Kinetic energy of a hollow cylinder

[lilphy]

Hello
1. Homework Statement
In a uniform solid cylinder of radius r, mass m and height h we emptied a cylindrical cavity of radius r/4 tangent and parallel to the symmetrical axis z of the original cylinder.
The cylinder is rolling without slipping on a horizontal plane. Find the kinetic energy in terms of θ and θ' where θ is the angle between vertical and the line joining the center of the original solid cylinder and the center of the cavity.

Homework Equations

K= 1/2 I_cmw²+1/2mv_cm²
K=1/2 I w²

The Attempt at a Solution

I found the moment of inertia I=17/32 m r²
and I_cm=17/32 m r² - md²
To find d² I found the z component of cdm = h/2

I don't know which formula is easier to use the first one using Icm and vcm or the second one ?
It is rolling without slipping so w and v_cm are related...
But I don't know how yo find w, is it w=θ' * r ?
Thanks

 

[BvU]

Hello Lilphy,

Could you show the steps taken to find $I = {17\over 32} mr^2$.
I wonder what ##I_{\rm cm} is once you found I.

And I see variables in your equations and solution attempt that I don't see in the problem statement... d, h,

My side view of the cylinder (which you have drawn for yourself, I should hope) is as folllows:

$\omega$ is indeed $\dot \theta r$ as you expect.

[edit] oh boy! see post #3 ​

Do you think the mass of the rolling object is $m$ as in the problem statement, or is it ${15\over 16}m$ ?

 

[haruspex]

ω is indeed $\dot \theta r$ as you expect.

I think you mean $v_{cm}=r\omega=r\dot \theta$

 

[lilphy]

Hello Lilphy,

Could you show the steps taken to find $I = {17\over 32} mr^2$.
I wonder what ##I_{\rm cm} is once you found I.

And I see variables in your equations and solution attempt that I don't see in the problem statement... d, h,

My side view of the cylinder (which you have drawn for yourself, I should hope) is as folllows:

View attachment 94778

$\omega$ is indeed $\dot \theta r$ as you expect.

Do you think the mass of the rolling object is $m$ as in the problem statement, or is it ${15\over 16}m$ ?

thanks for the answer !

H is the height, d is just the variable used in the definition of the theorem of parallel axis.

To find 17/32 m r², I just calculated the triple integral
I= 2 π ρ h ∫ r³ dr for r from a/4 to a.
And replacing for ρ=M/(πhr² 15/16). But my M is false so ρ should be=m/(πhr²). So I would be now = 255 r²/512.

But i am not sure this is the good answer because the cylinder is not emptied in the center, so i don't know if this result holds.

To find the distance d to determine Icm, I don't know if it is correct but I calculated the z coordinate of the center of mass and found h/2 so d=h/2. coordinatez= ρ/M 2π ∫∫ zr dr dz r from a/4 to a and z from 0 to h. Then replaced M by ρ*h*π*r² 15/16.

 

[haruspex]

thanks for the answer !

H is the height, d is just the variable used in the definition of the theorem of parallel axis.

To find 17/32 m r², I just calculated the triple integral
I= 2 π ρ h ∫ r³ dr for r from a/4 to a.
And replacing for ρ=M/(πhr² 15/16). But my M is false so ρ should be=m/(πhr²). So I would be now = 255 r²/512.

But i am not sure this is the good answer because the cylinder is not emptied in the center, so i don't know if this result holds.

To find the distance d to determine Icm, I don't know if it is correct but I calculated the z coordinate of the center of mass and found h/2 so d=h/2.

You don't need to any integration for this problem, just apply standard results.
I would not bother finding the common mass centre. Just treat the motion as rotation about the rolling point, so find the moments of inertia about that point.

 

[lilphy]

You don't need to any integration for this problem, just apply standard results.
I would not bother finding the common mass centre. Just treat the motion as rotation about the rolling point, so find the moments of inertia about that point.

I am not sure I understand what you mean by rolling point ?
So I just have to calculate E=1/2 I w² with w=θ'
And the moment of inertia would be the moment of inertia of a solid cylinder but with the 15/16 of the mass ?

 

[haruspex]

I am not sure I understand what you mean by rolling point ?

I mean about the point of contact with the ground.

And the moment of inertia would be the moment of inertia of a solid cylinder but with the 15/16 of the mass ?

No, I didn't say that.
Treat the hollow as a negative mass cylinder, same density, combined with a normal complete cylinder.
Each has a moment of inertia about the point of contact with the ground, and you can find these using standard formulae.

 

[lilphy]

I mean about the point of contact with the ground.

No, I didn't say that.
Treat the hollow as a negative mass cylinder, same density, combined with a normal complete cylinder.
Each has a moment of inertia about the point of contact with the ground, and you can find these using standard formulae.

