124
3 Rotational Kinematics
Fig. 3.16
force experienced by the particle at B is mv 2 /R, where v is the velocity of the
particle at this point. Now the weight mg of the particle acts vertically down
so that its component along the radius BO is mg cos θ . So long as mg cos θ >
mv 2 /R the particle will stick to the surface. The condition that the particle
will leave the surface is
mg cos θ =
mv 2
R
(1)
or cos θ =
v 2
g R
(2)
Now, in descending from A to B, the potential energy is converted into kinetic
energy
mgh =
1
2
mv
2
(3)
or
v 2
g
= 2h
(4)
using (4) in (2)
cos θ =
2h
R
(5)
Drop a perpendicular BC on AO.
Now cos θ =
OC
OB
=
R − h
R
(6)
Combining (5) and (6), h =
R
3
Thus the particle will leave the sphere at a point whose vertical distance below
the highest point is
R
3
.
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