{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

exam4_page1

# exam4_page1 - and by rotation about its CG Let’s list the...

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

Exam 4 SAR Physics AM Kolchin 1. Linear speed ( v ) is directly proportional to the rota- tional speed ( ω ) and the distance ( r ) of the point from the axis of rotation: v . 2. Centripetal Force on an object: F c = mv 2 r 3. Torque ( Γ): Γ = F d . The distance ( d ), is from the point at which the force is applied to the axis about which the torque is exerted. 4. Rotational Inertia (I): A small object of mass m , a dis- tance r from its axis of rotation, has a rotational inertia: I = mr 2 . 5. Rotational Acceleration ( α ): defined as rate of change of rotational speed per unit time: α = Δ ω/ Δ t . 6. Newton’s Second Law for Rotational Motion : Γ = . 7. Angular Momentum ( L ): L = . 8. Kinetic Energy : Total K.E. = Translational K.E. + Rotational K.E. K.E. total = 1 2 Mv 2 + 1 2 2 1

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

View Full Document
9. Translational and Rotational Analogs : We’ve said that an object’s motion can be described by translation of its CG
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: and by rotation about its CG. Let’s list the translational and rotational quantities and laws that are counterparts of each other: Translational Rotational velocity ( v ) rotational velocity ( ω ) ( m/s ) (rpm, 1 /s ) acceleration ( a ) rotational acceleration ( α ) ( m/s 2 ) (rpm/ s , 1 /s 2 ) mass ( m ) rotational inertia ( I ) ( kg ) ( kg * m 2 ) Force ( F ) Torque ( Γ ) ( N ) ( N * m ) ( note: not joules; joules are reserved for energy only .) Momentum ( p ) Angular Momentum ( L ) kg * m/s kg * m 2 /s K.E. (trans) = 1 2 Mv 2 K.E. (rot) = 1 2 Iα 2 (joules) ( J ) (joules) ( J ) Newton’s Second Law F = m a Γ = Iα p is conserved L is conserved 2...
View Full 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