/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 The velocity along a circular pa... [FREE SOLUTION] | 91Ó°ÊÓ

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The velocity along a circular pathline is given by the relation \(V=s^{5} t^{4}\), where \(V\) is the velocity in the direction of fluid motion in meters per second, \(s\) is the coordinate along the pathline in meters, and \(t\) is the time is seconds. The radius of curvature of the pathline is \(0.8 \mathrm{~m}\). Determine the components of the acceleration in the directions tangential and normal to the pathline at \(s=2.5 \mathrm{~m}\) and \(t=1.5 \mathrm{~s}\).

Short Answer

Expert verified
Tangential acceleration is approximately 1317.97 m/s². Normal acceleration is approximately 305542.00 m/s².

Step by step solution

01

Calculate the Tangential Acceleration

The tangential component of acceleration (abla_t) is given by the formula: \[ a_t = \frac{dV}{dt} \] First, express the velocity equation as a function of `t`:\[ V = s^5 t^4 \] Since the problem states we need the tangential acceleration at specific values of `s` and `t`, substitute the given value of `s = 2.5` m into the velocity equation:\[ V = (2.5)^5 t^4 = 97.65625 t^4 \] Next, take the derivative of `V` with respect to `t`:\[ \frac{dV}{dt} = \frac{d}{dt}(97.65625 t^4) = 390.625 t^3 \]Substitute `t = 1.5` s into the derivative:\[ a_t = 390.625 \times (1.5)^3 = 390.625 \times 3.375 = 1317.96875 \] Therefore, the tangential acceleration is approximately 1317.97 m/s².
02

Calculate the Normal Acceleration

The normal component of acceleration (abla_n) is given by the formula: \[ a_n = \frac{V^2}{r} \] Where `r` is the radius of curvature. Using the velocity equation:\[ V = s^5 t^4 = (2.5)^5 \times (1.5)^4 = 97.65625 \times 5.0625 = 494.384765625 \] Substitute the obtained velocity `V = 494.384765625` m/s and `r = 0.8` m into the normal acceleration equation:\[ a_n = \frac{(494.384765625)^2}{0.8} \approx \frac{244433.5938}{0.8} \approx 305541.99225 \] Therefore, the normal acceleration is approximately 305542.00 m/s².

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Key Concepts

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

Tangential Acceleration
Tangential acceleration refers to the component of acceleration that acts in the direction tangent to the path of a moving object. This acceleration occurs when there is a change in the magnitude of velocity along the pathline. In fluid mechanics, especially with curved pathlines, it's critical to understand that this component affects how quickly the velocity of the fluid particle increases or decreases along the path.

The formula for tangential acceleration is given by the derivative of velocity with respect to time, denoted as \( a_t = \frac{dV}{dt} \). This implies that any change in velocity over time directly translates to tangential acceleration. Given the relation \( V = s^5 t^4 \), once you differentiate with respect to time and plug in specific values of \( s = 2.5 \ \text{m} \) and \( t = 1.5 \ \text{s} \), you obtain a concrete measure of how rapidly the fluid's speed is changing along its path. In our example, the calculated tangential acceleration is approximately \( 1317.97 \ \text{m/s}^2 \), indicating a significant change in speed over time.
Normal Acceleration
When discussing motion along a curved path, normal acceleration becomes crucial. This type of acceleration acts perpendicular to the velocity of the pathline and directs toward the center of curvature. Unlike tangential acceleration, normal or centripetal acceleration does not contribute to a change in speed but rather changes the direction of the velocity.

The normal component is calculated using the formula \( a_n = \frac{V^2}{r} \), where \( V \) represents velocity and \( r \) the radius of curvature. Here, the relation \( V = s^5 t^4 \) was evaluated at the specific values \( s = 2.5 \ \text{m} \) and \( t = 1.5 \ \text{s} \) to find \( V \), and substituting into the formula gives us the normal acceleration. The result, a remarkably high \( 305542.00 \ \text{m/s}^2 \), highlights how sharp the curvature or how fast the movement would impact directional changes. This measure ensures that we consider how a fluid particle's velocity direction changes due to curvature.
Curved Pathline Dynamics
Curved pathline dynamics involve understanding the motion of particles along a non-linear trajectory. In fluid mechanics, pathlines describe the individual paths followed by fluid particles, and these paths often curve due to various influences like boundaries or pressured bends.

