{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Exploding projec6le similarly we can dene the

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

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

Unformatted text preview: kg, m2 = 2.5 kg, and m3 = 3.2 kg, that form an equilateral triangle with edge length a = 140 cm. xcm 1 = M ∑m x i =1 3 ii = m1 x1 + m2 x2 + m2 x3 = 83.0cm m1 + m2 + m2 ycm ˆ ⇒ rcm = (83cm )i + (58cm ) ˆ j 1 = M ∑m y i =1 i 3 i = 58.0cm What if the the distribu6on of par6cles is con6nuous? What is the center of mass of this object? Ans: We replace the summa6on by integra6on. 1 rcm = M ∑mr i =1 n ii 1 rcm = r dm M∫ In components: xcm = 1 x dm, M∫ ycm = 1 M ∫ y dm, 1 V zcm = 1 z dm, M∫ If the object has uniform density (made from same materials throughout its body): xcm = 1 x dV , V∫ ycm = ∫ y dV , zcm = 1 z dV V∫ Newton’s second law revisited Ques0on: How could we write down Newton’s second law for this object ? Answer: We relate everything to the center of mass and treat the object as if all the mass of it were concentrated there. Fnet = Macm Here: M: total mass of system or object. Fnet: net, external force ac6ng on system. Internal forces betwee...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern