Answer: Ф = 17.2657 ā 17°
Explanation:
we simply apply ET =0 about the ending of the rod
so In.g.L/2sinФ - In.a.L/2cosФ = 0
g.sinФ - a.cosФ = 0
g.sinФ = a.cosФ
ⓠtanФ = a/g
Ф = Ā tanā»Ā¹ a / g
Ф = tanā»Ā¹ ( 10 / 32.17405)
Ф = tanā»Ā¹ 0.31080948777
Ф = 17.2657 ā 17°
Therefore the angle of rotation of the rod during this acceleration is 17.2657 ā 17°