Conservation of Angular Momentum

Posted on 2006-04-23
Last Modified: 2008-02-01
You have a ballistic pendulum. You shoot a ball of mass m horizontally from a spring gun with speed v. The ball is immediately caugt a distance r below a frictionless pviot by a pivoted cather assembly of mass M. the moment of inertia of this assembly about its rotation axis through the pivot is I. the distance r is much greater than the radius of the ball.

a) use conservation of angular momentum to show that the anguluar speed of the ball and catcher just after the ball is caught is w = mvr/(mr^2 + I)

b) after the ball is caught, the center of mass of the ball - catcher assembly system swings upp with the maximum height increase h. Use conservation of energy to show that w = sqrt[ 2(M + m)gh / (mr^2) + i) ].

c) your lab partner says that linear momentum is conserved in the collision and derives the expression mv = (m + M)V where V is the speed of the ball immediately after the collision, she then uses conservation of energy to derivate that V = sqrt(2gh), so that mv = (m + M)sqrt[2gh]. Use the results of parts (a) and (b) to show that this equation is satisfied only for the special case when r is given by I = Mr^2.

No idea how to start.
Question by:BrianGEFF719
    LVL 26

    Accepted Solution



    Treat the two questionbs separately.
    Question 1
    What is the angular momentum before the collision.
    (The ball is a distance r below the pivot going a speed v)
    That will be the angular momentum after the collision with the ball caught.

    question 2

    The energy (KE) of the ball is the total energy. That energy is converted to PE by the apparatus
    LVL 19

    Author Comment

    Thanks aburr, these quesions are acctually pretty easy you know where you need to end up :)


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