{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

quiz20_solution

# quiz20_solution - b Find the angular velocity as a function...

This preview shows page 1. Sign up to view the full content.

04/18/08 Physics 201 Quiz 20 A hula-hoop of radius R and mass M is attached at its rim to a horizontal rod, about which it can pivot. (if the hoop is in the yz plane, the rod is along the x-axis). The hoop is held horizontal, which we define to be theta=90. (theta=0 at bottom) The hoop is released. a) Find the angular momentum as a function of the angle. cons of energy gives -MgR cos(theta)+L^2/(2I)=0, and I=MR^2+MR^2 (parallel axis thm), so L=sqrt(4 g M^2 R^3 cos(theta)). (Check units)
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: b) Find the angular velocity as a function of the angle. L=I omega so omega=L/I=sqrt(2 Mg R cos(theta)/I)=sqrt(g cos(theta)/R) c) Now find the angular frequency another way: Find the torque by differentiating the angular momentum, and set it equal to the torque due to gravity. tau=L0/2 sin(theta)/sqrt(cos(theta)) *omega, where L0=2 sqrt(M^2 g R^3). tau=mgR sin(theta) which gives omega=2 MgR sqrt(cos(theta))/L0=sqrt(g cos(theta)/R)...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online