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Center of Mass: Boat and regular pentagon

  1. Jan 7, 2012 #1
    Six problems due Monday, and I have no idea what I'm doing on either of these.

    Problem 1:

    1. The problem statement, all variables and given/known data

    A group of people has a total mass of 1500kg and are standing on one end of a 20,000kg boat. They walk 6.5m to the other end of the boat. How much does the boat move? The water is frictionless.

    2. Relevant equations

    rcom = (1/M)*[itex]\int[/itex](r)dm

    3. The attempt at a solution

    I really have no idea where to begin, let alone how to try to solve it.

    Problem 2:

    1. The problem statement, all variables and given/known data

    A regular pentagon has sides of length a. Find the center of mass if you remove the triangle formed by the geometric center and the two vertices on the bottom of the pentagon.

    2. Relevant equations

    rcom = (1/M)[itex]\int[/itex](r)dm

    3. The attempt at a solution

    The book says to split the rest of the pentagon into four equal triangles, and I found the center of mass of the removed triangle. Since the five triangles that make up the pentagon are the same, their centers of mass should be the same distance from the geometric center of the pentagon, which I'm using as the origin.
    Also, the book says the answer to this one is ".115a above the vertex of the removed triangle."
     
  2. jcsd
  3. Jan 7, 2012 #2
    For the first question the center of mass of the boat and people must remain the same since there are no outside forces. So first of all find the center of mass with the people standing on one end then shift them over to the other end and find the center of mass again, the difference between the two is the distance the boat would have to move to keep the center of mass in the same spot.

    For the second one you have four triangles and you know the center of mass of each one, so you essentially have just four point masses and you need to find the center of mass of those.
     
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