120
3 Rotational Kinematics
Dividing the two equations
tan θ =
ω 2 R
g
=
ω 2 r sin θ
g
ω =
g
r cos θ
3.17 Since the wire is continuous, tension in the parts AB and BC will be identical.
Equating the horizontal and vertical components of forces separately
mv 2
r
= T sin 30
◦
+ T sin 60
◦
(1)
mg = T cos 30
◦
+ T cos 60
◦
(2)
As the right-hand sides of (1) and (2) are identical
mv 2
r
= mg
or v =
√ gr
3.18 Resolve the centripetal force along and normal to the funnel surface, Fig. 3.13.
When the funnel rotates with maximum frequency, the cube tends to move up
the funnel, and both the weight (mg) and the frictional force (μN ) will act
down the funnel surface, Fig. 3.13. Now
N = mg cos θ + mω
2 r sin θ
Taking the upward direction as positive, equation of motion is
mω
2 r cos θ − mg sin θ − μ(mg cos θ + mω
2 r sin θ) = 0
∴ f max =
ω
2π
=
1
2π
g
r
(sin θ + μ cos θ)
(cos θ − μ sin θ)
Fig. 3.13