What are the conditions for stable equilibrium of a cube on top of a hemisphere?

[robb_]

Homework Statement

A cube of sides B sits ontop of a hemisphere of radius R. Determine the conditions such that the cube is in a stable equilibrium.

Homework Equations

A stable equilibrium is one in that the second derivative of the potential energy function is positive. U = mgh.

The Attempt at a Solution

I figure that I need to write out the potential energy of the center of mass of the cube as it is displaced from the eq. pt. Also, as the cube is perturbed, the arclength of the circle traversed on the hemisphere is equal to the length from the middle of the one cube's side (originaly contacting the hemisphere) to the point of contact. I am having trouble writing down the vertical position of the c.m. as a function of displacement. Is it just this geometry that I need to think about or is something else missing? thanks

 

[OlderDan]

Homework Statement

A cube of sides B sits ontop of a hemisphere of radius R. Determine the conditions such that the cube is in a stable equilibrium.

Homework Equations

A stable equilibrium is one in that the second derivative of the potential energy function is positive. U = mgh.

The Attempt at a Solution

I figure that I need to write out the potential energy of the center of mass of the cube as it is displaced from the eq. pt. Also, as the cube is perturbed, the arclength of the circle traversed on the hemisphere is equal to the length from the middle of the one cube's side (originaly contacting the hemisphere) to the point of contact. I am having trouble writing down the vertical position of the c.m. as a function of displacement. Is it just this geometry that I need to think about or is something else missing? thanks

You just need to work out the geometry. If tilting the cube without slipping makes the CM rise, it will be stable.

 

[robb_]

thanks, i'll give it a shot.

 

[robb_]

Okay, I am stuck staring at this thing. I am trying to solve by drawing right triangles with one vertex at the c.m., another at the middle of one side of the cube, and the last at either the point of contact or such that the one leg of the triangle is vertical. As the cube is perturbed, I can see that the arclength between the original point and the point of contact can be constructed as one leg of my triangles. This will bring in the radius of the hemisphere into the equation. I am not sure where to go from here? thanks

 

[OlderDan]

Okay, I am stuck staring at this thing. I am trying to solve by drawing right triangles with one vertex at the c.m., another at the middle of one side of the cube, and the last at either the point of contact or such that the one leg of the triangle is vertical. As the cube is perturbed, I can see that the arclength between the original point and the point of contact can be constructed as one leg of my triangles. This will bring in the radius of the hemisphere into the equation. I am not sure where to go from here? thanks

You can write the height of the center of the cube relative to the center of the sphere as the sum of 3 terms involving the rotation angle: the height of the contact point, the height of the center of the bottom face of the cube relative to the contact point and the height of the center of the cube relative to the center of the bottom face. The difference between this height and its value when the angle is zero must be positive to be stable. You can expand the trig functions about zero angle and use small angle approximations.

 

[robb_]

Okay I solved it! R > b/2. I wrote the height of the center of the cube w.r.t. the center of the sphere as: h = R cos(theta) +R*theta*sin(theta) +b/2*cos(theta). I then employed the small angle approx. for sin and cos. Then I took the derivative of h w.r.t. theta and made that greater than zero. I tried at first to take the derivative of h then use the small angle approx and this did not work. Why? I promise to learn latex soon.

 

[OlderDan]

Okay I solved it! R > b/2. I wrote the height of the center of the cube w.r.t. the center of the sphere as: h = R cos(theta) +R*theta*sin(theta) +b/2*cos(theta). I then employed the small angle approx. for sin and cos. Then I took the derivative of h w.r.t. theta and made that greater than zero. I tried at first to take the derivative of h then use the small angle approx and this did not work. Why? I promise to learn latex soon.

I got the same result. I was a bit surprised how big the cube could be. I had initially just looked at the difference h - h(0) but taking the derivative before doing the approximation works as well. If you do it that way, you can get a (sinθ)/θ ratio that goes to 1 and cosθ goes to 1 in the small angle limit. You don't even need to expand the cosine that way.

 

[robb_]

Doh, I forgot the product rule of differentiation for the R*theta*sin(theta) term. I just took the deriv of sin(theta). Will I ever be a physicist??

 

[OlderDan]

Doh, I forgot the product rule of differentiation for the R*theta*sin(theta) term. I just took the deriv of sin(theta). Will I ever be a physicist??

Just keep learning from the mistakes you do make.