Momentum of a ball an pendulum

[Dell]

this is a question i have previoulsly posted but had no replies to, after a few changes it looks like this, but still not right

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

http://picasaweb.google.com/devanlev...59076394966178



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

u1=omega*L [angular velocity*length of pole, balls velocity after collision]
u2=omega*(L/2) [the angular velcocity*distance to centre of mass]

u=u1=(1/2)u2


the momentum is conserved during the collision so

mv=mu1+Mu2=u(m+M/2)

v=u(1+M/2m)

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+0.5I(omega)^2-mgL-0.5MgL=-mgLcos(alpha)-Mg(L/2)cos(alpha)

0.5mu^2+(1/6)Mu^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 I've gone wrong and don't want to continue to the next part till i know this is right. please help

 

[jbsteele]

! thanks You have an error in your calculation of u2. You assumed that the velocity of the ball after the collision is equal to the angular velocity of the pole times the length of the pole, but this is not correct. After the collision, the balls velocity will be the same as the velocity of the center of mass of the pole. Since the center of mass is located at a distance of L/2 from the bottom end of the pole, then the velocity of the ball after the collision will be equal to the angular velocity of the pole times the distance from the center of mass to the bottom end of the pole, which is L/2. So, instead of u2 = omega*L, it should be u2 = omega*(L/2). You can then use the conservation of momentum and energy to solve the question. The momentum of the system before the collision is simply the momentum of the ball, mv. The momentum of the system after the collision is the sum of the momentum of the ball and the momentum of the pole, which is mv + Mu2. By equating these two, you get v = (m+M/2)u2. Then, you can use conservation of energy to calculate the velocity of the ball before the collision, v. The total energy before the collision is the kinetic energy of the ball, 0.5mv^2. The total energy after the collision is the kinetic energy of the ball and the pole plus the potential energy of the pole, 0.5mu^2 + 0.5I(omega)^2 - mgL - 0.5MgL. By equating these two and solving for v, you should get the same answer as the textbook.

 

[thomate1]

Dear writer,

Thank you for sharing your question and calculations with me. After reviewing your work, I can see that you have made a few mistakes in your calculations.

Firstly, in your calculation of the initial velocity of the ball, you have used the incorrect value for the angular velocity. The correct value would be u1=omega*(L/2), as the ball is attached to the pole at its center of mass.

Secondly, in your calculation of the final velocity, you have not taken into account the rotational inertia of the pole after the collision. The correct equation should be v^2=u^2*[1+(M/3m)L]^2+I/mL^2.

Lastly, in your final equation, you have incorrectly substituted the value for the angular velocity. The correct equation should be v^2=2gL(1-cos(alpha))(1+(M/2m))(1+(I/mL^2)).

I would recommend double-checking your calculations and equations to ensure accuracy. It is also helpful to label all variables and units to avoid any confusion.

I hope this helps and please let me know if you have any further questions. Keep up the good work in your scientific studies!