6.3 Solutions
271
Potential energy of the spring
U =
1
2
kx
2
=
1
2
kr
2
θ
2
Total energy
E = K + U =
1
2
(mr
2
+ I ) ˙
θ
2
+
1
2
kr
2
θ
2
= constant
Differentiating with respect to time
dE
dt
= (mr
2
+ I ) ˙
θ · ¨
θ + kr
2
θ · ˙
θ = 0
Cancelling ˙
θ
¨
θ +
kr 2 θ
mr 2 + I
= 0
which is the equation for angular SHM with
ω
2
=
kr 2
mr 2 + I
. Therefore
ω =
kr 2
mr 2 + I
6.38 Let at any instant the centre of the cylinder be displaced by x towards right.
Then the spring at C is compressed by x while the spring at P is elongated by
2x. If v = ˙
x is the velocity of the centre of mass of the cylinder and ω = ˙
θ its
angular velocity, the total energy in the displaced position will be
E =
1
2
m ˙
x
2
+
1
2
I C ˙
θ
2
+
1
2
k 1 x
2
+
1
2
k 2 (2x)
2
(1)
Substituting x = r θ , ˙
x = r ˙
θ, and I C =
1
2
mr 2 , where r is the radius of the
cylinder, (1) becomes
E =
3
4
mr
2 ˙
θ
2
+
1
2
r
2
(k 1 + 4k 2 )θ
2
= constant
dE
dt
=
3
2
mr
2 ˙
θ ¨
θ + r
2
(k 1 + 4k 2 )θ ˙
θ = 0
∴ ¨
θ +
2
3m
(k 1 + 4k 2 )θ = 0