How Fast Must a Bullet Travel to Tip a Cube on Its Edge?

[gaborfk]

Can someone please help me with the following problem:

A solid cube of wood of side 2a and mass M is resting on a horizontal surface. The cube is constrained to rotate about an axis AB. A bullet of mass m and speed v is shot at the face opposite ABCD at a height of 4a/3. The bullet becoes embedded in the cube. Find the minimum value of v required to tip the cube so that it falls on face ABCD. Assume m<<M.

This is what I got so far:

I understand that it is a perfectly inelastic collision and that

mv=(m+M)V

L(initial)=I*Omega where I is the combined Inertia for both object. The I for a cube rotating around one of its edge is I=(8Ma^2)/3 and the bullet is a point mass with an r of sgrt((4a/3)^2+(2a)^2).

omega=V/R where V is the final speed and R is the radius of rotation, which is ...?

Thank you in advance!

 

[Doc Al]

This is what I got so far:

I understand that it is a perfectly inelastic collision and that

mv=(m+M)V

This won't help. Since the cube is constrained, translational momentum is not conserved. But angular momentum is conserved. (Hint: How much energy is needed to topple the "cube + bullet"?)

 

[arildno]

"mv=(m+M)V"
Careful about this one!
You cannot use this with respect to the C.M. of the bullet+cube system, since that system is affected by an external force comparable in magnitude of the internal forces.
That external force acts from the axis to which the cube is constrained, ensuring that the cube's contact velocity (with the axis) remains zero throughout the collision.

Do you agree so far?

Note:
it's only fair that Doc Al beats me occasionally..

 

[gaborfk]

So then

K(initial)=K(final)

Bullet K(initial)=1/2mv^2
Bullet + Block= K(final)=1/2I*omega^2

1/2mv^2=1/2I*omega^2

where I is the combination of both bullet and block, I(block)=(8M(2a)^2)/3 and I(bullet)=mr^2 where r is sgrt[(4a/3)^2+(2a)^2]. Also, omega is=V/R where V is the final speed of a point on the block.

Can you please double check...

But how about R and how do I figure out how much Energy is needed in order to topple?

 

[arildno]

Why do you think kinetic energy is conserved in the collision?
Doc Al wrote that the system's angular momentum is conserved (with respect to a point lying on the fixed axis)

 

[Doc Al]

During the collision, angular momentum is conserved; after the collision, mechanical energy is conserved.

 

[gaborfk]

How about this:

Angular momentum right before the collision for the bullet is

mvr where r is the distance from the rotation point.

After the collision for the bullet

mr^2*omega

The block's angular momentum after the collision is

L=I*omega

In both cases omega is the angular velocity of the block, which breaks down to omega=V/R

So is it mvr=mr^2*omega+I*omega?

mvr=mr^2*(V/R)+I(V/R)

I=(8M(2a)^2)/3

Is R SQRT[8a^2]? How about V?

 

[Doc Al]

Angular momentum right before the collision for the bullet is

mvr where r is the distance from the rotation point.

Careful: \vec{L} = \vec{r}\times m\vec{v}

After the collision for the bullet

mr^2*omega

The block's angular momentum after the collision is

L=I*omega

OK. What's the rotational inertia of the cube about its edge?

Since m << M, you can probably ignore the moment of inertia of the bullet.

You should be able to find \omega in terms of the bullet speed v. What minimum value of \omega is required to get the cube to tip? (That's where energy conservation will come in.)

 

[gaborfk]

so L=rx\omega so it is the cross product...

that means that I need to take the perpendicular component to r which is the \sin[\theta] of the bullet's angle of impact to r?

Rotational Inertia of the block around its edge is:

I=\frac{8M(2a)^2}{3}?

 

[Doc Al]

so L=rx\omega so it is the cross product...

that means that I need to take the perpendicular component to r which is the \sin[\theta] of the bullet's angle of impact to r?

Sure. Or you can find the perpendicular distance between the velocity vector and the the axis---that's given to you.

Rotational Inertia of the block around its edge is:

I=\frac{8M(2a)^2}{3}?

Doesn't look right to me. I'd say the rotational inertia would be I = 2/3 Ms^2, where s is the edge length, in your case s = 2a, but you'd better check it.

 

[gaborfk]

I thought that L=\frac{2}{3}Ms^2is if the center of rotation is in the center of the cube. I will use the parallel axis formula to move it's rotational axis to see what I get...

 

[Doc Al]

With the axis in the center of the cube, the rotational inertia would be I = 1/6 Ms^2

 

[gaborfk]

Still in need of understanding

I am still having trouble finding \omega in terms of the bullet's speed v. I do not understand how conservation of energy comes into play after the collision.

BTW. You were right about the rotational inertia of the cube around it's edge.

I=\frac{2}{3}Ms^2 since s=2a I=\frac{2}{3}M(2a)^2<br /> I=\frac{2}{3}M4a^2=\frac{8}{3}Ma^2

Thank you, and Happy Thanksgiving!

 

[Doc Al]

I am still having trouble finding \omega in terms of the bullet's speed v.

This is just conservation of angular momentum: initial angular momentum (due to the bullet, in terms of v) equals angular momentum of rotating cube (in terms of \omega) immediately after the collision.

I do not understand how conservation of energy comes into play after the collision.

