The plate OAB forms an equilateral triangle which rotates counterclockwise with increasing speed about point O. If the normal and tangential components of acceleration of the centroid C at a certain instant are 68 m/s2 and 22 m/s2, respectively, determine the values of and at this same instant. The angle θ is the angle between line AB and the fixed horizontal axis and the distance b = 160 mm.

the angular velocity θ' = 27.125 rad/s

the angular acceleration is θ'' = 238.044 rad/s²

Explanation:

The plate OAB forms an equilateral triangle which rotates counterclockwise with increasing speed about point O.

If the normal and tangential components of acceleration of the centroid C at a certain instant are 68 m/s2 and 22 m/s2

From the distance (radius r) of the centroid C in the centre of the equilateral triangle to the point of rotation O; the position OC can be calculated as:

r = OC

r = 0.667 × 138.564

r =  92.42 mm

r = 0.09242 m

However; the angular velocity can be determined by using the expression:

where;

= normal acceleration = 68 m/s²

r = 0.09242 m

angular velocity  = ???

θ'² = 735.771478

θ' =

θ' = 27.125 rad/s

Thus; the angular velocity θ' = 27.125 rad/s

Similarly ; the angular acceleration can be determined as by the following relation:

where;

= tangential components of acceleration = 22 m/s²

r = 0.09242 m

= angular acceleration

θ'' = 238.044 rad/s²

Thus; the angular acceleration is θ'' = 238.044 rad/s²

trig 15.3 cos

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