Math / Calculus

QuestionProblems 12.34 and 12.35 In the mechanism shown, block $B$ is fixed and has a profile described by the following relation: $y=h\left[1+\frac{1}{2}\left(\frac{x}{d}\right)^{2}-\frac{1}{4}\left(\frac{x}{d}\right)^{4}\right]$
The follower moves with the shuttle $A$ , and the tip $C$ of the follower remains in contact with $B$ . Problem 12.34 Assume that $h=0.25$ in., $d=1$ in., and the horizontal position of $C$ is $x=d \sin (w t)$ , where Page 666 $w=2 \pi \mathrm{rad} / \mathrm{s}$ , and $t$ is time in seconds. Determine an analytical expression for the speed of $C$ as a function of $x$ and the parameters $d, h$ , and $\omega$ . Then, evaluate the speed of $C$ for $x=0, x=0.5$ in., and $x=1 \mathrm{in}$ . Express your answers in $\mathrm{ft} / \mathrm{s}$ . Answer $v=w \sqrt{d^{2}-x^{2}} \sqrt{1+h^{2}\left(\frac{x}{d^{2}}-\frac{x^{3}}{d^{4}}\right)^{2}},\left.v\right|_{x=0}=0.5236 \mathrm{ft} / \mathrm{s},\left.v\right|_{x=0.5}=0.4554 \mathrm{ft} / \mathrm{s},\left.v\right|_{x=1} ^{\prime}=0$

Studdy Solution

STEP 1

1. Block $B$ follows the profile described by $y = h\left[1 + \frac{1}{2}\left(\frac{x}{d}\right)^2 - \frac{1}{4}\left(\frac{x}{d}\right)^4\right]$ .
2. The horizontal position of $C$ is given by $x = d \sin(\omega t)$ .
3. We need to find the speed of $C$ as a function of $x$ and the parameters $d$ , $h$ , and $\omega$ .
4. Given $h = 0.25$ in., $d = 1$ in., and $\omega = 2\pi$ rad/s.
5. The final speeds need to be expressed in ft/s.

STEP 2

1. Determine an expression for the horizontal velocity $\dot{x}$ of point $C$ .
2. Determine an expression for the vertical velocity $\dot{y}$ of point $C$ .
3. Combine the horizontal and vertical velocities to find the total speed of $C$ .
4. Evaluate the speed of $C$ for $x = 0$ , $x = 0.5$ in., and $x = 1$ in.

STEP 3

Determine the horizontal velocity $\dot{x}$ of point $C$ .
Given $x = d \sin(\omega t)$ , differentiate with respect to $t$ :
$\dot{x} = \frac{d}{dt} \left( d \sin(\omega t) \right) = d \omega \cos(\omega t)$
Since $\cos(\omega t) = \sqrt{1 - \sin^2(\omega t)}$ and $\sin(\omega t) = \frac{x}{d}$ , we get:
$\dot{x} = \omega \sqrt{d^2 - x^2}$

STEP 4

Determine the vertical velocity $\dot{y}$ of point $C$ .
Given $y = h \left[1 + \frac{1}{2}\left(\frac{x}{d}\right)^2 - \frac{1}{4}\left(\frac{x}{d}\right)^4\right]$ , differentiate with respect to $t$ :
$\dot{y} = \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$
First, find $\frac{dy}{dx}$ :
$\frac{dy}{dx} = h \left[ \frac{d}{dx} \left( 1 + \frac{1}{2}\left(\frac{x}{d}\right)^2 - \frac{1}{4}\left(\frac{x}{d}\right)^4 \right) \right]$
$\frac{dy}{dx} = h \left[ \frac{1}{2} \cdot \frac{2x}{d^2} - \frac{1}{4} \cdot \frac{4x^3}{d^4} \right] = h \left[ \frac{x}{d^2} - \frac{x^3}{d^4} \right]$
Now, multiply by $\dot{x}$ :
$\dot{y} = h \left[ \frac{x}{d^2} - \frac{x^3}{d^4} \right] \cdot \omega \sqrt{d^2 - x^2}$

STEP 5

Combine the horizontal and vertical velocities to find the total speed of $C$ .
The total speed $v$ is given by:
$v = \sqrt{\dot{x}^2 + \dot{y}^2}$
Substitute $\dot{x}$ and $\dot{y}$ :
$v = \sqrt{\left( \omega \sqrt{d^2 - x^2} \right)^2 + \left( h \left[ \frac{x}{d^2} - \frac{x^3}{d^4} \right] \cdot \omega \sqrt{d^2 - x^2} \right)^2}$
Simplify:
$v = \omega \sqrt{d^2 - x^2} \sqrt{1 + h^2 \left( \frac{x}{d^2} - \frac{x^3}{d^4} \right)^2}$

STEP 6

Evaluate the speed of $C$ for $x = 0$ , $x = 0.5$ in., and $x = 1$ in.
For $x = 0$ :
$v = \omega \sqrt{1 + h^2 \left( \frac{0}{d^2} - \frac{0}{d^4} \right)^2} = \omega d = 2\pi \cdot 1 = 2\pi \text{ in/s} = 0.5236 \text{ ft/s}$
For $x = 0.5$ in:
$v = 2\pi \sqrt{1 - (0.5)^2} \sqrt{1 + 0.25^2 \left( \frac{0.5}{1^2} - \frac{(0.5)^3}{1^4} \right)^2}$
$v \approx 0.4554 \text{ ft/s}$
For $x = 1$ in:
$v = 2\pi \sqrt{1 - 1^2} \sqrt{1 + 0.25^2 \left( \frac{1}{1^2} - \frac{1}{1^4} \right)^2} = 0$
Solution:
$v = \omega \sqrt{d^2 - x^2} \sqrt{1 + h^2 \left( \frac{x}{d^2} - \frac{x^3}{d^4} \right)^2}$ $\left. v \right|_{x=0} = 0.5236 \text{ ft/s}, \left. v \right|_{x=0.5} = 0.4554 \text{ ft/s}, \left. v \right|_{x=1} = 0 \text{ ft/s}$

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