162
4 Rotational Dynamics
4.20 Equation of motion is
Ma = Mg − T
(1)
Torque τ = T R = I α =
1
2
M R
2 a
R
(2)
∴ T =
1
2
Ma
(3)
Solving (1) and (3) a =
2
3
g T =
Mg
3
4.21 (a) Obviously m 1 moves down and m 2 up with the same acceleration ‘a’ if
the string is taut. Let the tension in the string be T 1 and T 2 (Fig. 4.5). The
equations of motion are
m 1 a = m 1 g − T 1
(1)
m 2 a = T 2 − m 2 g
(2)
Taking moments about the axis of rotation O
T 1 R − T 2 R = I α =
M R 2
2
α
(3)
where α is the angular acceleration of the pulley and I is the moment of
inertia of the pulley about the axis through O.
But α =
a
R
∴ T 1 − T 2 =
Ma
2
(4)
Adding (1) and (2)
(m 1 + m 2 )a = T 2 − T 1 + (m 1 − m 2 )g
(5)
Using (4) in (5) and solving for ‘a’, we find
a =
(m 1 − m 2 )g
m 1 + m 2 + (1/2) M
(6)
(b) α =
a
R
=
(m 1 − m 2 )g
(m 1 + m 2 + (1/2) M) R
(c) Using (5) in (1) and (2), the values of T 1 and T 2 can be obtained from
which the ratio T 1 /T 2 can be found.
4 Rotational Dynamics
4.20 Equation of motion is
Ma = Mg − T
(1)
Torque τ = T R = I α =
1
2
M R
2 a
R
(2)
∴ T =
1
2
Ma
(3)
Solving (1) and (3) a =
2
3
g T =
Mg
3
4.21 (a) Obviously m 1 moves down and m 2 up with the same acceleration ‘a’ if
the string is taut. Let the tension in the string be T 1 and T 2 (Fig. 4.5). The
equations of motion are
m 1 a = m 1 g − T 1
(1)
m 2 a = T 2 − m 2 g
(2)
Taking moments about the axis of rotation O
T 1 R − T 2 R = I α =
M R 2
2
α
(3)
where α is the angular acceleration of the pulley and I is the moment of
inertia of the pulley about the axis through O.
But α =
a
R
∴ T 1 − T 2 =
Ma
2
(4)
Adding (1) and (2)
(m 1 + m 2 )a = T 2 − T 1 + (m 1 − m 2 )g
(5)
Using (4) in (5) and solving for ‘a’, we find
a =
(m 1 − m 2 )g
m 1 + m 2 + (1/2) M
(6)
(b) α =
a
R
=
(m 1 − m 2 )g
(m 1 + m 2 + (1/2) M) R
(c) Using (5) in (1) and (2), the values of T 1 and T 2 can be obtained from
which the ratio T 1 /T 2 can be found.