Mass point sliding down a sphere

[Cepterus]

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 Equations

The Attempt at a Solution

We can determine the speed $v$ at any given angle $\theta$ by using the conservation of energy:
$$\begin{align*}E_{\rm{kin}}+E_{\rm h}&=E_{\text{h,0}}\\
\frac12mv^2+mgr\cos\vartheta&=mgr\\
\Rightarrow v&=\sqrt{2gr(1-\cos\vartheta)}\end{align*}$$ and geometrical considerations lead to horizontal and vertical components of $v_x = v\cos\vartheta$ and $v_y = v\sin\vartheta$.

I need some characterization of the spot on the surface at which the lift-off is going to occur, but I don't have any good ideas.

 

[TSny]

Your energy analysis looks good to me.

Consider the forces that act on the particle while it is in contact with the surface. There's something special that happens to one of the forces as the particle approaches the point where it leaves the surface.

 

[Cepterus]

Your energy analysis looks good to me.

Consider the forces that act on the particle while it is in contact with the surface. There's something special that happens to one of the forces as the particle approaches the point where it leaves the surface.

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 $F_N$ in a formula.

 

[TSny]

Draw a free body diagram for the particle at an arbitrary point before it lifts off.

 

[Cepterus]

Draw a free body diagram for the particle at an arbitrary point before it lifts off.

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 $$\vec{F_N} = \begin{pmatrix} mg\sin\vartheta\\mg\cos\vartheta\end{pmatrix},$$however this doesn't make too much sense since $\vec {F_N}$ would never be zero.

 

[TSny]

Before liftoff the particle is moving in circular motion. How would you set up Newton's second law for this circular motion?

 

[Cepterus]

Before liftoff the particle is moving in circular motion. How would you set up Newton's second law for this circular motion?

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

 

[TSny]

Net force along a certain direction equals mass times acceleration in that direction. What is the direction of the acceleration a_Z? Is the normal force equal to the net force in this direction?

 

[Cepterus]

Net force along a certain direction equals mass times acceleration in that direction. What is the direction of the acceleration a_Z? Is the normal force equal to the net force in this direction?

$a_Z$ 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 $F_N$ to use this information.

 

[TSny]

Yes, $a_Z$ is toward the center. Thus, the net force must also be toward the center. Since the normal force points away from the center, there is no way that the normal force could represent the net force toward the center.

Look at your free body diagram and see if there is any force that has a component toward the center.

 

[Cepterus]

Yes, $a_Z$ is toward the center. Thus, the net force must also be toward the center. Since the normal force points away from the center, there is no way that the normal force could represent the net force toward the center.

Look at your free body diagram and see if there is any force that has a component toward the center.

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 $\begin{pmatrix} mg\tan\vartheta \\ 0\end{pmatrix}$

 

[TSny]

The symbol F_Z represents the net force acting toward the center. It is not one of the physical forces acting on the particle. F_N and F_G are the two physical forces acting on the particle. F_Z is what you get when you add together the "toward-the-center components" of F_N and F_G.

I don't follow your expressions for the components of F_G. When the particle is at angle $\theta$ as shown in the diagram, what is the component of F_G toward the center?

 

[Cepterus]

I don't follow your expressions for the components of F_G. When the particle is at angle $\theta$ as shown in the diagram, what is the component of F_G toward the center?

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, $-mg\tan\vartheta$ because the horizontal components in total have to cancel out.

 

[TSny]

The red vector shows the projection of mg onto the centripetal direction; that is, it represents the component of mg toward the center. From the triangle, you should be able to get an expression for the magnitude of the red vector.

 

[Cepterus]

The red vector shows the projection of mg onto the centripetal direction; that is, it represents the component of mg toward the center. From the triangle, you should be able to get an expression for the magnitude of the red vector.
View attachment 211648

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

 

[TSny]

Yes. Good. How would you express the net force toward the center, F_Z?

 

[Cepterus]

Yes. Good. How would you express the net force toward the center, F_Z?

Centripetal force of $F_Z = ma_Z = \frac{mv^2}r = 2mg(1-cos\vartheta)$.

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?

 

[TSny]

So now we know that $2mg(1-\cos\vartheta) = mg\cos\vartheta + F_N$

This is not quite correct. On the right side you should be writing down the net force toward the center. But, you know that the normal force is directed away from the center.

 

[Cepterus]

This is not quite correct. On the right side you should be writing down the net force toward the center. But, you know that the normal force is directed away from the center.

Then we are left only with the component of F_G 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?

 

[TSny]

The normal force needs to be included in getting the net force toward the center. Is the centripetal component of the normal force positive or negative?

 

[Cepterus]

The normal force needs to be included in getting the net force toward the center. Is the centripetal component of the normal force positive or negative?

I would say negative, since it points in the opposite direction (away from the center).

This is not quite correct. On the right side you should be writing down the net force toward the center. But, you know that the normal force is directed away from the center.

Did you just want me to change the sign before $F_N$ into a $-$?

 

[TSny]

I would say negative, since it points in the opposite direction (away from the center).

Yes.

Did you just want me to change the sign before $F_N$ into a $-$?

Yes, as well as understand why.

 

[Cepterus]

Thanks for your help!

 

[TSny]

OK. Good work.