Potential energy in a physical pendulum

[Quadrat]

Hey PF!

1. Homework Statement
If I have a pendulum; a vertically hanging rod with (length ##L## and mass ##m##) which can rotate freely about a point ##p## on the upper edge of the rod. Now I fire a bullet (also with mass ##m##) into it (strictly horizontal on the lower end of the rod).

I want to understand how to get the potential energy from the rod in this case when the whole system have rotated to a maximal angle ##@## and comes to a rest just before it starts to swing back. I treat the bullet as a point mass and can understand that the potential energy of that point particle is given by ##PE_{bullet}=mgL(1-cos(@)##. I know that the potential energy will be equal to ##(1/2)mgL(1-cos(@)## for the rod. But can someone explain to me how to get this result?

Homework Equations

I figure it's some kind of a integral to get this answer but I'm not able to get a grip on this.

The Attempt at a Solution

I'm assuming it's some integral with the limits 0 and L for all dx but as I stated - I'm having troubles understandning this. It would so helpful if someone could explain this to me in a clear way. :)

Cheers

 

[gneill]

Hey PF!

1. Homework Statement
If I have a pendulum; a vertically hanging rod with (length ##L## and mass ##m##) which can rotate freely about a point ##p## on the upper edge of the rod. Now I fire a bullet (also with mass ##m##) into it (strictly horizontal on the lower end of the rod).

I want to understand how to get the potential energy from the rod in this case when the whole system have rotated to a maximal angle ##@## and comes to a rest just before it starts to swing back. I treat the bullet as a point mass and can understand that the potential energy of that point particle is given by ##PE_{bullet}=mgL(1-cos(@)##. I know that the potential energy will be equal to ##(1/2)mgL(1-cos(@)## for the rod. But can someone explain to me how to get this result?

Homework Equations

I figure it's some kind of a integral to get this answer but I'm not able to get a grip on this.

The Attempt at a Solution

I'm assuming it's some integral with the limits 0 and L for all dx but as I stated - I'm having troubles understandning this. It would so helpful if someone could explain this to me in a clear way. :)

Cheers

For the change in potential energy you can treat the rod as a point mass located at its center of mass (L/2 for a uniform rod).

 

[ehild]

I want to understand how to get the potential energy from the rod in this case when the whole system have rotated to a maximal angle ##@## and comes to a rest just before it starts to swing back. I treat the bullet as a point mass and can understand that the potential energy of that point particle is given by ##PE_{bullet}=mgL(1-cos(@)##. I know that the potential energy will be equal to ##(1/2)mgL(1-cos(@)## for the rod. But can someone explain to me how to get this result?

The easy way is, that the potential energy is the same as if all mass was concentrated in the centre of mass.
The more basic way is to divide the rod into small pieces of mass dm=ρdl ( ρ is the linear density) and integrate their potential energy. dU=dm gl (1-cos(θ))
[tex]U=\int_0^L{\rho (1-\cos(\theta))gldl}[/tex]

 

[Quadrat]

The easy way is, that the potential energy is the same as if all mass was concentrated in the centre of mass.
The more basic way is to divide the rod into small pieces of mass dm=ρdl ( ρ is the linear density) and integrate their potential energy. dU=dm gl (1-cos(θ))
[tex]U=\int_0^L{\rho (1-\cos(\theta))gldl}[/tex]

Oh I see. So the linear mass density, ##ρ## can be written as ##\frac{dx}{L}## and when I integrate from ##x=0## to ##x=L## I get ##\frac{1}{L}(1-cos\theta)gM\frac{L^2}{2}## which simplifies to what I was looking for. Did I get it right? I'm still having some issues with integrals (but I'm trying to master them).

Thanks ehild and gneill

 

[ehild]

Oh I see. So the linear mass density, ##ρ## can be written as ##\frac{dx}{L}## and when I integrate from ##x=0## to ##x=L## I get ##\frac{1}{L}(1-cos\theta)gM\frac{L^2}{2}## which simplifies to what I was looking for. Did I get it right? I'm still having some issues with integrals (but I'm trying to master them).

It is right

 

[Quadrat]

It is right

Great, thanks!