Physics II Lab w answers_Part_39

# Physics II Lab w answers_Part_39 - d dt This is related to...

This preview shows pages 1–2. Sign up to view the full content.

89 dt d Φ . This is related to the maximum velocity v rms of the magnet. I.2. Calculation of v rms : The maximum velocity of the magnet is clearly obtained at the equilibrium point i.e. at the bottom of the swing .The velocity v rms can be easily calculated. If M is the mass of the frame and magnet, l is the distance of the center of mass from the point of suspension of frame and magnet , and ° O is the initial release angle , then by conservation of energy, we have 2 sin 2 ) cos 1 ( 2 2 max 2 1 O O l g M l Mg I ° ° ± = - = --------------------(1) Thus 2 sin 2 max O I l g M ° ± = ----------------(2) Thus the quantity I l g M is actually the natural fig 2 frequency of small oscillations .Thus if T is the time period of small oscillations , then l g M I T ² 2 = --------------------------------------------------(3) Using eq (3) and v max = R ± max (R is the radius of the arc) we get 2 sin T R 4 v O max ° ± = ------------------------------------(4) I.4. We now give a rough argument that the maximum value of the emf ³ max = ³ o is proportional to v max .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern