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Chapter 4: Problem 4

The coordinates of an object moving in the \(x y\) plane vary with time according to the equations \(x=-(5.00 \mathrm{m}) \sin (\omega t)\) and \(y=(4.00 \mathrm{m})-(5.00 \mathrm{m}) \cos (\omega t),\) where \(\omega\) is a constant and \(t\) is in seconds. (a) Determine the components of velocity and components of acceleration at \(t=0 .\) (b) Write expressions for the position vector, the velocity vector, and the acceleration vector at any time \(t>0 .\) (c) Describe the path of the object in an \(x y\) plot.

Short Answer

Expert verified
The velocity components at \(t=0\) are \(v_x=\omega(5.00 m)\), \(v_y=-\omega (5.00 m)\) and the acceleration components are \(a_x=0\) and \(a_y=\omega^2 (5.00 m)\). The position, velocity, and acceleration vectors at any time \(t>0\) are \(r=t\), \(v=-\omega (5.00m)\cos(\omega t)*i-\omega (5.00m)\sin(\omega t)*j\), and \(a=\omega^2 (5.00 m)sin(\omega t)*i + \omega^2 (5.00 m)cos(\omega t)*j\) respectively. The path of the object is a semi-circle at origin with radius 5.00 m.

Step by step solution

01

Express the given equations

The equations are \(x=-(5.00 m) \sin (\omega t)\) and \(y=(4.00 m)-(5.00 m) \cos (\omega t)\). Where \(x\) and \(y\) are the coordinates of the moving object in the \(x y\) plane. \(\omega\) is a constant and \(t\) is time.
02

Calculating components of velocity

Velocity is a derivative of position with respect to time. So, we differentiate \(x\) and \(y\) w.r.to time. The derivative of \(\sin(\omega t)\) is \(\omega cos(\omega t)\) and of \(\cos(\omega t)\) is \(-\omega \sin(\omega t)\). Thus, \(v_x=dx/dt=-\omega(5.00 m)cos(\omega t)\) and \(v_y=dy/dt=-\omega (5.00 m)sin(\omega t)\). At time \(t=0\), \(v_x=\omega(5.00 m)\) and \(v_y=-\omega (5.00 m)\) due to sin(0)=0 and cos(0)=1.
03

Calculating components of acceleration

Acceleration is a derivative of velocity with respect to time. So, we differentiate \(v_x\) and \(v_y\) w.r.to time. The derivative of \(cos(\omega t)\) is \(-\omega sin(\omega t)\) and of \(\sin(\omega t)\) is \(\omega cos(\omega t)\). Thus, \(a_x=dv_x/dt= \omega^2 (5.00 m)sin(\omega t)\) and \(a_y=dv_y/dt=\omega^2 (5.00 m)cos(\omega t)\). At time \(t=0\), \(a_x=0\) and \(a_y=\omega^2 (5.00 m)\) due to sin(0)=0 and cos(0)=1.
04

Writing position, velocity, and acceleration vectors

For any time \(t>0\), The position vector \(r=t\), the velocity vector \(v=v_x*i+v_y*j=rand(-\omega *(5.00m)cos(\omega t)*i-\omega *(5.00m)sin(\omega t)*j)\), and the acceleration vector \(a=a_x*i+a_y*j=\omega^2 (5.00 m)sin(\omega t)*i + \omega^2 (5.00 m)cos(\omega t)*j \).
05

Describing the path of the object

To describe the path of the object, substitute \(sin(\omega t)\) in the equation \(y=4.00m-5.00m cos(\omega t)\) by using the trigonometric identity \(cos(\omega t)=sqrt[1- sin^2(\omega t)]\). The final equation \(y= 5.00 m - sqrt[(5.00 m)^2-(x)^2]\) looks like the equation for the top half of a circle at origin with radius \(5.00 m\). Hence, the path is a semicircular trajectory.

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Key Concepts

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

Velocity Components
In kinematics, velocity components are vital for understanding how an object moves along a path. These components, derived from position equations, help in breaking down the motion along the coordinate axes, here the x and y axes. By knowing the position functions, we can differentiate these with respect to time to obtain the velocity components. For instance, given the equations \(x = -(5.00 \, \text{m}) \sin(\omega t)\) and \(y = (4.00 \, \text{m}) - (5.00 \, \text{m}) \cos(\omega t)\), the velocity components are obtained by differentiating these positions concerning time.

