128
3 Rotational Kinematics
From energy conservation, mgh =
1
2
mv
2
∴ v =
2gh = s
g
L
3.32 In coming down from angular displacement of 60 ◦ to 45 ◦ , loss of potential
energy is given by
mg(h 1 − h 2 ) = mgL(1 − cos 60
◦
) − mgL(1 − cos 45
◦
)
= 0.207 mgL
Gain in kinetic energy =
1
2
mv
2
∴
1
2
mv
2
= 0.207 mgL = 0.207 mg (∵ L = 1 m)
or v = 2.014 m/s
The tension in the string would be
T =
mv 2
L
+ mg cos 45
◦
= mg
4.056
gL
+ 0.707
N = 1.12 mg N
3.33 When the bob is displaced through angle θ , the potential energy is mgL(1 −
cos θ). At the lowest position the energy is entirely kinetic
1
2
mv
2
= mg L (1 − cos θ)
(1)
The tension in the string will be
T = mg +
mv 2
L
= mg + 2mg (1 − cos θ)
(2)
where we have used (1)
By problem
T = 2mg
(3)
From (2) and (3) we find cos θ =
1
2
or θ = 60 ◦
3 Rotational Kinematics
From energy conservation, mgh =
1
2
mv
2
∴ v =
2gh = s
g
L
3.32 In coming down from angular displacement of 60 ◦ to 45 ◦ , loss of potential
energy is given by
mg(h 1 − h 2 ) = mgL(1 − cos 60
◦
) − mgL(1 − cos 45
◦
)
= 0.207 mgL
Gain in kinetic energy =
1
2
mv
2
∴
1
2
mv
2
= 0.207 mgL = 0.207 mg (∵ L = 1 m)
or v = 2.014 m/s
The tension in the string would be
T =
mv 2
L
+ mg cos 45
◦
= mg
4.056
gL
+ 0.707
N = 1.12 mg N
3.33 When the bob is displaced through angle θ , the potential energy is mgL(1 −
cos θ). At the lowest position the energy is entirely kinetic
1
2
mv
2
= mg L (1 − cos θ)
(1)
The tension in the string will be
T = mg +
mv 2
L
= mg + 2mg (1 − cos θ)
(2)
where we have used (1)
By problem
T = 2mg
(3)
From (2) and (3) we find cos θ =
1
2
or θ = 60 ◦