Answer:
v = 3.81 m/s
Explanation:
First, we will find out the radius (r)
sinΘ = r / L
or
r = LsinΘ
on substituting the values, we get
r = 1.2 × sin18°
or
r = 0.370 m
Now,
The tension (T) can be calculated as
ΣF(y) = 0
or
Tcos18° - mg = 0
or
T = mg / cos18.8°
on substituting the values, we have
T = (2kg x 9.8m/s²) / cos18°
or
T = 20.60 N
Now, applying the equilibrium condition in horizontal direction. we have
[tex]\Sigma F_x = ma_c = Tsin\theta = 0[/tex]
also, the centripetal acceleration [tex]a_c=\frac{v}{r}[/tex]
on substituting the values, we get
m(v² / r) = Tsin18°
v = (rTsin18° / m)
v = [(1.2m)(20.60 × sin18°) / 2kg]
v = 3.81 m/s
