Introductory Mathematics for Engineering Applications
Introductory Mathematics for Engineering Applications
1st Edition
ISBN: 9781118141809
Author: Nathan Klingbeil
Publisher: WILEY
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Textbook Question
Chapter 6, Problem 1P

The tip of a one-link robot is located at θ = 0 at time t = 0 s as shown in Fig. P6.1.It takes 1 for the robot to move from θ = 0 ,to θ = 2 π rad If l = 5 in., plot the x and y components as a function of the. Also find the amplitude, frequency, period, phase angle, and time shift.

Chapter 6, Problem 1P, The tip of a one-link robot is located at =0 at time t=0 s as shown in Fig. P6.1.It takes 1 for the

FIGURE P6.1 Rotating one-link robot starting at θ = 0 ° .

Expert Solution & Answer
Check Mark
To determine

To plot:

The x -component and y -component as a function of time and determine the amplitude, frequency, period, phase angle and time shift.

Answer to Problem 1P

The graph for x -componentas a function of timeis plotted is as shown in Figure 1.

The graph for y -component as a function of time is plotted is as shown in Figure 2.

The amplitude of sine and cosine function is 5, the frequency is 1 Hz, the period of the tip of a one-link robot is 1 s, the phase angle is 0 rad and the time shift of the one-link robot is 0 s.

Explanation of Solution

Given:

The tip of a one-link robot is initially located at θ=0 attime t=0 s. The given figure is shown below,

Introductory Mathematics for Engineering Applications, Chapter 6, Problem 1P , additional homework tip  1

Time taken for the robot to move form θ=0 to θ=2π rad is 1 s.

A one-link robot is length of 5 in.

Concept used:

Write the expression for the linear frequency.

  f=1T ....... (1)

Here, f is the linear frequency and T is the time period.

Write the expression for the angular frequency.

  ω=2πT ....... (2)

Here, ω is the angular frequency.

Write the expression for the time period.

  T=2πω ....... (3)

Write the expression for x -component at any time,

  x(t)=lcos(θ) ....... (4)

Here, x(t) is the expression for x -component at any time, l is the length of one-link robot, θ is the angle.

Write the expression for x -component at any time.

  y(t)=lsin(θ) ....... (5)

Here, y(t) is the expression for y -component at any time.

Write the expression for the time shift.

  t0=ϕω ....... (6)

Here, t0 is the expression for the time shift.

Calculation:

The one-link robot completes one revolution in 1 s.

Substitute 1 s for T in equation (2).

  ω=2π1 s=2π rad/s

Substitute 2π rad/s for ω in equation (3).

  T=2π2π s=1 s

Therefore, the period of the tip of a one-link robot is 1 s.

Substitute 1 s for T in equation (1).

  f=11 s=1 Hz

Therefore, the frequency is 1 Hz.

Since the one-link robot initially start form θ=0 and the phase angle is 0 rad.

Substitute ωt+ϕ for θ in equation (4).

  x(t)=lcos(ωt+ϕ) ....... (7)

Here, t is any given time and ϕ is the phase angle.

Substitute 5 in for l, 0 rad for ϕ and 2π rad/s for ω in equation (7).

  x(t)=(5)cos(( 2π)t+0)=5cos2πt

The plot for the x -component as a function of time is shown in Figure 1.

Introductory Mathematics for Engineering Applications, Chapter 6, Problem 1P , additional homework tip  2

Substitute ωt+ϕ for θ in equation (5).

  y(t)=lsin(ωt+ϕ) ....... (8)

Substitute 5 in for l, 0 rad for ϕ and 2π rad/s for ω in equation (8).

  x(t)=(5)sin(( 2π)t+0)=5sin(2πt+0)=5sin(2πt)

The plot for the y -component as a function of time is shown in Figure 2.

Introductory Mathematics for Engineering Applications, Chapter 6, Problem 1P , additional homework tip  3

Figure 2

Substitute 0 rad for ϕ and 2π rad/s for ω in equation (6).

  t0 =0 rad2π rad/s=0 s

Therefore, the time shift of the one-link robot is 0 s.

The amplitude of sine and cosine function is 5.

Conclusion:

Thus, the graph for x -component is plotted in Figure 1 and y -component as a function of time is plotted in Figure 2. The amplitude of sine and cosine function is 5, the frequency is 1 Hz, the period of the tip of a one-link robot is 1 s, the phase angle is 0 rad and the time shift of the one-link robot is 0 s.

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Chapter 6 Solutions

Introductory Mathematics for Engineering Applications

Ch. 6 - A spring-mass system is displaced x=10 cm and let...Ch. 6 - Suppose the spring-mass system of problem P6-11 is...Ch. 6 - Figure P.13 A spring-mass system for problem...Ch. 6 - The position of a spring-mass system shown in Fig....Ch. 6 - The position of a spring-mass system shown in Fig....Ch. 6 - A simple pendulum of length L=100 cm is shown in...Ch. 6 - Repeat problem P6-16 if L=10 in, and...Ch. 6 - A sinusoidal current i(t)=0.1sin(100t) amps is...Ch. 6 - A sinusoidal current i(t)=200cos(120t+14.86) mA is...Ch. 6 - A series RL circuit is subjected to a sinusoidal...Ch. 6 - A series RL circuit is subjected to a sinusoidal...Ch. 6 - A sinusoidal voltage (t)=10sin(1000t)V is applied...Ch. 6 - A parallel RL circuit is subjected to a sinusoidal...Ch. 6 - A parallel RL circuit is subjected to a sinusoidal...Ch. 6 - Consider the RC circuit shown in Fig. P6.25, where...Ch. 6 - Consider the RC circuit shown in Fig.P6.25, where...Ch. 6 - Two voltages 1(t)=10sin(100t45)V and...Ch. 6 - Repeat problem P6-27 if 1(t)=10cos(100t+90)V and...Ch. 6 - Two voltages 1(t)=102sin(500t34)V and...Ch. 6 - A pair of springs and masses vibrate under simple...Ch. 6 - Suppose the positions of the masses in problem...Ch. 6 - Two oscillating masses are connected by a spring...Ch. 6 - Now assume that the positions of two masses in...Ch. 6 - A manufacturing plant employs a heater and a...Ch. 6 - Repeat problem P6-34 if H(t)=100sin(120t)V and...Ch. 6 - In the three-phase circuit shown in Fig.P.36....Ch. 6 - In the three-phase circuit shown in Fig.P.36....Ch. 6 - In the three-phase circuit shown in Fig.P.36....Ch. 6 - The Hip implant shown in Fig. P6.39 is subjected...Ch. 6 - Repeat problem P6-3 if F(t)=15sin(10t)+75N.
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