M7 mechanics: mass on smooth sphere

Mechanics problem, TeeEm style.

A mass $m$ is placed on the surface of a smooth hemisphere, radius $R$, at a height $R \frac{\sqrt{3}}{2}$ from the base of the hemisphere and is given an initial horizontal velocity of $U$.

Find an expression for the normal reaction $N(\theta)$ of the hemisphere on the mass when it has fallen to a height $R\cos\theta$, where $\theta$ is the polar angle measured from the vertical. (So that $\theta=0$ corresponds to the top of the hemisphere)

(I think that I've got the correct result for this - it was messier than I'd anticipated)

sorry teaching today

I'm happy to wait.

Mechanics problem, TeeEm style.

A mass $m$ is placed on the surface of a smooth hemisphere, radius $R$, at a height $R \frac{\sqrt{3}}{2}$ from the base of the hemisphere and is given an initial horizontal velocity of $U$.

Find an expression for the normal reaction $N(\theta)$ of the hemisphere on the mass when it has fallen to a height $R\cos\theta$, where $\theta$ is the polar angle measured from the vertical. (So that $\theta=0$ corresponds to the top of the hemisphere)

(I think that I've got the correct result for this - it was messier than I'd anticipated)

I got

N = mU²/R +mg( √3-cosθ )

Correct

He likes his excuses

I got

N = mU²/R +mg( √3-cosθ )

Oh dear. I get something quite different:

$\displaystyle N = 2mg \cos\theta -\frac{mg\sqrt{3}}{2}+\frac{mU^2}{R}(\frac{3}{4\sin^2\theta}-1)$

we may be solving a very different problem ...

what I see being described here is a very simple problem and I suspect you mean something a lot more complicated which I interpreted in a simplistic way

Did you notice that the initial velocity is horizontal i.e. initially parallel to the base?

I did not ...
then it leaves the surface?

Not sure that I follow. It will leave the surface for a sufficiently large value of $U$, I guess, otherwise it'll spiral round the sphere until it falls off, no?

Now I see what you are describing
The velocity is horizontal and tangential.
will try it again if not tonight tomorrow when I am off

Right. That is a key point that I didn't make clear - it is also tangential. [I've edited the original post]

I spent over 2 hours this afternoon
I feel I got something (I can post my attempt) but the parameterization or connection between z and r is a total mess

Not sure that I follow. It will leave the surface for a sufficiently large value of $U$, I guess, otherwise it'll spiral round the sphere until it falls off, no?

Sure, put it up. I'll post my attempt at a solution tomorrow - I took a Lagrangian approach again.