Cylinder pulled by string on flat surface w/out slipping

[highroller]

1. The problem statement: a 100kg homogeneous cylinder of radius .3m. Starts at rest. Is pulled by a string wrapped around the cylinder coming off the top with a force of 500N.

Find the angular velocity after the cylinder has rolled one revolution?



The attempt at a solution: I= 4.5kg*m^2

circumferance=1.885m

work= 500*1.885= .5m(wr)^2+.5Iw^2 (using w for omega)

answer I get ... w=11.81rad/s

supposedly correct answer is 16.71

can someone please explain. I suppose the force could be doing rotational work as well but I'm having a hard time believing that the force does double the work because of where it is applyed.

My question is which answer is correct and WHY? thanks for any help

 

[tiny-tim]

Welcome to PF!

Hi highroller! Welcome to PF!

(have an omega: ω and try using the X² tag just above the Reply box )

a 100kg homogeneous cylinder of radius .3m. Starts at rest. Is pulled by a string wrapped around the cylinder coming off the top with a force of 500N.

Find the angular velocity after the cylinder has rolled one revolution?
…
work= 500*1.885= .5m(wr)^2+.5Iw^2 (using w for omega)

Yes intial energy is zero, so work done = final energy.

But what is your .5m(ωr)² supposed to be?

There's no extra mass … it's only a cylinder, with energy 1/2 Iω²

 

[highroller]

the cylinder has both linear and rotational ke so the .5m(wr)^2 comes from .5mv^2 with v=wr

 

[tiny-tim]

the cylinder has both linear and rotational ke so the .5m(wr)^2 comes from .5mv^2 with v=wr

ah, sorry … i looked at your 500*1.885 and assumed that the axle was fixed.

Hint: how fast is the string moving?

 

[Doc Al]

work= 500*1.885= .5m(wr)^2+.5Iw^2 (using w for omega)

Imagine someone pulling the string. In order to have the cylinder roll through one revolution, for what distance must they pull the string? (Don't forget that the string is unwinding as it's being pulled.)