Answer:
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
[tex]r = \dfrac{2}{3} \sqrt {160^2 - 80^2}[/tex]
r = 0.667 × 138.564
r = 92.42 mm
r = 0.09242 m
However; the angular velocity can be determined by using the expression:
[tex]a_n = \theta'^2 r[/tex]
where;
[tex]a_n[/tex] = normal acceleration = 68 m/s²
r = 0.09242 m
[tex]\theta'^2 =[/tex] angular velocity = ???
[tex]68 = \theta'^2 * 0.09242[/tex]
[tex]\dfrac{68}{0.09242} = \theta'^2[/tex]
θ'² = 735.771478
θ' = [tex]\sqrt{735.771478}[/tex]
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:
[tex]a_t = r \theta"[/tex]
where;
[tex]a_t[/tex] = tangential components of acceleration = 22 m/s²
r = 0.09242 m
[tex]\theta"[/tex] = angular acceleration
[tex]22= 0.09242* \theta"[/tex]
[tex]\dfrac{22}{0.09242}= \theta"[/tex]
[tex]\theta"= \dfrac{22}{0.09242}[/tex]
θ'' = 238.044 rad/s²
Thus; the angular acceleration is θ'' = 238.044 rad/s²