Okℤ
For the solid cylinder K=1/2(1/2Mr²θ'²)
I think that for the hollow cylinder there will be a dependence in θ, but in the sketch of bvu, the center of the solid cylinder and of the cavity are aligned with the vertical, so i don't "see" what theta represents ?
Thank you !
Edit: Now i know what theta is i forgot that the cylinder is rolling

 

[haruspex]

Okℤ
For the solid cylinder K=1/2(1/2Mr²θ'²)

No, you need the moment of inertia about the point of contact with the ground.

 

[lilphy]

No, you need the moment of inertia about the point of contact with the ground.

Right, so I have to use parallel axis theorem.
I=1/2Mr²+Mr² = 3/2 M r² ?

 

[lilphy]

and for the hollow it would be
I=1/2m(r/4)²+mr/4sinθ with m=-ρlπr²/16 ?

 

[haruspex]

Right, so I have to use parallel axis theorem.
I=1/2Mr²+Mr² = 3/2 M r² ?

Where M is the mass of the completed cylinder, yes.

and for the hollow it would be
...+mr/4sinθ ... ?

A few problems with that term. How far is the centre of the cavity from the point of contact?
(It's not clear from the problem statement whether theta is measured from the upper radius or the lower radius. I.e. at theta=0, is the cavity above or below the middle of the cylinder?)

 

[lilphy]

Where M is the mass of the completed cylinder, yes.

A few problems with that term. How far is the centre of the cavity from the point of contact?
(It's not clear from the problem statement whether theta is measured from the upper radius or the lower radius. I.e. at theta=0, is the cavity above or below the middle of the cylinder?)

The centre of the cavity using cos law is at a distance of sqrt(5r²/4 -r²/2 cosθ)
So I=m/2 (r/4)^2+ msqrt(5r²/4 -r²/2 cosθ)

Then the total inertia moment at the point of contact is the sum of both.
And K=1/2 I_totθ'²

 

[haruspex]

The centre of the cavity using cos law is at a distance of sqrt(5r²/4 -r²/2 cosθ)

That's not what I get. Please show your working.

 

[lilphy]

That's not what I get. Please show your working.

Oups I made an error in calculating the distance.
It is sqrt(15r²/16-r²/2 cosθ)

 

[haruspex]

Oups I made an error in calculating the distance.
It is sqrt(15r²/16-r²/2 cosθ)

That's nearer to what I get, but still two differences.
Please clarify whether you are measuring theta from the radius that goes up from the centre or from the radius that goes down from the centr.

 

[lilphy]

That's nearer to what I get, but still two differences.
Please clarify whether you are measuring theta from the radius that goes up from the centre or from the radius that goes down from the centr.

http://imageshack.com/a/img911/1145/CawDxl.jpg
This is an approximate sketch of the situation and how i calculate the distance. C is the contact point.
BD= r/4 sinθ
AD=r/4cosθ
So BC=sqrt( (AC-AD)^2+BD^2 ) = sqrt( (r-r/4cosθ)^2 + (r/4 sinθ)^2 ) = sqrt( r²-2r²/4 cosθ + (r/4)²) = sqrt( 17r²/16 -r²/2 cosθ)

 

[haruspex]

http://imageshack.com/a/img911/1145/CawDxl.jpg
This is an approximate sketch of the situation and how i calculate the distance. C is the contact point.
BD= r/4 sinθ
AD=r/4cosθ
So BC=sqrt( (AC-AD)^2+BD^2 ) = sqrt( (r-r/4cosθ)^2 + (r/4 sinθ)^2 ) = sqrt( r²-2r²/4 cosθ + (r/4)²) = sqrt( 17r²/16 -r²/2 cosθ)

Good, you got 17/16 this time, which is what I have. The other difference was because I was measuring theta from the upper vertical radius, whereas you show it from the lower.
On to the next step.

 

[lilphy]

Good, you got 17/16 this time, which is what I have. The other difference was because I was measuring theta from the upper vertical radius, whereas you show it from the lower.
On to the next step.

The total moment of inertia at the contact point is Itot= 3/2Mr²+mr²/32+17mr²/16-mr²/2 cosθ = 3/2Mr²+35mr²/32-mr²/2cosθ
m=-1/16M
Itot=3/2Mr²-35Mr²/512+Mr²cosθ/32
Is my Itot correct ?

 

[TSny]

Itot=3/2Mr²-35Mr²/512+Mr²cosθ/32
Is my Itot correct ?

I think that's right. Of course, you can combine the first two terms.

 

[lilphy]

And so the kinetic energy at the contact point is just K=1/2 Itot w^2
With w=θ' ?

 

[TSny]

Yes, that's the total KE of the cylinder. I'm not sure what you mean by "at the contact point".

 

[lilphy]

Yes, that's the total KE of the cylinder. I'm not sure what you mean by "at the contact point".

Oh right, it is of course the moment of inertia at the contact point but it is the kinetic energy of the cylinder..
Thanks for your answer
And thanks to haruspex for his help