Minimum Velocity for Cube to Fall Off Table - Rotational Motion Question

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SUMMARY

The discussion centers on determining the minimum velocity required for a solid cube of side 2a and mass M to fall off a frictionless table after hitting an obstacle. The cube must rotate about its edge AB, which remains on the table, until it completely rotates off the edge. The critical condition for this rotation is established by analyzing the forces and torques acting on the cube at the tipping point.

PREREQUISITES
  • Understanding of rotational motion and torque
  • Familiarity with the concepts of center of mass and angular momentum
  • Knowledge of basic physics principles related to frictionless surfaces
  • Ability to apply conservation laws in mechanics
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  • Study the principles of rotational dynamics and torque calculations
  • Learn about the conditions for tipping and stability in rigid bodies
  • Explore the concept of center of mass and its application in rotational motion
  • Investigate the effects of angular momentum conservation in dynamic systems
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Physics students, mechanical engineers, and anyone interested in understanding the dynamics of rigid body motion and tipping conditions in mechanics.

semc
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Solid cube of side 2a and mass M is sliding on a frictionless table with velocity v. It hits a small obstacle at the end of the table causing it to tilt. What is the minimum velocity to cause the cube to fall off the table?

Totally no clue where to start I don't even know what's the condition for the cube to fall off...

A bola basically consist of three stones at the end of three cord so when it is thrown in air they will spread out with constant 120 degree angles between the cord. Assume three stones with mass m and cord with length l. The hunter takes one stone and swing the bola so 2 stones move together in a horizontal circle of radius 2l and speed v0. Calculate the angular speed of the bola about the center of mass.

Using COM 2mv0=3mvf so vf=[tex]\frac{2}{3}[/tex]v0 and v=l[tex]\omega[/tex] so [tex]\omega[/tex]=[tex]\frac{2}{3l}[/tex]v0
However if I use conservation of angular momentum,
L=I[tex]\omega[/tex]
mrvf=[tex]\frac{4}{3}[/tex]lmv0=I[tex]\omega[/tex]
[tex]\omega[/tex]=[tex]\frac{4}{9l}[/tex]v0

Why are the angular velocity different?
 
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Hi semc! :wink:
semc said:
Solid cube of side 2a and mass M is sliding on a frictionless table with velocity v. It hits a small obstacle at the end of the table causing it to tilt. What is the minimum velocity to cause the cube to fall off the table?

Totally no clue where to start I don't even know what's the condition for the cube to fall off...

A lower edge AB of the cube stops at the edge of the table. The cube was going so fast that the opposite lower edge starts to lift up, and the whole cube rotates about edge AB. How fast does it have to be for the rotation to go so far that the cube rotates right round, and goes off the table? :smile:
 

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