Physics Help Forum angular momentum

 Jan 3rd 2009, 01:29 PM #1 Junior Member   Join Date: Nov 2008 Posts: 21 angular momentum a pole with a mass of M, and a length of L is hung on a nail through its top end, so that it can swing freely. a ball with a mass of m moving horizontally hits the bottom end of the pole and sticks to it, the pole rises in circular motion to an angle of "alpha". there is no friction with the air or between the nail and the pole Picasa Web Albums - Devan - Drop Box 1)what is the velocity of the ball before the collision? 2)how much energy is lost during the collision. what i did was I=(M/3)L^2 u=omega*L the momentum is conserved during the collision so P1=P2+L2 mv=mu+(M/3)L^2(omega) mv=mu+(M/3)Lu v=u[1+(M/3m)L] now using conservation of energy after th collision until the max height, saying that my gravitational potential energy is equal to 0 at the top of the pole, therefore my height, h, is negative from the top of the pole and is -L*cos(alpha) for the ball and -(L/2)*cos(alpha) for the pole[measured from the centre of mass, at the centre of the pole] 0.5mu^2+I(omega)^2-mgL-0.5MgL=-mgLcos(alpha)-Mg(L/2)cos(alpha) 0.5mu^2+(M/6)u^2=mgL+0.5MgL-mgLcos(alpha)-Mg(L/2)cos(alpha) u^2[(m/2)+(M/6)=gL[m+(M/2)](1-cos(alpha)) u^2=2gl(m+(M/2))(1-cos(alpha))(3/(3m+M)) u^2=3gL[(2m+M)/(3m+M)](1-cos(alpha)) v^2=u^2*[1+(M/3m)L]^2 v^2=[3gL[(2m+M)/(3m+M)](1-cos(alpha))]*[1+(M/3m)L]^2 does this seem correct?? the answer in my textbook is v^2=2gL(1-cos(alpha))(1+(M/2m))(1+(I/mL^2)) cant see where ive gone wrong and dont want to continue to the next part till i know this is right. please help

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