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 &lt;&lt; \omega and that they are nearly constant. If I use my expression for impulse along with \phi &lt;&lt; \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.