One of the curators at the art museum is tilting a large cylinder backward. At what angle (theta) will the cylinder of height h and radius r will tumble?

My thinking is that the

length = height = h
2r = w
It seems that this cylinder will tumble when the angle is large enough to cause the center of gravity to go beyond the supporting edge.

I am not sure how to go about this...?

Your assumption is correct. The CG is along the central axis midway between top and bottom. It will be above the edge (point of contact with the ground) when the tilt angle is arctan (2r/h)

A square of side 2L is removed from one corner of a square sandwich that has sides of length L. The center of mass of the remainder of the sandwich moves from C to C’. The displacement of the x coordinate of the center of mass (from C to C’) is?

In the picture, C' moves up and to the right. So positive x and positive y direction.

To determine the angle at which the cylinder will tumble, you can analyze the equilibrium condition of the cylinder when it is tilted backward.

1. First, consider the forces acting on the cylinder. The main forces involved are the gravitational force acting downward and the normal force exerted by the ground (supporting edge) upward.

2. At the point of tumbling, the center of gravity of the cylinder will be directly above the supporting edge. This means that the gravitational force will act through the supporting edge.

3. The gravitational force can be decomposed into two components: one perpendicular to the supporting edge (normal force) and one parallel to it. The perpendicular component will be balanced by the normal force, while the parallel component will cause a torque.

4. The torque exerted by the parallel component of the gravitational force will try to rotate the cylinder about its edge, causing it to tumble.

5. To find the angle at which the torque is sufficient to make the cylinder tumble, you can equate the torque to the maximum torque that the cylinder can resist without tumbling.

6. The maximum torque the cylinder can resist without tumbling is given by the product of the perpendicular component of the gravitational force and the radius of the cylinder (r):

Torque max = F_perpendicular * r

7. The perpendicular component of the gravitational force can be calculated as the weight of the cylinder (mg) times the sine of the angle (θ) at which it is tilted:

F_perpendicular = mg * sin(θ)

8. Since the cylinder is in equilibrium just before tumbling, the maximum torque should be equal to the torque exerted by the parallel component of the gravitational force:

Torque parallel = F_parallel * h/2

9. The parallel component of the gravitational force is given by the weight of the cylinder (mg) times the cosine of the angle (θ):

F_parallel = mg * cos(θ)

10. equating the maximum torque and the torque exerted by the parallel component, we have:

mg * sin(θ) * r = mg * cos(θ) * h/2

11. Simplifying the equation, we can cancel out the weight (mg):

sin(θ) * r = cos(θ) * h/2

12. Rearranging the equation, we get:

tan(θ) = (h/2) / r

13. Finally, solving for θ, we have:

θ = arctan((h/2) / r)

So, the angle at which the cylinder will tumble is given by arctan((h/2) / r). Substitute the values of the height (h) and radius (r) of the cylinder to find the specific angle.

To determine the angle at which the cylinder will tumble, we need to consider the position of the center of gravity and the conditions under which it will move beyond the supporting edge.

The center of gravity of the cylinder is located at a height equal to half the height of the cylinder, which is h/2. As the cylinder tilts backward, the center of gravity will shift and if it moves beyond the supporting edge, the cylinder will tumble.

To find the angle at which this occurs, we can use trigonometry. Let's imagine a right triangle formed by the radius of the cylinder (r), the distance from the center of gravity to the supporting edge (d), and the height of the cylinder (h).

Using this triangle, we can write the equation:

tan(theta) = d / r

We know that the length of the base of the triangle is equal to the radius of the cylinder, so d can be expressed as r - w, where w is the width of the base. Since you mentioned that the width is equal to twice the radius (2r), we can substitute that into the equation:

tan(theta) = (r - 2r) / r
tan(theta) = -r / r
tan(theta) = -1

Taking the arctan (inverse tangent) of both sides of the equation, we find:

theta = arctan(-1)

The arctan of -1 is -π/4 or -45 degrees.

Therefore, the cylinder will start to tumble when it is tilted at an angle of -45 degrees.