118
3 Rotational Kinematics
Fig. 3.10
tan θ 0 =
ω 2 r
g
=
4π 2 n 2 r
g
(3)
where n = number of rotations per second. From the geometry of Fig. 3.10,
r = R + L sin θ 0
n =
1
2π
g tan θ 0
R + L sin θ 0
3.13 The equilibrium condition requires that the centripetal force = the frictional
force, mω 2 r = μ mg
∴ f max =
ω
2π
=
1
2π
μg
r
3.14 Let the spring length be stretched by x. Equating the centripetal force to the
spring force
mω
2
(L 0 + x) = kx
∴ x =
mω 2 L 0
k − mω 2
Therefore, the new length L will be
L = L 0 + x =
k L 0
k − mω 2
and the tension in the spring will be
m ω
2 L =
m ω 2 k L 0
k − mω 2
3 Rotational Kinematics
Fig. 3.10
tan θ 0 =
ω 2 r
g
=
4π 2 n 2 r
g
(3)
where n = number of rotations per second. From the geometry of Fig. 3.10,
r = R + L sin θ 0
n =
1
2π
g tan θ 0
R + L sin θ 0
3.13 The equilibrium condition requires that the centripetal force = the frictional
force, mω 2 r = μ mg
∴ f max =
ω
2π
=
1
2π
μg
r
3.14 Let the spring length be stretched by x. Equating the centripetal force to the
spring force
mω
2
(L 0 + x) = kx
∴ x =
mω 2 L 0
k − mω 2
Therefore, the new length L will be
L = L 0 + x =
k L 0
k − mω 2
and the tension in the spring will be
m ω
2 L =
m ω 2 k L 0
k − mω 2