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Chapter 2: Problem 123

During a short interval the slotted guides are designed to move according to \(x=16-12 t+4 t^{2}\) and \(y=2+15 t-3 t^{2},\) where \(x\) and \(y\) are in millimeters and \(t\) is in seconds. At the instant when \(t=2 \mathrm{s}\) determine the radius of curvature \(\rho\) of the path of the constrained pin \(P\)

Short Answer

Expert verified
The radius of curvature at \( t = 2\text{s} \) is approximately 2.604 mm.

Step by step solution

01

Differentiate x and y with respect to t

To find the radius of curvature \( \rho \) at the given time \( t = 2 \text{s} \), we must first calculate the velocity components, \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). Differentiating \( x = 16 - 12t + 4t^2 \) with respect to \( t \), we get \( \frac{dx}{dt} = -12 + 8t \). Similarly, differentiating \( y = 2 + 15t - 3t^2 \) gives \( \frac{dy}{dt} = 15 - 6t \).
02

Differentiate velocity components to find acceleration components

Next, differentiate \( \frac{dx}{dt} = -12 + 8t \) with respect to \( t \) to find \( \frac{d^2x}{dt^2} = 8 \). Similarly, for \( \frac{dy}{dt} = 15 - 6t \), differentiating gives \( \frac{d^2y}{dt^2} = -6 \). These are the acceleration components.
03

Evaluate velocity and acceleration components at t = 2s

Plug \( t = 2 \text{s} \) into the velocity and acceleration expressions to calculate their values at this specific time. For velocity: \( \frac{dx}{dt} = -12 + 8(2) = 4 \) mm/s and \( \frac{dy}{dt} = 15 - 6(2) = 3 \) mm/s. For acceleration: \( \frac{d^2x}{dt^2} = 8 \) mm/s虏 and \( \frac{d^2y}{dt^2} = -6 \) mm/s虏.
04

Use radius of curvature formula

Compute the speed \( v \) using \( v = \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } = \sqrt{4^2 + 3^2} = \sqrt{25} = 5 \) mm/s. The formula for the radius of curvature \( \rho \) is \( \rho = \frac{v^3}{| \frac{dx}{dt} \frac{d^2y}{dt^2} - \frac{dy}{dt} \frac{d^2x}{dt^2} |} \). Substituting the values: \( \rho = \frac{5^3}{|4(-6) - 3(8)|} = \frac{125}{| -24 - 24 |} = \frac{125}{48} = 2.604 \text{ mm} \).

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus that helps us understand how things change. It's like having a mathematical magnifying glass to see how a function's output moves as its input changes. Here, we are dealing with two equations that describe the motion of a pin along a path defined by:
  • \(x = 16 - 12t + 4t^2\)
  • \(y = 2 + 15t - 3t^2\)
To compute how fast these are changing with respect to time \(t\), we differentiate each equation with respect to \(t\).
For the \(x\) equation, differentiation gives us the velocity component in the \(x\) direction, \(\frac{dx}{dt} = -12 + 8t\).
For the \(y\) equation, it provides the \(y\) direction velocity component, \(\frac{dy}{dt} = 15 - 6t\).
Differentiating gives us insights into the behavior of each component of the path the pin is taking. Differentiation helps locate critical points, understand curve shapes, and analyze motion.
Velocity Components
In this exercise, the velocity components represent how fast the pin is moving along the \(x\) and \(y\) axes at any given time. Once we differentiate each component of the path, we are essentially finding these velocity components.
For \(t = 2 \text{s}\), substituting this into the velocity expressions gives us:
  • \(\frac{dx}{dt} = -12 + 8(2) = 4\) mm/s
  • \(\frac{dy}{dt} = 15 - 6(2) = 3\) mm/s
This means our pin moves 4 millimeters per second along the \(x\) axis and 3 millimeters per second along the \(y\) axis at exactly 2 seconds.
Understanding these components is crucial as they are direct inputs for more complex calculations of dynamics, like finding the radius of curvature and determining how different forces affect the object in motion.
Acceleration Components
Acceleration components give us information about how the velocity of the pin is changing over time. It's like a second-level investigation, where we differentiate the velocity equations we obtained from the first differentiation.By differentiating the velocity components with respect to \(t\), we arrive at the acceleration components:
  • \(\frac{d^2x}{dt^2} = 8\) mm/s虏
  • \(\frac{d^2y}{dt^2} = -6\) mm/s虏
For \(t = 2 \text{s}\), this tells us that along the \(x\) direction, the pin's speed is increasing at 8 mm/s虏, while along the \(y\) direction, its speed is decreasing at 6 mm/s虏.
These components are pivotal for calculating dynamics effectively, allowing us to understand how forces impact movement and enabling precise computations like finding the radius of curvature through more advanced mathematical formulas.

