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So I was looking at this post and I started thinking about this.

Let's say I have a cube-like object on the floor. Friction is not zero. Suddenly, I apply a momentary force at a place that will create torque. Consider this diagram:

enter image description here

Given enough momentum/force/torque, is it possible that the square will gain height such that it will stop touching the ground? If so, which is the last point that will touch the ground? Maybe the bottom right corner(edge)?

My thoughts: I think it will takeoff. Because if not, it would have to keep doing the weird square spin at high velocity...this is not realistic. I guess the vertical acceleration caused by both the bullet and the ground can be high enough? I also guess a minimum torque will be needed, else the object would just slide....

George Menoutis
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1 Answers1

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enter image description here

The figure shows that you would require to rotate the cube at least till highest point to tip it off. So assume that after collision the acquire $\omega$ as angular speed such that $\omega=\frac{v_{cm}}{a/2}$. So using energy conservation $$\frac{1}{2}I\omega^2=Mg(\frac{a}{\sqrt{2}}-\frac{a}{2})$$ where $I$ is the moment of inertia about the edge. So $I=\frac{ma^2}{6}+m(\frac{a}{\sqrt{2}})^2=\frac{2Ma^2}{3}$. So minimum $V_{cm}$ required is, $$\frac{2Ma^2}{2*3}\frac{4v_{cm}^2}{a^2}=Mga(\frac{1}{\sqrt{2}}-\frac{1}{2})$$ $$v_{cm}=\frac{\sqrt{3ga(\frac{1}{\sqrt{2}}-\frac{1}{2})}}{4}$$ Now use conservation of momentum and conservation of energy to calculate the momentum required by bullet to tip it off, $$\frac{1}{2}mv_1^2=\frac{1}{2}I\omega^2$$ $$mv_1=Mv_{cm}+mv_2$$ where $v_1$ and $v_2$ are the final velocity of the bullet.

sslucifer
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  • A bullet fired into a block does not conserve energy. With friction at the lower surface it also does not conserve linear or angular momentum. – R.W. Bird May 04 '20 at 21:22
  • OP has mentioned that the collision is 100% elastic. Since the block is not moving then the friction is not acting at the instant of collision. I know this is quite subtle approach but I can't think of another analysis. – sslucifer May 04 '20 at 21:24
  • Pardon me for not being clear.The question is not whether the cube will rotate at least 90 degrees - it is whether it will "fly" (stop touching the ground with any points) – George Menoutis May 05 '20 at 05:34
  • I don't that's possible just from the horizontal force of bullet. To fly off, the cube needs some external vertical force to overcome gravity. It may be possible when you shoot the bullet at some angle, so that there will be a component of force in the vertical direction, but in the case stated not possible. – sslucifer May 05 '20 at 13:53
  • As the block tries to rotate, it pushes down on the lower right corner. The surface responds by pushing back. This is the force that lifts the center of mass, and may give it enough vertical velocity for the block to lose contact with the surface. – R.W. Bird May 05 '20 at 14:21
  • What do you mean by it pushes the ground? The cube is a rigid body it can't be compressed so it will not store any energy that will later be used to lift itself from the ground. – sslucifer May 05 '20 at 14:24
  • It has kinetic energy, both linear and rotational, as a result of the impulse from the bullet. – R.W. Bird May 05 '20 at 15:24