
347Motion in Non-Inertial Systems
There are three forces acting on the particle:
(1) The weight of the particle
(2) The tension in the string
(3) The centrifugal force
−××=−××−
mrmk kb ib k[( )] (
ˆ
)(
ˆ
)(sin
ˆ
cos
ˆ
)
ωω ωω
= mb iωθ
2
sin
ˆ
.
When the particle is in equilibrium, the net force acting on it is zero:
−− ++ =mgkT iT km i
ˆ
sin
ˆ
cos
ˆ
sin
ˆ
θθωθ
2
0
or, by rearrangement,
(sin sin)
ˆ
(cos )
mb Ti
gkωθ θθ
2
0−+
from which it follows that
mω
2
b sin θ − T sin θ = 0
T cos θ − mg = 0.
Solving these two equations, we find
T = mω
2
b
and
θ = cos
–1
(g/ω
2
b).
Example 11.3: Surface of a Rotating Liquid
A bucket of liquid spins with ...