132
3 Rotational Kinematics
From (1) and (2)
h = 2.5R
3.41 Let the nail be located at D at distance x vertically below A, the point of
suspension, Fig. 3.21. Initially the bob of the pendulum is positioned at B at
height h above the equilibrium position C.
h = L(1 − cos θ) = L(1 − cos 60
◦
) =
1
2
L
(1)
Fig. 3.21
where L = 1 m is the length of the pendulum. When the pendulum is released
its velocity at C will be
v =
2gh =
gL
(2)
The velocity needed at C to make complete revolution in the vertical circle
centred at the nail and radius r is
v =
5gr
(3)
From (2) and (3)
r =
1
5
L
(4)
Therefore x = AD = AC − DC = L −
L
5
= 0.8 L
= 0.8 × 1 m = 80 cm
3.42 If M and m are the mass of the test tube and cork, respectively, and their
velocity V and υ respectively, momentum conservation gives
3 Rotational Kinematics
From (1) and (2)
h = 2.5R
3.41 Let the nail be located at D at distance x vertically below A, the point of
suspension, Fig. 3.21. Initially the bob of the pendulum is positioned at B at
height h above the equilibrium position C.
h = L(1 − cos θ) = L(1 − cos 60
◦
) =
1
2
L
(1)
Fig. 3.21
where L = 1 m is the length of the pendulum. When the pendulum is released
its velocity at C will be
v =
2gh =
gL
(2)
The velocity needed at C to make complete revolution in the vertical circle
centred at the nail and radius r is
v =
5gr
(3)
From (2) and (3)
r =
1
5
L
(4)
Therefore x = AD = AC − DC = L −
L
5
= 0.8 L
= 0.8 × 1 m = 80 cm
3.42 If M and m are the mass of the test tube and cork, respectively, and their
velocity V and υ respectively, momentum conservation gives