ME2302 Quiz 7 and 8 Work in groups up to three people Do not

ME2302: Quiz 7 and 8 Work in groups up to three people. Do not work wi th other groups. Due Tuesday Nov 29th in my mailbox by close of business. This assignment will be counted for quizzes 7 and 8. Observe the four bar mechanisan given below. This is a used for couverting rotary motion into oscillatory augular motiou. 2. Using voctor loops, find symbolic functious for the motioa of bar Cas a fuactiou of and its derivatives. Use L for the length of baa A, for the lesgth of bar B, L for the length of bar C, and d for the separation between the mounting brackets. Both Band 6 are measured poeitively CCW from the positive borizontal akis. Give functions for 6.0, in terms of unresolved dot products. Do not resolve the dot products in this step. 3. Write out the vector frame transforms for all of the frames in your analysis. Make a list of the unrewolved dot products found in your function in step 2, and use your transforms to fully resolve theee dot products. You don\'t need to rewrite the functious for aand its derivatives, just the resolutioa of the dot products found Cos(a) Sinn) only make herein. EG: h a list. Don\'t plug these dot products into your functious for and its derivatives until Step 4 mputer to generate plots of and using the following 4. Use a com Bar A haa a length of 3 inches, bar B has a leagth of 5 inches, and bar C has a length of 6 inches. The mounting brackets ated by 5 inches. There is a are separ motor on the left bracket that drives bar A at constant CCW angular velocity of 45 RPM. Let 6000 90. Run your plots from t 0 to 3 conds. Comment ou the shape of the graplu ia light of the function described in Part 1. 6. Extra Credit: Make a parametric position plot for a point located at the halfwaw point of Bar B

Solution

solution:

1)here for given four bar mechanism loop closure equation is for ABCD is given as in cloclwise numbering is as follows

DA\'=AB\'+BC\'+CD\'

on differentiating we get velocity of CD link as follows

LaWa(La^*K^)+LbWb(Lb^*K^)+LcWc(Lc^*K^)=0

where as both velocity Wb andWc are unkown hence multiplied by (Lb^*K^) and (Lc^*K^),we get both velocity as follow equation

Wc=-[La*Wa*(La^*K^)*(Lb^*K^)*K^]/[Lc*(Lc^*K^)*(Lb^*K^)*K^]

for Wb we get

Wc=-[La*Wa*(La^*K^)*(Lc^*K^)*K^]/[Lb*(Lb^*K^)*(Lc^*K^)*K^]

where accelaration obtain by differentiating velocity equation and on multiplying by (Lb^*K^)

ac=[[La*Wa^2*(La^)*(Lb^*K^)*K^+Lc*Wc^2*(Lc^)*(Lb^*K^)*K^]/[Lc*(Lc^*K^)*(Lb^*K^)*K^]]

2)for getting angle beta b and theta we have solve by chace solution here we get

DA\'-AB\'=BC\'+CD\'

5i-3j=BC\"+CD\'=R\'

here by chace method angle is given from point of intersection of two circle

y=R^2+BC^2-CD^2/2*R=5.8309^2+5^2-6^2/2*5.8309=1.9720 cm

x=4.594

from

x^2+y^2=BC^2

so BC\' and CD\' is given as

BC\'=3.937i-3.374j

BC^=.759i-.65j

CD\'=1.063i+.374j

CD^=.837i+.2945j

5)in this way angular velocity for input angle 90 and velocity

Wa=4.712 rad/s

is as follows

Wc=-3*4.712*cos(49.42+90)/6cos(49.42-19.38)=2.067 rad/s

Wb=-3*4.712*cos(-70.61+90)/5cos(-70.61+40.57)=-3.080 rad/s

where accelaration is given as

ac=3*4.712^2*cos(49.42+90)+6*2.067^2*cos(49.42-19.38)/6cos(49.42-19.38)=-5.467 rad/s2

6)in this way by using input angle for various time we can calculate particular velocity and accelaration from given equation by vector method

Wa=phi-90/time

7)in this way we can calculate input angle and respective parameter