4.3 Solutions
159
Fig. 4.23
Using (4) in (1)
a = ¨
x =
5
7
g sin θ
(5)
Thus the centre of the sphere moves with a constant acceleration. The
assumption made in the derivation is that we have pure rolling without
sliding
(b) v =
√
2as =
2 ×
5
7
× 9.8 × sin 30 ◦ × 3 = 4.58 m/s
(c) L = I ω =
2
5
M R 2 v
R
=
2
5
Mv R
=
2
5
× 0.1 × 4.58 × 0.25 = 0.0458 kg m
2
/T
(d) Using (5) in (1)
F =
2
7
Mg sin θ
(4.9)
∴
F
N
=
2
7
tan θ
(4.10)
For no slipping F/N must be less than μ, the coefficient of friction
between the surfaces in contact. Therefore, the condition for pure rolling
is that μ must exceed (2/7) tan θ .
μ =
2
7
tan θ
(4.11)
∴ tan θ =
7μ
2
=
7
2
×0.26 = 0.91
(4.12)
or θ = 42.3
◦
(4.13)
4.15 (a) Equation of motion of the cylinder for sliding down the incline is
ma s = mg sin θ − μmg cos θ
(1)
or a s = g(sin θ − μ cos θ)
(2)
159
Fig. 4.23
Using (4) in (1)
a = ¨
x =
5
7
g sin θ
(5)
Thus the centre of the sphere moves with a constant acceleration. The
assumption made in the derivation is that we have pure rolling without
sliding
(b) v =
√
2as =
2 ×
5
7
× 9.8 × sin 30 ◦ × 3 = 4.58 m/s
(c) L = I ω =
2
5
M R 2 v
R
=
2
5
Mv R
=
2
5
× 0.1 × 4.58 × 0.25 = 0.0458 kg m
2
/T
(d) Using (5) in (1)
F =
2
7
Mg sin θ
(4.9)
∴
F
N
=
2
7
tan θ
(4.10)
For no slipping F/N must be less than μ, the coefficient of friction
between the surfaces in contact. Therefore, the condition for pure rolling
is that μ must exceed (2/7) tan θ .
μ =
2
7
tan θ
(4.11)
∴ tan θ =
7μ
2
=
7
2
×0.26 = 0.91
(4.12)
or θ = 42.3
◦
(4.13)
4.15 (a) Equation of motion of the cylinder for sliding down the incline is
ma s = mg sin θ − μmg cos θ
(1)
or a s = g(sin θ − μ cos θ)
(2)