Circular Motion

Quote:

Originally Posted by free_to_fly

I would like some help on this question please, just spent the last hour trying to figure it out.

A smooth hemispherical bowl of radius a has its rim horizontal. A particle P os projected downwards with speed u from a point on the surface of the bowl 0.5a below the level of the rim. If P remains in contact with the bowl during the subsquent motion, show that $\ u \le \sqrt {ag} \$

Let $\ R_1 \$ be the reaction force from the point of projection.
$\ R_2 \$ be the reaction force at the lowest point, eg bottom of the bowl.
Let v be the speed of P when passing the lowest point.

I've used PE lost=KE gained and F=ma to come up with 3 equations:
$\ \begin{array}{l} ag = v^2 - u^2 \\ R_1 = \frac{{mu^2 }}{a} + \frac{{mg}}{2} \\ R_2 = \frac{{mv^2 }}{a} + mg \\ \end{array} \$
But I'm not sure what to do after this. Any help would be loved.

The initial KE is (1/2)mu^2. If it remains in contact with the bowl the gain
in PE at the rim is greater than or equal to the initial KE (that is it has insufficient energy
to pass the rim).

The additional PE at the rim is mga/2, so we have:

mg (a/2) >= (1/2) m u^2

or:

g a >= u^2

or u <= sqrt(g a).

RonL