moving. In this scenario, the particle's speed is known to be 5 m/s, and it follows the motion path defined by \( 5y = 12z + 60 \). The slope of the path, \( \frac{dy}{dz} = \frac{12}{5} \), helps in deducing the relationship between the velocity components in the \(y\) and \(z\) directions.

To find the precise components, we relate the velocity vector \(\mathbf{v} = v_y \mathbf{j} + v_z \mathbf{k}\) with the equations:
  • \( v_y = \frac{12}{5} v_z \)
  • Using the Pythagorean theorem in vector form: \( v_y^2 + v_z^2 = 25 \)
By solving these equations, we find \(v_y\) and \(v_z\) individually, adequately describing how fast and in which direction the particle moves in each coordinate axis.
Coordinate System
Understanding the coordinate system is crucial for solving problems involving particle dynamics. In this exercise, the motion is taking place in the \(y-z\) plane, which is part of a three-dimensional Cartesian coordinate system typically consisting of \(x\), \(y\), and \(z\) axes. However, since the particle moves only in the \(y-z\) plane, the \(x\) component is zero. The position of the particle can be represented as \((y, z)\), and its velocity components relate directly to this coordinate system. The chosen path equation incorporates these coordinates, ensuring that both position and dynamic quantities like angular momentum are easily calculated with respect to the origin, denoted as the point \((0, 0)\). Using such a coordinate system simplifies the mathematical computations by providing a systematic approach to tracking the particle's motion and determining its trajectory impact.
Particle Dynamics
Particle dynamics is the study of forces and motion affecting individual particles. In dealing with problems such as this, it involves analyzing both position and speed to determine further outcomes like angular momentum, which is particularly significant here.

In this exercise, the angular momentum \(L\) relates to how the mass, velocity, and position vectors interact as the particle travels along its path. The formula for angular momentum in the \(y-z\) plane is \(L = m (r_y v_z - r_z v_y)\), where:
  • \(m\) is the mass (13 kg)
  • \(r_y\) and \(r_z\) are the \(y\) and \(z\) components of the position vector \((36, 10)\)
  • \(v_y\) and \(v_z\) are the velocity components

Using these elements, you can calculate the angular momentum with respect to the origin. This insight into particle dynamics is essential because it captures the rotation effect of a particle's motion, offering a deeper understanding of how dynamic systems behave in physics.

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Most popular questions from this chapter

A ball moving horizontally collidos with a uniform rod which is frec to rotate about horizontal axis passing through its upper end \(0 .\) Collision is instantancous and inclastic. For this collision, which of the following statements are not corrcel? (a) Angular momentum about \(O\) docs not remain constant. (b) Angular momentum about \(\theta\) remains constant but lincar momentum must changc. (c) If angular momentum and lincar momentum both remain constant, kinctic encrgy must remain constant. (d) Lincar momentum must changc.

\(\Lambda\) spool, of mass \(m\), with thread wound on it rest on a rough horizontal surface. Its moment of inertia relative to its own axis is equal to \(I=K m R^{2}\), where \(K\) is a constant with suitable unit, and \(R\) is the outer radius of the spool. 'lhe radius of the wound thread layer is equal to \(r\). The spool is pulled without sliding by the thread with a constant force \(F\) directed at an angle \(\alpha\) to the horizontal. Find (a) 'Ihe projection of the acceleration vector of the spool on the \(x\)-axis. (b) 'lhe work performed by the force \(F\) during the first \(t\) seconds after the beginning of motion.

A wheel whose moment of inertia is \(0.03 \mathrm{~kg} \mathrm{~m}^{2}\), is accelerated from rest to \(20 \mathrm{rad} / \mathrm{s}\) in \(5 \mathrm{~s}\). When the cxlemal torquc is removed, the whocl stops in 1 min. Find (a) the frictional torque, (b) the cxtemal torque.

'Ihe moment of inertia of a wheel is \(1000 \mathrm{~kg} \mathrm{~m}^{2} . \Lambda\) t a given instant, its angular velocity is \(10 \mathrm{rad} / \mathrm{s}\). After the wheel rotates through an angle of 100 radians, the wheel's angular velocity is \(100 \mathrm{rad} / \mathrm{s}\). Calculate (a) the torque applied on the wheel (b) the increase in rotational kinetic energy

A uniform rod of length \(l\) lies on a smooth horizontal table. A particle moving on the table strikes the rod perpendicularly at an cnd and stops. Find the distanec, travelled by the ecntre of the rod by the time it turns through an angle \(\theta\)
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