Answer to Question #200279 in Mechanical Engineering for Muhammad Idrees

Question #200279

The engine mechanism shown in Fig. 8.38 has crank OB = 50 mm and length of connecting rod A B = 225 mm. The centre of gravity of the rod is at G which is 75 mm from B. The engine speed is 200r.p.m. For the position shown, in which OB is turned 45° from O A, Find 1. the velocity of G and the angular velocity of A B, and 2. the acceleration of G and angular acceleration of A B.


1
Expert's answer
2021-06-07T05:43:03-0400

BG=75 mm

"N_B=200 rpm \\implies \\omega_{BO}=\\frac{2 \\pi N_D}{60}=\\frac{2 \\pi*200}{60}=20.94 rad\/s"

"V_{BO}=\\omega _{BO}*BO=20.94*0.05=1.0472 m\/s"

The velocity of link BO (crank) on velocity diagram will be "BO=v_{BO}*50=1.0472*50=52.36 mm"

"\\frac{bg}{ba}=\\frac{BG}{BA} \\implies bg=\\frac{75}{225}*37.5=12.5 mm"

This can be drawn on the diagram considering the following values.

"oa =42.92; ba=37.5; V_{oa}=\\frac{oa}{50}=\\frac{42.92}{50}=0.858 m\/s"

"v_g=\\frac{46.14}{50}=0.923m\/s; V_{BA}=\\frac{37.5}{50}=0.75 m\/s"

Radial accelertion of the components

"OB =a^rob=\\frac{1.0472^2}{0.05}=21.93 m\/s^2 ; AB =a^rab=\\frac{0.75^2}{0.22}=2.5 m\/s^2"

"\\frac{b'g'}{b'a'}=\\frac{BG}{BA} \\implies b'g'= \\frac{75}{225}*77.35=25.784 mm"

For acceleration diagram

"o'g' =93.31 mm " then "ag=\\frac{93.31}{5}=18.662 m\/s^2"

Angular accleartion of AB

"a^t_{AB}=\\frac{76.33}{5}=15.266 m\/s ^2"

"\\alpha_{AB}=\\frac{15.266}{0.255}=67.85 rad\/s"


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