Cube tipping on horizontal surface

[raving_lunatic]

Homework Statement

A cube of mass m and side 2a is held on one of its edges on a horizontal surface. It is released from this position and allowed to tip. Find an expression for the angular velocity of the cube as its face strikes the surface of the table in the following cases:

a) the surface is sufficiently rough to stop the edge from moving
b) the edge of the cube can slide freely on the table

Homework Equations

G = r x F
G = dJ/dt (G torque, J angular momentum)

I = 2/3 m a² - moment of inertia for cube about its center of mass

I = 2/3 m a ² + 2ma² - moment of inertia for a cube about one of its edges (using the parallel axis theorem)

1/2 I w² = rotational kinetic energy

The Attempt at a Solution

Okay. So, first I tried to work out the torque of the weight about the rotation axis on one of the cube's edges, which I determined to be mga tan θ in the direction of increasing θ, where θ is defined as the angle between the vertical and the line between the center of mass, and the edge about which the cube tips. The reaction force acts through this point, so it exerts no torque about this axis. I wasn't able to integrate the equation for torque to find angular momentum (as θ is a function of time) so I tried a conservation of energy approach, reasoning that the center of mass moved by (√2-1)a and equating the loss in gravitational potential energy to the rotational kinetic energy. This gave me an answer for the case where the motion is purely rotational (which I think is the first case), but I wasn't sure how to deal with the case when the block is sliding freely. Possibly a term should be added to compensate for the translational kinetic energy, but I can't find the relation between this and the angular velocity? Any help with this problem would be greatly appreciated

 

[BvU]

Hello Looney,

At first I was a little surprised by your statement that in the first case "the motion is purely rotational", but I think you are correct there, because your rotation is around an edge and you take that into account.

So the hint for the other case jumped on me: where is the axis of rotation ?

(Not an expert, but intrigued by this 'simple' elegant problem!)

 

[raving_lunatic]

I'm guessing the axis of rotation is the one that passes through the center of mass, parallel to the surface? (There's only two that seem to make sense to me.)

Okay - so we have a combination of translational motion of the CM and rotational motion about the CM in the case where the edge of the block is free to slide, is that correct? So we need to take into account the velocity of the system's center of mass as well as the rotation about it? I can see how that would be possible (we can get it in terms of the angle of rotation, at least, defining a suitable origin...)

 

[BvU]

Again, at first I had a different idea: there's a gravitational force acting at the CM axis and a normal force at the supporting edge. So my axis was halfway, but that leads to nonsense (I don't think the CM can be moved sideways). Your approach looks a lot better. And indeed, CM dropping straight down imparts some kinetic energy. But due to lower moment of inertia the final answer might well come out the same. Or more. Or less .
Let me (and other listeners) know what you find!

 

[HalfLostAndSearching]

!



Hello,

Thank you for sharing your approach to this problem. First, it is important to note that the torque on the cube will change as it tips, since the angle θ will also change. So, integrating the torque equation may not give you the correct answer. Instead, you can use the conservation of angular momentum, which states that the angular momentum of a system remains constant unless an external torque acts on it.

For the first case, when the surface is rough and the edge is stopped, the angular momentum of the cube will be conserved. This means that the initial angular momentum, which is zero, will be equal to the final angular momentum of the cube when it strikes the surface. We can use the parallel axis theorem to calculate the moment of inertia of the cube about the edge as I = 2/3 ma^2 + 2ma^2. The angular velocity can then be calculated using the equation L = Iω, where L is the angular momentum and ω is the angular velocity.

For the second case, when the edge can slide freely on the surface, we need to consider both the rotational and translational kinetic energies. The initial kinetic energy of the cube will be equal to the final kinetic energy when it strikes the surface. This means that we can equate the initial rotational kinetic energy, which is 1/2Iω^2, to the final translational kinetic energy, which is 1/2mv^2. Here, v is the velocity of the cube at the moment it strikes the surface. We can use the parallel axis theorem again to calculate the moment of inertia and then solve for the angular velocity ω.

I hope this helps to guide you in the right direction. Remember to always consider conservation laws when solving physics problems. Keep up the good work!