Deriving MOI of a Radially-Dependent Density Cylinder on an Incline

[Stickboy10]

Hello there I am really struggling with this and another problem and I was hoping you guys could help me out. The problem is as follows:

"the cylinder shown has a radially-dependent density with mass M and radius R. The cylinder starts from rest and rolls without slipping down the incline with height, H. At the bottom of the plane its translational speed is: (8gH/7)^1/2

Show MOI of the object is 3/4MR^2."

(this is not a typical shape used in problems. it is going to be something more like a can and I just can't seem how to include the velocity in deriving the equation for I)

My attempt is here:
I = r²dm \leftarrow I don't know what dm becomes. I am thinking that \sigma is involved in some way but I don't see where the velocity comes in from there. It may have something to do with density also, but again I can't find a way to connect velocity into that. I may totally wrong and probably am though.

 

[Andrew Mason]

Hello there I am really struggling with this and another problem and I was hoping you guys could help me out. The problem is as follows:

"the cylinder shown has a radially-dependent density with mass M and radius R. The cylinder starts from rest and rolls without slipping down the incline with height, H. At the bottom of the plane its translational speed is: (8gH/7)^1/2

Show MOI of the object is 3/4MR^2."

(this is not a typical shape used in problems. it is going to be something more like a can and I just can't seem how to include the velocity in deriving the equation for I)

What is the change in potential energy of the cylinder (use change in height of the centre of mass). What is its total kinetic energy then? Work out the translational kinetic energy. Is that the total kinetic energy? How much is missing? In what form is the missing kinetic energy? How is that related to the moment of inertia?

AM

 

[Stickboy10]

U=MgH \leftarrow since I don't have numbers I can't calculate that. but I know
Ui + Ki = Uf + Kf (assuming no E loss) \leftarrow where Ki is zero and Uf is zero. I can split kf into K_rot and K_trans
so
U + 0 = 0 + K_rot + K_trans
K_trans = \frac{1}{2}Mv²\leftarrow inserting my v_trans in gives me
K_trans = \frac{8gHM}{14}

You asked how much is missing but I don't know how to calculate the missing K_rot because I can't calculate the potential energy.

K_rot = I \omega²\leftarrow So I solve for I: \frac{2K_rot}{\omega²}

Am I on the right track now?

 

[AlexChandler]

you should be able to relate the velocity and the angular velocity by a circular to angular conversion. Something similar to

v = r \omega

 

[Stickboy10]

I see the relationship in the equation between v and \omega but from there I don't see how I am relating the r with the work that I have done so far.

 

[AlexChandler]

You don't need to worry about work. Just use the conservation of mechanical energy. If you use the relation between v and \omega you will be able to find the total kinetic energy in terms of the mass, the final velocity, and the rotational inertia. Set this equal to the change in gravitational potential energy mg \Delta h and you should be able to solve for I.

 

[Stickboy10]

Ah, ok I think I see what you are saying and I see how the relationship brings in the r which something I have to have in my final equation.

So I have:
MgH = \frac{1}{2}I\omega² + \frac{1}{2}Mv²
and
\omega= v/r
so
MgH = \frac{1}{2}I\frac{v}{r}² + \frac{1}{2}Mv²
I don't know if this is good math or not but this is where I am at now:
\frac{r²\cdot2MgH - \frac{8ghM}{14}}{\frac{8gh}{7}} = I

 

[AlexChandler]

\omega= v/r
so
MgH = \frac{1}{2}I\frac{v}{r}² + \frac{1}{2}Mv²

Looks good except for you should have

\omega ^2 = v^2 / r^2

(and when you are using latex, to make a superscript you just have to write r^2, not [SUP.])

 

[Stickboy10]

why are they all squared when the original equation was v = r\omega? where does the squaring come from?

oh, omega is sqaured in the conservation equation I set up so you just moved it into the other equation to show an easy variable switch.

 

[AlexChandler]

Ah, ok I think I see what you are saying and I see how the relationship brings in the r which something I have to have in my final equation.


I don't know if this is good math or not but this is where I am at now:
\frac{r^2 \cdot 2MgH - \frac{8ghM}{14}}{\frac{8gh}{7}} = I

And actually your units do not work in this equation. Try doing it again.