Question

Robots with telescoping arms are sometimes used to perform tasks (e.g., welding or placing screws) where access may be difficult for other robotic types. During a test run, a robot arm is programmed to extend according to the relationship r = 3 + 0.5cos(4θ) and the arm rotates according to the relationship θ=−π4t2+πt , where r is in feet, θ is in radians, and t is in seconds. Determine the velocity and acceleration of the robot tip A

Answer 1

Answer:

a) $v=\frac{dr}{dt}=2(\pi-8)sin(4\theta)$

b) $a=8\pi^2(1-8t)^2cos(4\theta)$

Explanation:

a) You have the following motion equation:

$r=3+0.5cos(4\theta)\\\\\theta=-\pi4t^2+\pi t$

The velocity, in polar coordinates, is given by:

$\frac{dr}{dt}=\frac{dr}{d\theta}\frac{d\theta}{dt}$

You calculate the derivatives:

$\frac{dr}{d\theta}=0.5(4)sin(4\theta)=-2sin(4\theta)\\\\\frac{d\theta}{dt}=-8\pi t+\pi\\\\v=\frac{dr}{dt}=-2\pi(1-8t)sin(4\theta)$

b) the acceleration is:

$a=\frac{dv}{dt}=\frac{dv}{d\theta}\frac{d\theta}{dt}\\\\\frac{dv}{d\theta}=-8\pi(1-8t)cos(4\theta)\\\\\frac{d\theta}{dt}=-8\pi t+\pi \\\\a=8\pi^2(1-8t)^2cos(4\theta)$