see the attached figure carefully to get an idea of the forces acting on the ball and why they have such magnitudes.. (ask if there is any problem in understanding this !!)
NOTE:- I am discussing this problem from the frame of the ball (accelerated frame), so i have taken the centrifugal force (pseudo force) on it..
Here symbols denote usual meaning and f = friction. I have taken the direction of friction upwards 'coz i m taking the ball to have a tendency to fall down..
NOTE:- Slipping may take place upwards or downwards according as the value of
changes ... and this is the base of this question..
For tendency to slip downward, frction acts upwards and taking this particular case, we write the equation for the body to remain at rest..
mgsin
= m
2Rsin
.cos
+
( mgcos
+ m
2Rsin2
)
note: normal reaction = mgcos
+ m
2Rsin2
Simplify this and u'll get
= { g( sin
-
cos
) / Rsin
( cos
+
sin
) }1/2.
Now, if
starts increasing, centrifugal force increases, and so the forces trying to slip the body upwards increases.. This causes tendency to slip upwards and thus friction this time acts downwards.
For tendency to slip upward, frction acts downwards and taking this particular case, we write the equation for the body to remain at rest..
mgsin
+
( mgcos
+ m
2Rsin2
) = m
2Rsin
.cos
Simplify this and u'll get
= { g( sin
+
cos
) / Rsin
( cos
-
sin
) }1/2.
ALTERNATIVELY:- Do it like this..
When
increases to its maximum value for which the ball remains stationary, only the direction of friction is reversed. So replace
by ( -
). This will give u the answer.. (no need write another equ^n in this case)