Moving reference frame kinematics Given Arm AB is pinned to

Moving reference frame kinematics

Given: Arm AB is pinned to ground at point A. A block that is pinned to AB at B is constrained to slide within a slot on a disk. The disk is rotating about a shaft at O with a constant rate of omega. At the instant shown, the slot is horizontal. Find: At this position, determine the angular velocity and angular acceleration of arm AB. Write your answers as vectors. Use the following parameters in your analysis: omega = 10 rad/s (CW), theta = 53.13 degree, L = 3 ft and d = 0.5 ft.

Solution

solution: here problem can be solved by vector algebra method

where link AB=R1\'=r1.r1^

where r1=3 ft and theta 1=53.13

for link Bo=R2\'=r2.r2^

where r2=.5 ft and theta 2=0

where for lonk OA=r3\'=r3.r3^

where r3=3.324 ft from trianle form and theat3=46.21

3) loop closure equation is given by

R3\'=R1\'+R2\'

4) for finding velocity we have take derivative wrt time

r3\'.r3^+r3w3(k^*r3^)=r1\'.r1^+r1w1(k^*r1^)+r2\'.r2^+r2w2(k^*r2^)

as here r1=r3= constant hence

r1\'=r3\'=0 andw3=0

hence eqquation become

r1w1(cos53.13i+sin53.13j)*k^=r2\'.r2^+r2w2(r2^*k^)

on putting value we get i part as

3*w1(cos53.13i+sin53.13j)*k=-(r2\'(cos0i+sin0j)+.5*10*(cos0i+sin0j)*k^)

where i*k=j

j*k=-i

so finally we get

1.8w1j-2.4w1i=-r2\'i-5j

1.8w1=-5

w1=-2.777 rad/s

2.4w1=r2\'

r2\'=6.666 ft/sec

4) for accelaration we have take derivative of velocity equation as follows

r1a1(k^*r1)-r1w1^2r1^=-(r2\'\'r2^+r2\'r2\'^-r2w2^2r2^)

herea2=0 and w2=constant,r2\'=0

so equation become

3a1(cos53.13i+sin53.13j)*k^-3*2.777^2(cos53.13i+sin53.13j)

=-[r2\'\'(cos0i+sin0j)+6.66(-sin0i+cos0j)-.5*10^2*(cos0i+sin0j)]

finally we get

1.8a1j-2.39a1i-13.88i-18.50j=-r2\'\'i+6.666j+50i

on equating part we get

-2.39a1-13.88=-r2\'\'+50

and 1.8a1-18.50=6.66

so we get a1=13.98 rad/sec2

r2\'\'=-97.28 ft/sec2

5) all angle measured cw as positive