4.3 Solutions
161
(b) K
K 1
=
K 2 − K 1
K 1
=
K 2
K 1
− 1
=
(1/2) I 2 ω 2
2
(1/2) I 1 ω 2
1
− 1 = (0.8)
5π
4π
2
− 1 =
1
4
4.18 At the bottom of the incline translational energy is (1/2) Mv 2
0 while the rotational energy is
1
2
I ω
2
=
1
2
×
2
5
M R
2 v 2
0
R 2 =
1
5
Mv
2
0
Total initial kinetic energy =
1
2
Mv
2
0 +
1
5
Mv
2
0 =
7
10
Mv
2
0
Let the sphere reach a distance s up the incline or a height h above the bottom
of the incline. Taking potential energy at the bottom of the incline as zero, the
potential energy at the highest point reached is Mgh. Since the entire kinetic
energy is converted into potential energy, conservation of energy gives
7
10
Mv
2
0 = Mgh
But h = s sin θ , so that
s =
7
10
v 2
0
g sin θ
4.19 Equation of motion is
Ma = mg − T
(1)
The resultant torque τ on the wheel is TR and the moment of inertia is
(1/2) M R 2 .
Now τ = I α
∴ T R =
1
2
M R
2 a
R
or T =
1
2
Ma
(2)
Solving (1) and (2)
a =
2mg
M + 2m
T =
Mmg
M + 2m