Recent content by Cepterus

Homework Statement Given an isolated system of 2 particles in space, we can express the motion of both particles as follows: $$m_1\ddot{\vec{x_1}}=-\frac\partial{\partial \vec{x_{1}}} V(\vec{x_1},\vec{x_2})\\ m_2\ddot{\vec{x_2}}=-\frac\partial{\partial \vec{x_2}} V(\vec{x_1},\vec{x_2}),$$ where...

Thanks for your help!

I would say negative, since it points in the opposite direction (away from the center). Did you just want me to change the sign before ##F_N## into a ##-##?

Then we are left only with the component of FG toward the center on the right side: ##2mg(1-\cos\vartheta) = mg\cos\theta \Rightarrow 0 = mg(2-3\cos\vartheta)##. But which role does the normal force play?

I think that calculation should still hold. So now we know that ##2mg(1-\cos\vartheta) = mg\cos\vartheta + F_N \Rightarrow F_N = mg(2-3\cos\vartheta)##. For this to be zero, it must hold ## \cos\vartheta = \frac23 \Rightarrow \vartheta = 48.19^\circ##. Correct?

Ok, then it's ##mg\cos\vartheta##.

I am still sticking with my original answer: I first calculate the horizontal component by using ##\tan\vartheta = \frac{F_x}{mg}\Rightarrow F_x = mg\tan\vartheta ## and the component toward the center is the sum of the vertical component ##-mg## and the opposite of the horizontal component...

Apart from F_Z and F_N, there is only F_G left. I guess we could divide that into a component toward the center ##\begin{pmatrix} -mg\tan\vartheta \\ -mg\end{pmatrix}## and a component to the right ##\begin{pmatrix} mg\tan\vartheta \\ 0\end{pmatrix}##

##a_Z## is always directed inwards, to to the center of the sphere. The same applies for ##F_Z##. Since ##F_N## is pointed in the exact opposite direction, I would say yes, at the point of lift-off the normal force is equal to the centripetal force. I would still need to find a formula for...

Centripetal force of ##F_Z = ma_Z = \frac{mv^2}r = 2mg(1-cos\vartheta)##. Should this be equal to the normal force?

There is an upward component of ##F_N\cos\vartheta## and a component to the right of ##F_N\sin\vartheta##. I thought that for ##\vartheta = 0##, the particle is in rest which means ##mg = F_G = F_N\cos\vartheta = F_N##, so $$\vec{F_N} = \begin{pmatrix}...

There is the gravitational force which is constant, and another one which basically "pulls" the particle along the surface of the sphere (normal force?). That normal force will get smaller and smaller until it is zero, which would be the point of lift-off. But I don't know how to express...

Homework Statement A mass point is sliding down from the top of a hemisphere, with an initial speed of zero. I want to determine the angle ##\theta## at which the mass point will lift off the surface of the hemisphere. Homework EquationsThe Attempt at a Solution We can determine the speed...

Thanks! I get the result ##\rho(x,y,z)=\epsilon_0(2ax+c)##. If this is correct, what does it actually mean? Like, to create ##\vec E##, should I start with zero charge on the ##yz##-plane and then place linearly more and more charges in space the larger my distance to the ##yz##-plane gets...

This isn't correct: The ball is dropped from rest, but by the time it hits the sand, it already has a certain speed, as it is accelerated by gravity. So, what is ##v_i## when the ball reaches the ground?