How Much Work Is Done by the Force on the Rolling Cylinder?

[scratchfish]

Homework Statement

Upon application of the 10-N force F to the cord in Figure P5.3, the cylinder begins to roll to the right. After C has moved 5 m, how much work has been done by F?

do not have picture so let me decribe the picture: 1) cylinder with radius .5m is on the ground 2) the cord is attached on the very top of the cylinder (symmetrical to the center) and is pulled to the right at 10N force

Homework Equations

Translational KE: 1/2mv²
Rotational KE: 1/2Iw²

The Attempt at a Solution

I found translational KE, which is equal to Fd = (10N)(5m) = 50 joule
I don't know how to find the rotational energy of this problem.
The answer should be 100J.
Help me! I am prparing for the final

 

[Lok]

http://en.wikipedia.org/wiki/Rotational_energy

Is the cylinder full or hollow?

 

[scratchfish]

cylinder is full the mass of it is unknown. This seems to require more math than normal intro physics would require, where it briefly touches on the rotational motion

 

[Lok]

cylinder is full the mass of it is unknown. This seems to require more math than normal intro physics would require, where it briefly touches on the rotational motion

Not really. I've only read the article. If full or hollow the difference is KErotational equals half or exactly the KEtranslational. Seems logical.

 

[scratchfish]

uhm.., but how do you assume that it's exactly half? any work?
I mean how did you solve 1/2Iw^2??
How did you find w? where did this derive from?
By the way, this is dyanamics problem, which you take after physics2 and statics,,,,
It can't be that simple.

 

[Lok]

oh that... well ... didn't work ... I assumed you would :P

If the Cyl is empty them you have KE of the whole Cyl as it gains speed after 5 m, that would be translational. And KE of the mass as it rotates with the same speed as the translational motion.( When I say the same speed then the rotational can be likened to translational and you get a result like 1/2mv2)

Same for a full Cylinder just that the outside has the trans speed and the middle has 0, so you get only half of it.

Basicaly Rot KE is equal to 1/2mv^2 with:

m=mass

v=speed of center of gravity

The latter is the geometrical center of gravity of the crosssection of the cylinder. If full the the center of gravity is r/2 from the middle and if it spins with W then we get half of the trans KE. If hollow them the center of gravity is at r from the middl and we get KEtrans=KErot.

 

[2Tesla]

Not really. I've only read the article. If full or hollow the difference is KErotational equals half or exactly the KEtranslational. Seems logical.

This is incorrect. If the cylinder rolls without slipping, then

w=v/r.

If it is full, then moment of inertia

I=(1/2)mr^2

and we have rotational energy = (1/2)*translational kinetic energy:

R=(1/2)Iw^2 = (1/2)*((1/2)mr^2)*(v/r)^2 = (1/4)mv^2 = K/2

If, however, the cylinder is hollow, then moment of inertia

I=mr^2

and we have rotational energy = translational kinetic energy:

R=(1/2)Iw^2 = (1/2)*(mr^2)*(v/r)^2 = (1/2)mv^2 = K

 

[2Tesla]

The latter is the geometrical center of gravity of the crosssection of the cylinder. If full the the center of gravity is r/2 from the middle and if it spins with W then we get half of the trans KE. If hollow them the center of gravity is at r from the middl and we get KEtrans=KErot.

The center of mass is at the center of the cylinder regardless of whether it is full or hollow. This is obvious by the rotational symmetry of the cylinder.

 

[Lok]

I don't see the wrong in my assumptions ... :P

Considering the final answer 100J, is the Cylinder full or hollow?

 

[Lok]

The center of mass is at the center of the cylinder regardless of whether it is full or hollow. This is obvious by the rotational symmetry of the cylinder.

I'm talking about the center of gravity of the crossection. That is as if the whole mass of the cylinder would be in a point that rotates at a distance around the middle of the cylinder.

 

[Lok]

If it is hard to understand them imagine a donut or torus. The crossection would be a full circle and the center of gravity of it would be in it's center.

Most round object Volume and Inertia formulas can be found like this.

 

[scratchfish]

Thanks Telsa! that's just it!
Why couldn't I think of that lol lol!
exactly you are right.!