Chapter 15.2, Problem 15.59P

The test rig is shown was developed to perform fatigue testing on fitness trampolines. A motor drives the 9-in.-radius flywheel AB, which is pinned at its center point A, in a counterclockwise direction. The flywheel is attached to slider CD by the 18-in. connecting rod BC. Knowing that the a “feet” at D should hit the trampoline twice every second, at the instant when $\theta =0°$ , determine (a) the angular velocity of the connecting rod BC, (b) the velocity of D, (c) the velocity of midpoint CB.

Fig. P15.59

To determine

(a)

The angular velocity of the connecting rod BC

Answer to Problem 15.59P

The angular velocity of connecting rod BC is $3.485rad/s$ clockwise

Explanation of Solution

Given information:

Radius of flywheel is $9in$

The length of connecting rod BC is $18in$

The absolute value of point A

$v_A=v_B+v_{A/B}$

The relative velocity of A with respect to B is defined as

$v_{A/B}=\omega k\times r_{A/B}$

$r_{A/B}$ - Position vector of A with respect to B

$\omega$ - Angular velocity

Calculation:

The angular velocity of AB

${\omega }_{AB}=\frac{2rev}{s}\left(\frac{2\pi }{rev}\right)=4\pi rad/s$

According to the geometry

$9\cos \theta =18\cos \alpha$

Therefore

$\alpha =64.34°$

The position vector of B relative to A

$r_{B/A}=\left(9\cos \theta \right)i+\left(9sin\theta \right)j$

Substitute

$r_{B/A}=7.794i+4.5j$

The position vector of C relative to B

$r_{C/B}=-\left(18\cos \alpha \right)i+\left(18sin\alpha \right)j$

Substitute

$r_{C/B}=-7.794i+16.224j$

The absolute velocity of $v_B$

$v_B=v_A+{\omega }_{AB}k\times r_{B/A}$

The absolute velocity of $v_C$

$v_C=v_B+{\omega }_{BC}k\times r_{C/B}$

But we know that

$\begin{array}{l}{\left(v_C\right)}_x=0\\ v_D=v_C\\ v_A=0\end{array}$

According to above equations

$v_C={\omega }_{AB}k\times r_{B/A}+{\omega }_{BC}k\times r_{C/B}$

Substitute

${\left(v_C\right)}_{y}j=4\pi k\times \left(7.794i+4.5j\right)+{\omega }_{BC}k\times \left(-7.794i+16.224j\right)\to \left(1\right)$

Equate components

$i:0=-18\pi -16.225{\omega }_{BC}$

Therefore

${\omega }_{BC}=-3.485rad/s$

Conclusion:

The angular velocity of connecting rod BC is $3.485rad/s$ clockwise

To determine

(b)

The velocity of D

Answer to Problem 15.59P

$v_D=125.107in/s\uparrow$

Explanation of Solution

Given information:

Radius of flywheel is $9in$

The length of connecting rod BC is $18in$

The absolute value of point A

$v_A=v_B+v_{A/B}$

The relative velocity of A with respect to B is defined as

$v_{A/B}=\omega k\times r_{A/B}$

$r_{A/B}$ - Position vector of A with respect to B

$\omega$ - Angular velocity

Calculation:

According to sub part a

We have found

${\omega }_{BC}=-3.485rad/s$

And

${\left(v_C\right)}_{y}j=4\pi k\times \left(7.794i+4.5j\right)+{\omega }_{BC}k\times \left(-7.794i+16.224j\right)\to \left(1\right)$

Equate components

$j:{\left(v_C\right)}_y=97.945-{\omega }_{BC}\left(-7.794\right)$

Therefore

${\left(v_C\right)}_y=125.107in/s$

We know that

$\begin{array}{l}{\left(v_C\right)}_x=0\\ v_D=v_C\end{array}$

Therefore

$v_D=125.107in/s\uparrow$

Conclusion:

The velocity of point D is $125.107in/s\uparrow$

To determine

(c)

The velocity of mid-point CB

Answer to Problem 15.59P

The velocity of mid-point of CB is $115.047in/s$ at $75.77°$ with horizontal.

Explanation of Solution

Given information:

Radius of flywheel is $9in$

The length of connecting rod BC is $18in$

The absolute value of point E

$v_E=v_C+v_{E/C}$

The relative velocity of E with respect to C is defined as

$v_{E/C}={\omega }_{BC}k\times r_{E/C}$

$r_{E/C}$ - Position vector of E with respect to C

$\omega$ - Angular velocity

Calculation:

Assume E as the midpoint of CB

The position vector of E relative to C

$r_{E/C}=\left(9\cos \alpha \right)i-\left(9sin\alpha \right)j$

Substitute

$r_{E/C}=3.897i-8.112j$

According to sub part a

${\omega }_{BC}=3.485rad/s$

The velocity of point E

$v_E=v_C+{\omega }_{BC}k\times r_{E/C}$

Substitute

$\begin{array}{l}{v}_E=125.107j-3.485k\times \left(3.897i-8.112j\right)\\ =-28.27i+111.52j\end{array}$

The magnitude of the velocity

$\begin{array}{l}{v}_E=\sqrt{{\left(-28.27\right)}^2+{\left(111.52\right)}^2}\\ v_E=115.047in/s\end{array}$

The angle $\beta$

$\begin{array}{l}\tan \beta =\frac{111.52}{28.27}\\ \beta =75.77°\end{array}$

Conclusion:

The velocity of mid-point of CB is $115.047in/s$ at $75.77°$ with horizontal.