74
2 Particle Dynamics
Distance travelled by either mass, s = 40 cm. Time taken
t =
2s
a
=
2 × 0.4
0.49
= 1.28 s.
2.18 Equations of motion are
ma 1 = mg sin θ − μmg cos θ
(rough incline)
ma 2 = mg sin θ
(smooth incline)
∴ a 1 = (sin θ − μ cos θ)g
a 2 = g sin θ
t 1 =
2s
a 1
t 2 =
2s
a 2
∴
t 1
t 2
=
4
3
sin θ
sin θ − μ cos θ
=
sin 45 ◦
sin 45 ◦ − μ cos 45 ◦ =
1
√
1 − μ
∴ μ =
7
16
2.19 The normal reaction N = mg cos θ
Resultant downward force F = mg sin θ − μmg cos θ
Given that N = 2F
mg cos θ = 2 mg(sin θ − 0.5 cos θ)
∴ tan θ = 1 → θ = 45
◦
2.20 By prob. (2.17), each mass will have acceleration
a =
(m 1 − m 2 )g
m 1 + m 2
The heaver mass m 1 will have acceleration a 1 vertically down while the lighter
mass m 2 will have acceleration a 2 vertically up:
a 2 = −a 1
The acceleration of the centre of mass of the system will be
a CM =
m 1 a 1 + m 2 a 2
m 1 + m 2
=
(m 1 − m 2 )a 1
m 1 + m 2
∴ a CM =
(m 1 − m 2 ) 2 g
(m 1 + m 2 ) 2
2 Particle Dynamics
Distance travelled by either mass, s = 40 cm. Time taken
t =
2s
a
=
2 × 0.4
0.49
= 1.28 s.
2.18 Equations of motion are
ma 1 = mg sin θ − μmg cos θ
(rough incline)
ma 2 = mg sin θ
(smooth incline)
∴ a 1 = (sin θ − μ cos θ)g
a 2 = g sin θ
t 1 =
2s
a 1
t 2 =
2s
a 2
∴
t 1
t 2
=
4
3
sin θ
sin θ − μ cos θ
=
sin 45 ◦
sin 45 ◦ − μ cos 45 ◦ =
1
√
1 − μ
∴ μ =
7
16
2.19 The normal reaction N = mg cos θ
Resultant downward force F = mg sin θ − μmg cos θ
Given that N = 2F
mg cos θ = 2 mg(sin θ − 0.5 cos θ)
∴ tan θ = 1 → θ = 45
◦
2.20 By prob. (2.17), each mass will have acceleration
a =
(m 1 − m 2 )g
m 1 + m 2
The heaver mass m 1 will have acceleration a 1 vertically down while the lighter
mass m 2 will have acceleration a 2 vertically up:
a 2 = −a 1
The acceleration of the centre of mass of the system will be
a CM =
m 1 a 1 + m 2 a 2
m 1 + m 2
=
(m 1 − m 2 )a 1
m 1 + m 2
∴ a CM =
(m 1 − m 2 ) 2 g
(m 1 + m 2 ) 2