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Most popular questions from this chapter

A particle \(P\) is launched from point \(A\) with the initial conditions shown. If the particle is subjected to aerodynamic drag, compute the range \(R\) of the particle and compare this with the case in which aerodynamic drag is neglected. Plot the trajectories of the particle for both cases. Use the values \(v_{0}=\) \(65 \mathrm{m} / \mathrm{s}, \theta=35^{\circ},\) and \(k=4.0 \times 10^{-3} \mathrm{m}^{-1}\). (Note: The acceleration due to aerodynamic drag has the form \(\mathbf{a}_{D}=-k v^{2} \mathbf{t},\) where \(k\) is a positive constant, \(v\) is the particle speed, and \(t\) is the unit vector associated with the instantaneous velocity \(\mathbf{v}\) of the particle. The unit vector \(t\) has the form \(t=\frac{v_{x} \mathbf{i}+v_{y} \mathbf{j}}{\sqrt{v_{x}^{2}+v_{y}^{2}}},\) where \(v_{x}\) and \(v_{y}\) are the instantaneous \(x\) -and \(y\) -components of particle velocity, respectively.)

At a football tryout, a player runs a 40 -yard dash in 4.25 seconds. If he reaches his maximum speed at the 16 -yard mark with a constant acceleration and then maintains that speed for the remainder of the run, determine his acceleration over the first 16 yards, his maximum speed, and the time duration of the acceleration.

The main elevator \(A\) of the CN Tower in Toronto rises about \(350 \mathrm{m}\) and for most of its run has a constant speed of \(22 \mathrm{km} / \mathrm{h}\). Assume that both the acceleration and deceleration have a constant magnitude of \(\frac{1}{4} g\) and determine the time duration \(t\) of the elevator run.

A drop of water falls with no initial speed from point \(A\) of a highway overpass. After dropping \(6 \mathrm{m}\) it strikes the windshield at point \(B\) of a car which is traveling at a speed of \(100 \mathrm{km} / \mathrm{h}\) on the horizontal road. If the windshield is inclined \(50^{\circ}\) from the vertical as shown, determine the angle \(\theta\) relative to the normal \(n\) to the windshield at which the water drop strikes.

A block of mass \(m\) rests on a rough horizontal surface and is attached to a spring of stiffness \(k .\) The coefficients of both static and kinetic friction are \(\mu\) The block is displaced a distance \(x_{0}\) to the right of the unstretched position of the spring and released from rest. If the value of \(x_{0}\) is large enough, the spring force will overcome the maximum available static friction force and the block will slide toward the unstretched position of the spring with an acceleration \(a=\mu g-\frac{k}{m} x,\) where \(x\) represents the amount of stretch (or compression) in the spring at any given location in the motion. Use the values \(m=5 \mathrm{kg}, k=150 \mathrm{N} / \mathrm{m}, \mu=0.40,\) and \(x_{0}=200 \mathrm{mm}\) and determine the final spring stretch (or compression \(x_{f}\) when the block comes to a complete stop.
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