  • The x-component of velocity, \(v_x\), is derived to be \(v_x = -\omega (5.00 \, \text{m}) \cos(\omega t)\).
  • The y-component of velocity, \(v_y\), is obtained as \(v_y = -\omega (5.00 \, \text{m}) \sin(\omega t)\).

At time \(t = 0\), these components provide specific values \(v_x = \omega (5.00 \, \text{m})\) and \(v_y = 0\), aiding in the analysis of the object's speed and movement direction.
Acceleration Components
Acceleration, another crucial concept in kinematics, is the rate of change of velocity over time. Similar to velocity components, acceleration can be dissected into components along the axes. From the velocities derived, we differentiate them with respect to time to obtain the acceleration components. Given the velocity equations, the x and y components of acceleration are calculated as follows:

  • The x-component of acceleration, \(a_x\), is \(a_x = \omega^2 (5.00 \, \text{m}) \sin(\omega t)\).
  • The y-component, \(a_y\), is derived as \(a_y = \omega^2 (5.00 \, \text{m}) \cos(\omega t)\).
These components describe how the object鈥檚 velocity changes in both directions at any given time.

At \(t = 0\), we find \(a_x = 0\) and \(a_y = \omega^2 (5.00 \, \text{m})\), indicating that initially, the acceleration acts entirely in the y-direction, affecting how the object accelerates according to these parameters.
Circular Motion
The motion of an object along a curved path, such as a circle, is intriguing in kinematics. When analyzing circular motion via kinematics, we explore how position, velocity, and acceleration relate to a circular trajectory. The given position equations lead us to discover that the path is part of a circle.

To describe the object's path, we consider both the trigonometric functions in the position equations. The equations transform through trigonometric identities to reveal the trajectory's shape.
  • The equation \(y = 4.00 \, \text{m} - 5.00 \, \text{m} \cos(\omega t)\) simplifies to show a semicircular path centered at \((0, 4.00 \, \text{m})\) with radius \(5.00 \, \text{m}\).
The periodic functions \(\sin\) and \(\cos\) indicate cyclical movement characteristic of circular motions.

These insights suggest the object鈥檚 trajectory in the xy-plane is a semicircle, emphasizing the intertwined relationship between kinematic equations and circular motion.

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Most popular questions from this chapter

The vector position of a particle varies in time according to the expression \(\mathbf{r}=\left(3.00 \hat{\mathbf{i}}-6.00 t^{2} \hat{\mathbf{j}}\right) \mathrm{m} .\) (a) Find expres- sions for the velocity and acceleration as functions of time. (b) Determine the particle's position and velocity at \(t=1.00 \mathrm{s}\)

The water in a river flows uniformly at a constant speed of \(2.50 \mathrm{m} / \mathrm{s}\) between parallel banks \(80.0 \mathrm{m}\) apart. You are to deliver a package directly across the river, but you can swim only at \(1.50 \mathrm{m} / \mathrm{s} .\) (a) If you choose to minimize the time you spend in the water, in what direction should you head? (b) How far downstream will you be carried? (c) What If? If you choose to minimize the distance downstream that the river carries you, in what direction should you head? (d) How far downstream will you be carried?

Young David who slew Goliath experimented with slings before tackling the giant. He found that he could revolve a sling of length \(0.600 \mathrm{m}\) at the rate of 8.00 rev/s. If he increased the length to \(0.900 \mathrm{m},\) he could revolve the sling only 6.00 times per second. (a) Which rate of rotation gives the greater speed for the stone at the end of the sling? (b) What is the centripetal acceleration of the stone at 8.00 rev/s? (c) What is the centripetal acceleration at \(6.00 \mathrm{rev} / \mathrm{s} ?\)

A train slows down as it rounds a sharp horizontal turn, slowing from \(90.0 \mathrm{km} / \mathrm{h}\) to \(50.0 \mathrm{km} / \mathrm{h}\) in the \(15.0 \mathrm{s}\) that it takes to round the bend. The radius of the curve is \(150 \mathrm{m} .\) Compute the acceleration at the moment the train speed reaches \(50.0 \mathrm{km} / \mathrm{h} .\) Assume it continues to slow down at this time at the same rate.

Two swimmers, Alan and Beth, start together at the same point on the bank of a wide stream that flows with a speed v. Both move at the same speed \(c(c>v),\) relative to the water. Alan swims downstream a distance \(L\) and then \(\mathrm{up}^{-}\) stream the same distance. Beth swims so that her motion relative to the Earth is perpendicular to the banks of the stream. She swims the distance \(L\) and then back the same distance, so that both swimmers return to the starting point. Which swimmer returns first? (Note: First guess the answer.)
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