How will a yo-yo on a horizontal table move when a force is applied at its edge?

[jaderberg]

Homework Statement

http://www.fas.harvard.edu/~scdiroff/lds/NewtonianMechanics/Yo-yo/Yo-yo002.gif
similar to the situation in b. except the string and force P act from the edge of the yoyo rather than from an inner spool

A force (P) is applied to the string horizontally which is attached to the yoyo directly below its center of mass, where the yoyo is sitting stationary on a horizontal table with friction coefficient (u). What will happen?

Homework Equations

Just F=ma and moments

The Attempt at a Solution

So F(maximum friction) = uR where R = mg
F=umg in the opposite direction to P.
So if P > umg then the yoyo will start to slide to the right.
However if P > umg and assuming P acts from the edge of the yoyo, then taking moments about the center of mass Pr > umgr so there will be a moment anticlockwise and the yoyo will spin anticlockwise.
But what I am thinking is that the torque will still be greater than the frictional force so it won't grip and it will still slide to the right but while spinning anticlockwise...or will it? surely it must spin, but whether that will cause it to roll i don't know?!

Also i was thinking, if P acts at a distance which is less than r, so r-x, then could a situation arise where the moments will be such that while P>umg, P(r-x) = umgr so the yoyo will slide to the right but won't spin at all. Also then could the situation arise where P(r-x) < umgr so the yoyo will slide to the right and start to roll to the right?

Please help (at least with the basic problem where P acts at a distance r from COM) as there is a high chance that this will come up in a uni interview on tuesday!

 

[Shooting Star]

At the point when the body starts to slide, P = F => Pr = Fr. So, in the limiting case, the body just tends to slide, and the net torque is zero. If P>F, then Pr>Fr and so the body tends to slide toward P and also tends to spin anticlockwise. Your analysis seems to be correct.

If the force P acts at a dist x below the centre, (what you'd written as r-x), then at the point of sliding, P = F => Px<Fr, since x<r, and so the body tends to slide toward P and tends to spin clockwise.

As soon as the body actually starts to spin clockwise, the direction of F will become opposite.

 

[jaderberg]

ok cool, but if P>F and Pr>Fr and the body slides towards P and starts to spin anticlockwise, then will it roll to the left or just continue to accelerate towards P while spinning (P is a constant applied force)?

 

[Shooting Star]

Since P>F, the body will move in the dircn of P.

If it slides in the dircn of P, F is in the opp dircn. The net torque is anticlockwise, since Pr>Fr. It should then slide in the dircn of P, and spin anticlockwise.