3.3 Solutions
117
(ii) v max =
√
0.85 × 9.8 × 150 = 35.35 m/s
(iii) a N =
v 2
R
=
(35.35) 2
150
= 8.33 m/s towards the centre of the circle
(iv) tan θ =
v 2
g R
=
(35.35) 2
9.8 × 150
= 0.85
∴ θ = 40.36
◦
3.10 Equating the horizontal component of the tension to the centripetal force
T sin α = mω
2 R
(1)
Furthermore, the bob has no acceleration in the vertical direction.
T cos α = mg
(2)
tan α =
R
H
=
ω 2 R
g
∴ ω =
g
H
3.11 Using the results of prob. (3.10), the difference in the level of the bob
H = H 1 − H 2 = g
1
ω 2
1
−
1
ω 2
2
(1)
ω 1 = 2π f 1 =
140
60
π
(2)
ω 2 = 2π f 2 =
160
60
π
(3)
Using (2) and (3) in (1) and g = 980 cm, H = 31.95 cm.
3.12 The centripetal force acting on the bob of the pendulum = mω 2 r , where r is
the distance of the bob from the axis of rotation, Fig. 3.10. For equilibrium, the
vertical component of the tension in the string of the pendulum must balance
the weight of the bob
∴ T cos θ 0 = mω
2 r
(1)
Further, the horizontal component of the tension in the string must be equal to
the centripetal force.
∴ T sin θ 0 = mω
2 r
(2)
Dividing (2) by (1)