# Homework Help: Sliding block transitions to rotation around pivot

1. Nov 4, 2013

### buridan

1. The problem statement, all variables and given/known data

This isn't strictly homework, but the nature of my problem is similar to homework problems:
A block with mass m and it's center of mass not necessarily at the center is sliding along a frictionless surface. The center of mass is distance r from the right front corner of the block. The block is instantaneously restrained by it's lower front corner.
a) Find an expression for the speed of the block required for the block to flip over the pivot.
b) Find the impulse the pivot exerted in the x direction (against v) when the block was restrained

2. Relevant equations

KE = .5*Iω^2
L = Iω
L = rxP
ω=vsin(θ) / r

3. The attempt at a solution

I tried tackling the problem two ways:

1. a) I used the pivot as my reference point for torque and angular momentum and said that the angular velocity would be v*sin(angle of v with r)/r. I plugged that value into the energy equation getting: .5 I v2 * sin2(θ)/r2. I then found the work done by gravity between the original position and the angle when the center of mass is directly over the pivot. This quantity is the force of gravity multiplied by the difference in position of the center of mass between those two angles. I wrote that as mg(r-hi). I then solved the expression: 0 = .5Iv2sin2(θ) - mg(r-h) for to get my answer to part a.

2. a)I used the center of mass as my reference point and said that it was experiencing an impulse of mv' at the front right corner. I don't know what v' should properly be. The block is also experiencing a second torque around the center from the normal force at the front right corner. My resulting expression was 0 = .5m2v2r2sin2(θ)/I - mg(r-h).

I had no idea if these were equivalent, so I found the moment of inertia for the case that the center of mass was in the geometric center to be I = mr2/3 around the center of mass and 4mr2/3 around the pivot. (Iuniform block = (l2+w2)/12 ). Plugging these moments of inertia into the equations resulted in different coefficients on the first term. Setting the first term equal to each other found that for this case, v' = 2v/3. More generally: v/v' = $\sqrt{\frac{m^{2}r^{4}}{I^{2}_{cm}+I_{cm}mr^{2}}}$.

Am I missing forces in the version around the center of mass? Are my expressions valid? Should v' = v?

2. Nov 4, 2013

### voko

I understand that the pivot is at the "lower front corner". However, the center of mass is given with regard to the "right front corner", and it is not clear how that corresponds to the "lower front corner". Moreover, just the distance from the "right front corner" does not specify unambiguously where CoM is.

3. Nov 4, 2013

### buridan

Sorry. "right front corner" was a typo. Should be "lower front corner". The center of mass is in the center of the block in the horizontal direction perpendicular to movement.

4. Nov 4, 2013

### voko

Hmm. Initially you said CoM was not at the center.

5. Nov 4, 2013

### buridan

not at the center along the horizontal axis parallel to the motion nor in the vertical axis.

6. Nov 4, 2013

### buridan

Does the total kinetic energy stay constant when you change reference frames? I think that was the main thing I wasn't sure of.

7. Nov 4, 2013

### haruspex

This is effectively a 2D problem, right? We only need to consider the vertical plane containing the CoM, direction of movement, and the pivot. Also need to assume the block is just a projection of its 2D shape in the third dimension. (Rotations of irregular 3D objects are very messy.)
You need to decide whether the collision is perfectly elastic. If it isn't you can't use energy like that. Since there will be an instantaneous upward impulse from the ground, elasticity there may also be an issue.
You have assumed no energy losses and that if the block just makes it into tipping over then it will be almost stationary as it does so. Those two assumptions might conflict. There may be a point where the CoM isn't going forward but the block is still rotating.