# Understanding Inertia

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^2

I 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

Homework Helper
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^2

I 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 MbulletL2.
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)

Homework Helper
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 :)