Steven W. answered • 01/03/17

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_{rot}= ½Iω

^{2}

Steven W.

_{rot }= mg(L/2)(1 - cos(60º)) = ½Iω

^{2}

^{2}in this case. This is usually given in tables of moments of inertia, but we can derive it if necessary.

^{o})) = (1/2)[(1/3)mL

^{2}](ω

^{2})

01/03/17

Steven W.

_{ra}= Iω

_{a }(with I the same as before, and ω

_{a}given in the problem)

_{b}= m

_{b}vL

_{b}= mass of the ball

_{after}= L

_{ra}+ L

_{b}= angular momentum before the collision (by conservation of angular momentum), which was calculated in Step 3

_{a}+ m

_{b}vL = Iω

01/03/17

Steven W.

_{rot-a}= (1/2)Iω

_{a}

^{2}

_{a}):

_{a}

^{2}= mg(L/2)(1 - cos(θ

_{a}))

_{a}. So you can solve for θ

_{a}here, answering the second question (note that taking inverse cosine will be involved in the solution).

01/03/17

Steven W.

_{a}, so the conservation of angular momentum equation now has two unknowns: v and ω

_{a}. So we cannot solve for either one with just that equation. We need a second one, which is provided by the statement that the collision is elastic.

^{2}= KE

_{rot-a}(for the rod) + KE

_{trans-a}(for the ball) = (1/2)Iω

_{a}

^{2}+ (1/2)m

_{b}v

^{2}

_{a}and v), so you can then solve for both using those equations.

_{a}into the linear speed at the end of the rod. Since the end of the rod is traveling in a circle of radius L, we can relate the tip's translational velocity to its rotational velocity using:

_{a}= Lω

_{a}

01/03/17

Steven W.

01/03/17

Steven W.

^{o})).01/03/17