Immediately after the collision, the cube+bullet has some kinetic energy. As the cube tips up, KE is transformed to gravitational PE. So, how much energy is needed to make it tip over? (If it doesn't have enough KE, it will just fall back down again. It's just like if you wanted to roll a ball over the top of a hill: how much energy do you have to give it at the bottom to make it all the way to the top and over?)

BTW. You were right about the rotational inertia of the cube around it's edge.

I know.

 

[gaborfk]

Initial Angular due to the bullet

L_{i} =mvr\sin{\theta}

Where r\sin{\theta}=d

Which is given as d=\frac{4a}{3}

so L_{i} =mv\frac{4a}{3}

Final Angular velocity of cube+bullet (bullet can be ignored, as you pointed out)

L_{f} =I\omega

Where I=\frac{8}{3}Ma^2

So L_{i} =L_{f}

Yields mv\frac{4a}{3}=\frac{8}{3}Ma^2\omega

Solve it for \omega and plug it into below...

KE+PE=0?

KE=\frac{1}{2}I\omega^{2}

It has to rise higher than when the cube is it's corner which is
2a=\sqrt{h^{2}+h^{2}} so h=\sqrt{2}a

That is how far I got... Is PE=Mgh? Or is it KE=PE so, KE-PE=0...

 

[Doc Al]

That is how far I got... I still do not know PE for the cube...

You're almost there. Consider that \Delta {KE} + \Delta{PE} = 0. The change in PE of the cube is Mg\Delta{h}, where \Delta{h} is the change in height of the cube's center of mass.

 

[gaborfk]

My last question for the week...

The cube has to rotate over it's edge. In order to do that I put the cube on it's corner figured out the height of a side corner to the groud..

So is h going to be 2a=\sqrt{h^{2}+h^{2}}

yields that h=\sqrt{2}a

Now I plug all that into the KE+PE=0 equation, solve for v and then make v larger than what I get on the other side. This way if v is larger it should have enough energy to rotate over a height h, which is the max height of a side corner if the cube is stands on one of it's corners.

 

[Doc Al]

So is h going to be 2a=\sqrt{h^{2}+h^{2}}

I assume you mean: h =\sqrt{a^{2}+a^{2}}

yields that h=\sqrt{2}a

That's the final height of the center of mass. What is the change in height?

Now I plug all that into the KE+PE=0 equation, solve for v and then make v larger than what I get on the other side. This way if v is larger it should have enough energy to rotate over a height h, which is the max height of a side corner if the cube is stands on one of it's corners.

Right! Be sure to use the change in height, not just the height.

 

[gaborfk]

The change in height is

h_{f}-h_{i}=\sqrt{2}a-a=(\sqrt{2}-1)a

Damn, that is going to be messy...

 

[gaborfk]

I am getting

v=\sqrt{-\frac{3(\sqrt{2}-1)Mga}{m}}

Which I think is wrong, since I got a negative under the SQRT.

Are we sure that it is \Delta KE+\Delta PE=0

and not \Delta KE=\Delta PE initial KE transfers to final PE

which is \Delta KE-\Delta PE=0 ?

 

[Doc Al]

I am getting

v=\sqrt{-\frac{3(\sqrt{2}-1)Mga}{m}}

Which I think is wrong, since I got a negative under the SQRT.

Yep, that's not right. (It's pretty close though; check over your work.)

Are we sure that it is \Delta KE+\Delta PE=0

That's just a statement of energy conservation. You are probably mixing up your signs somewhere. As the cube tips up, KE decreases and PE increases.

 

[gaborfk]

I can not see it...

So here it is.

I=\frac{8}{3}Ma^2

\omega=\frac{mv\frac{4a}{3}}{\frac{8}{3}Ma^2}=\frac{1}{2}\frac{mv}{Ma}

\Delta KE=\frac{1}{2}I\omega^2

\Delta PE=Mg(\sqrt{2}-1)a

\Delta KE+\Delta PE=0

\frac{1}{2}(\frac{8}{3}Ma^2)(\frac{1}{2}\frac{mv}{Ma}})^2+(Mg(\sqrt{2}-1)a)=0

(\frac{4}{3}Ma^2)(\frac{m^2v^2}{4M^2a^2})+(Mg(\sqrt{2}-1)a)=0

\frac{m^{2}v^{2}}{3M}+(Mg(\sqrt{2}-1)a)=0

\frac{m^{2}v^{2}}{3M}=-(Mg(\sqrt{2}-1)a)

v^{2}=-\frac{Mg(\sqrt{2}-1)a}{\frac{m^{2}}{3M}}

v=\sqrt{-\frac{3(\sqrt{2}-1)M^{2}ga}{m^{2}}


Is g negative, because it is acting againts to motion? That would fix the negative sign...

 

[Doc Al]

I can not see it...

So here it is.

I=\frac{8}{3}Ma^2

\omega=\frac{mv\frac{4a}{3}}{\frac{8}{3}Ma^2}=\frac{1}{2}\frac{mv}{Ma}

Right.

\Delta KE=\frac{1}{2}I\omega^2

No, \Delta {KE} = - 1/2 I\omega^2. KE decreases so \Delta {KE} is negative.

...

v=\sqrt{-\frac{3(\sqrt{2}-1)M^{2}ga}{m^{2}}

Exactly right... except for that minus sign!

 

[gaborfk]

Thank you very much! You guys, Doc Al and Arildno, always got the answers!

Bye