At What Angle Does the Object Fall Off the Hemisphere?

[kevin0960]

There is a small object, which mass is m on the top of hemisphere, with the mass of M.

the size of the object is neglectable. Also, the radius of hemisphere is R.

There is no friction between the hemisphere and the object, and the hemisphere and the surface. What if we slightly hit the object the object will be fall from the hemisphere.

At what point the mass will completely off from the hemisphere?

I attatched the the picture.
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I first tried to use energy conservation law to find the velocity of the object.
\frac{1}{2} mv^2 + mgR(cos \theta) + \frac{1}{2}MV^2 = mgR

Also by conservation of momentum on x axis, we can deduct

mv_x = MV

But I just stucked at here. Plz someone help me

 

[kuruman]

Let's say that θ in the first equation represents the angle at which the mass flies off. That's one unknown. You also have the final velocities, V and v_x that are unknown. So you have three unknowns and only two equations. You need one more equation. What is the condition that must be satisfied if the mass is to fly off? That's your third equation.

 

[kevin0960]

I found few more equations for that problem.

First, because the object moves on the hemisphere before it falls off It must satisfy following condition

\frac{v_y}{v_x + V} = tan \theta

Also, I am not quite sure about the equation but since the object is having on circular motion (at the frame of hemisphere),
\frac{m((v_x + V)^2 + v_y^2)}{R} = mgcos\theta - N

and at the time the object falls from the hemisphere, N will be 0

Is this right??

 

[kuruman]

In these two equations, you need the horizontal velocity component relative to the hemisphere. I would write that as v_x - V to keep V an algebraic quantity. Then when you substitute for V using the momentum conservation equation, you would get v_x+(m/M)v_x in the expression. This voids confusion.

So you have four equations (sorry I said three earlier) and four unknowns. You can find an expression for θ in terms of the given quantities.