3.3 Solutions
131
3.38 Gain in kinetic energy = loss of potential energy
1
2
mv
2
= mg(h − 2R)
v
2
= 2g (h − 2R)
(1)
The force exerted on the track at the top
F =
mv 2
R
− mg
(2)
By problem
F = mg
(3)
∴
mv 2
R
− mg = mg
or v
2
= 2g R
(4)
Using (4) in (1) we find h = 3R.
3.39 Let the particle velocity at the lowest position be u = 0.8944
√
5g R and v at
point P.
Loss in kinetic energy = gain in potential energy
1
2
mu
2
−
1
2
mv
2
= mg(R + R sin θ)
or v
2
=
0.8944
5g R
2 − 2g R(1 + sin θ)
(1)
The particle would leave at P (Fig. 3.7) when
mv 2
R
= mg sin θ
or v
2
= g R sin θ
(2)
Using (2) in (1) and solving
sin θ = 2/3 or θ = 41.8 ◦
3.40 Let the minimum height be h. The velocity of the block at the beginning of
the circular track will be
v =
2gh
(1)
For completing the circular track
v =
5g R
(2)