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 $\displaystyle \
u \le \sqrt {ag}
\$
Let $\displaystyle \
R_1
\$ be the reaction force from the point of projection.
$\displaystyle \
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:
$\displaystyle \
\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.
