What is the correct way to find the inertia of a bullet entering a door?

[Raisintoe]

I am having trouble with this one problem. I have tried multiple times to solve it, but come up with the same wrong answer every time. The mistake that I may be making is in finding the Inertia of the bullet entering the door: I figured that some mass at the given radius from the hinge of the door should have an inertia I = M*R^2, is this correct? I = ∫r^2*dm

1. Homework Statement
A 10 g bullet traveling at 350m/s strikes a 12kg , 1.2-m-wide door at the edge opposite the hinge. The bullet embeds itself in the door, causing the door to swing open.

What is the angular velocity of the door just after impact?

Homework Equations

Inertia of the door = ⅓*M*L^2 where L is the Width of the door: ⅓*(12kg)*(1.2m)^2

The Attempt at a Solution

I have taken the approach of finding the kinetic energy of the bullet: ½*M*V^2 and then converting that energy to rotational energy of the door. As I understand it, all energy should be rotational energy because the door swings on its end.

I find the angular velocity of the door (ω) by using: ω = √((M_bullet*V^2)/(⅓*M_door*L^2 + M_bullet*L^2)) = √((M_bullet*(V/L)^2)/(⅓*M_door + M_bullet)). I figured this equation from: KE_bullet = E_door = ½*M_bullet*V^2 = ½*(⅓*M_door*L^2 + M_bullet*L^2)*ω^2I think that I am going wrong when I try to find the Inertia of the bullet: Is it M*L^2? (0.01kg)*(1.2m)^2

 

[ehild]

I am having trouble with this one problem. I have tried multiple times to solve it, but come up with the same wrong answer every time. The mistake that I may be making is in finding the Inertia of the bullet entering the door: I figured that some mass at the given radius from the hinge of the door should have an inertia I = M*R^2, is this correct? I = ∫r^2*dm

1. Homework Statement
A 10 g bullet traveling at 350m/s strikes a 12kg , 1.2-m-wide door at the edge opposite the hinge. The bullet embeds itself in the door, causing the door to swing open.

What is the angular velocity of the door just after impact?

Homework Equations

Inertia of the door = ⅓*M*L^2 where L is the Width of the door: ⅓*(12kg)*(1.2m)^2

The Attempt at a Solution

I have taken the approach of finding the kinetic energy of the bullet: ½*M*V^2 and then converting that energy to rotational energy of the door. As I understand it, all energy should be rotational energy because the door swings on its end.I find the angular velocity of the door (ω) by using: ω = √((M_bullet*V^2)/(⅓*M_door*L^2 + M_bullet*L^2)) = √((M_bullet*(V/L)^2)/(⅓*M_door + M_bullet)). I figured this equation from: KE_bullet = E_door = ½*M_bullet*V^2 = ½*(⅓*M_door*L^2 + M_bullet*L^2)*ω^2I think that I am going wrong when I try to find the Inertia of the bullet: Is it M*L^2? (0.01kg)*(1.2m)^2

No, you are right, the moment of inertia of the bullet is M_bulletL².
But it is not true that the kinetic energy of the bullet converts to he rotational energy of the door(with bullet embedded). Energy is not conserved, but the angular momentum is.
What is the angular momentum of the bullet with respect to the hinge before collision?

 

[Raisintoe]

OK, Thank you, I got it now:

I_bullet = M_bullet*L^2:

P_bullet = I_bullet*V/L = M_bullet*L*V:

ω = P_bullet/(I_bullet + I_door)

 

[ehild]

OK, Thank you, I got it now:

I_bullet = M_bullet*L^2:

P_bullet = I_bullet*V/L = M_bullet*L*V:

ω = P_bullet/(I_bullet + I_door)

It is all right now :)

 

[HalfLostAndSearching]

Your approach is correct in finding the angular velocity of the door after impact. However, the mistake you are making is in finding the inertia of the bullet entering the door.

The inertia of an object is a measure of its resistance to changes in rotational motion. In this case, the bullet entering the door does not have its own rotational motion. Therefore, the correct way to find the inertia of the bullet would be to use the formula I = mr^2, where m is the mass of the bullet and r is the distance from the axis of rotation (in this case, the hinge of the door).

Since the bullet embeds itself in the door, its distance from the hinge can be taken as the radius of the door, which is 0.6 m (half of the door's width). Therefore, the inertia of the bullet would be I = (0.01 kg)(0.6 m)^2 = 0.0036 kg m^2.

Using this value for the inertia of the bullet in your calculation for the angular velocity of the door should give you the correct answer.