Rolling hoop hit with impulse. Find angle of deflection and max force.

[Dazed&Confused]

Homework Statement

A child's hoop of mass $M$ and radius $b$ rolls in a straight line with velocity $v$. Its top is given a light tap with a stick at right angles to the direction of motion. The Impulse of the blow is $I$.

a. Show that this results in a deflection of the line of rolling by angle $\phi = I/Mv$, assuming that the gyroscope approximation holds and neglecting friction with the ground.

b. Show that the gyroscope approximation is valid provided $F << \dfrac{2Mv^2}{b}$, where F is the peak applied force.

The Attempt at a Solution

I don't know for sure the direction of the impulse but assuming it is to the side, that would be the 'angular impulse' would be $Ib$ in the direction perpendicular to the angular momentum $Mbv$ of the wheel, so $\tan\phi = Ib/Mbv = I/Mv ≈ \phi$. The last part using the small angle approximation.

For b. I would assume that since $\phi << 1$, therefore $Fdt << Mv$.

I don't know if any of my attempt is correct.

Edit: if instead I approximate the impulse by a triangle height $F$ and base $dt$, then the impulse would be $\dfrac{Fdt}{2}$.

 

[Dazed&Confused]

Anyone?

 

[TSny]

For part a, I think your work is correct.

For part b, you need to find the meaning of the phrase "gyroscope approximation". I don't think it means $\phi << 1$. If this is not defined in your text or notes, then try a web search.

 

[Dazed&Confused]

For part a, I think your work is correct.

For part b, you need to find the meaning of the phrase "gyroscope approximation". I don't think it means $\phi << 1$. If this is not defined in your text or notes, then try a web search.

Ok so I checked online as it isn't in my book and it means $\Omega << \omega$ and that they are nearly constant. If I use my expression for impulse along with $\phi << \omega dt = \dfrac{v}{b}dt$ then I get the right answer.

Thanks for the tip and I hope this is correct.

 

[TSny]

OK, good. I think that's right.