3.3 Solutions
121
3.19 At the minimum frequency of rotation, the cube tends to go down the surface
and therefore the frictional force acts up the funnel. The equation of motion
becomes
mω
2 r cos θ − mg sin θ + μ(mg cos θ + mω
2 r sin θ) = 0
f min =
ω
2π
=
1
2π
g
r
(sin θ − μ cos θ)
(cos θ + μ sin θ)
3.20 Tension is provided by the weight Mg
T = Mg
(1)
Three forces, weight (Mg), tension (T ) and normal reaction (mω 2 r ), are to be
balanced:
T sin θ = mω
2 r
(2)
Further r = L sin θ
(3)
Combining (1), (2) and (3)
ω
2
=
Mg
m L
Frequency of rotation
f =
ω
2π
=
1
2π
Mg
m L
3.21 The two forces acting at right angles are (i) weight (mg) and (ii) reaction
(mv 2 /r ).
F =
(mg) 2 + (mv 2 /r ) 2 = mg
1 + (v 2 /gr) 2
Using v = 7 m/s, g = 9.8 m/s 2 , r = 100 m and m = 60 kg,
F = 60.075 kg wt.
3.22 Figure 3.14 shows the rear of the carriage speeding with v, negotiating a circular curve of radius r . ‘a’ is half of the distance between the wheels and h
is the height of the centre of gravity (CG) of the carriage above the ground.
The centripetal force mv 2 /r produces a counterclockwise torque about the
left wheel at A. The weight of the carriage acting vertically down through the