When particles move along a curved path, their dynamics are influenced by both tangential and normal accelerations. The tangential component controls the change in speed along the curve, while the normal component ensures the particle navigates the curvature itself.

In the given problem, we're working with a circular pathline motion characterized by the equation \( V = s^5 t^4 \), which captures these dynamics distinctly. By analyzing how tangential and normal components play roles, we grasp how each facet influences movement within a curved path context. This knowledge is especially useful in predicting fluid behavior when designing systems involving fluid flow, allowing engineers to account for acceleration effects in both the direction and curvature of the flow paths.

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Most popular questions from this chapter

A fire hose is to be used to direct a stream of water to a location on a building that is \(15 \mathrm{~m}\) above ground level, and the building is \(9 \mathrm{~m}\) away from the location of the nozzle. The hose has a diameter of \(100 \mathrm{~mm}\), and the nozzle exit has a diameter of \(25 \mathrm{~mm} .\) What is the minimum pressure that must exist in the hose at the entrance to the nozzle such that the water stream will be able to reach the target location?

A velocity field, \(\mathbf{v}\), is spatially uniform and varies with time according to the following relation: $$ \mathbf{v}=\left\\{\begin{array}{ll} 3 \mathbf{i}+\mathbf{j} \mathrm{m} / \mathrm{s}, & t \in[0 \mathrm{~s}, 8 \mathrm{~s}] \\ 5 \mathbf{i}-4 \mathbf{j} \mathrm{m} / \mathrm{s}, & t \in(8 \mathrm{~s}, 15 \mathrm{~s}] \end{array}\right. $$ If an injection point is located at the origin of a Cartesian coordinate system at ( \(0 \mathrm{~m}\), \(0 \mathrm{~m}\) ), sketch to scale the following at \(t=15 \mathrm{~s}\) along with their key coordinates: (a) the pathline of a particle released at the injection point at \(t=0 \mathrm{~s},(\mathrm{~b})\) the streakline of dye continuously released at the injection point starting at \(t=0 \mathrm{~s},\) and (c) the streamlines in the flow field at \(t=15 \mathrm{~s}\).

A closed \(15-\mathrm{cm}\) -diameter cylindrical tank is filled with water and is rotated at \(600 \mathrm{rpm} .\) What is the maximum pressure difference between the center and the wall of the \(\operatorname{tank} ?\)

A two-dimensional velocity field is given by $$ \mathbf{v}=5 x \mathbf{i}+2 y \mathbf{j} \mathbf{~ m} / \mathrm{s} $$ where \(x\) and \(y\) are in meters. Determine the equation of the pathline that originates at the point (1,2) . For what range of times since release does the released particle remain within the domain of \(0 \leq x \leq 100 \mathrm{~m}, 0 \leq y \leq 100 \mathrm{~m}\). Plot the pathline within this domain.

The velocity distribution for laminar flow between two infinite parallel plates (called Poiseuille flow) is given by $$ u(y)=\frac{1}{2 \mu} \frac{\mathrm{d} p}{\mathrm{~d} x}\left(y^{2}-h y\right) $$ where \(u(y)\) is the velocity at a distance \(y\) from the bottom plate, \(h\) is the vertical distance between the top and bottom plates, \(\mu\) is the dynamic viscosity, and \(\mathrm{d} p / \mathrm{d} x\) is the constant pressure gradient driving the flow, where \(\mathrm{d} p / \mathrm{d} x\) is negative. The density of the fluid is \(\rho .\) For a unit width perpendicular to the flow, determine (a) the volume flow rate, (b) the average flow velocity, and (c) the mass flow rate.

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