6.3 Solutions
267
d 2 x
dt 2 +
2
3
gx
(R − r )
= 0
which is the equation for SHM, with
ω
2
=
2
3
g
R − r
T =
2π
ω
= 2π
3(R − r )
2g
6.3.2 Physical Pendulums
6.31 If α is the angular acceleration, the torque τ is given by
τ = I α = I
d 2 φ
dt 2
(1)
The restoring torque for an angular displacement φ is
τ = −Mg D sin φ
(2)
which arises due to the tangential component of the weight. Equating the two
torques for small φ,
I
d 2 φ
dt 2 = −Mg D sin φ = −Mg D φ
or
d 2 φ
dt 2 +
Mg D
I
φ = 0
( 3 )
which is the equation for SHM with
ω
2
=
Mg D
I
(4)
T =
2π
ω
= 2π
I
Mg D
6.32 Equation for the oscillatory motion is obtained by putting I =
1
3
M L 2 and
D =
L
2
in (3) of prob. (6.31).