# Homework Help: Linear and angular momentum

1. Jan 4, 2009

### Dell

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

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

v1=omega*L

the momentum is conserved during the collision so

P1=P2+L2 (can i mix angular and linear momentum like that and say that the total momentum is conserved??)

mv0=mv1+(M/3)L^2(omega)

mv0=mv1+(M/3)Lv1

v0=v1[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]

0.5mv1^2+I(omega)^2-mgL-0.5MgL=-mgLcos(alpha)-Mg(L/2)cos(alpha)

0.5mv1^2+(M/6)v1^2=mgL+0.5MgL-mgLcos(alpha)-Mg(L/2)cos(alpha)

v1^2[(m/2)+(M/6)=gL[m+(M/2)](1-cos(alpha))

v1^2=2gl(m+(M/2))(1-cos(alpha))(3/(3m+M))

v1^2=3gL[(2m+M)/(3m+M)](1-cos(alpha))

v0^2=v1^2*[1+(M/3m)L]^2

v0^2=[3gL[(2m+M)/(3m+M)](1-cos(alpha))]*[1+(M/3m)L]^2

does this seem correct?? the answer in my textbook is

v0^2=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

Last edited by a moderator: May 3, 2017
2. Jan 4, 2009

### chaoseverlasting

You cant add linear and angular momentum like you have. Individually, linear and angular momentum are conserved which would give you two separate equations.

3. Jan 4, 2009

### Dell

so how do i work my equations?
can i say
mV0=(M+m)V1???

what about angular momentum? is it not conserved??? since at the beginning omega=0

how would you go about solving this ???

4. Jan 5, 2009

### Dell

i think what i need to do is this:

from now on V0=V and V1=u

mv=Mu+mu

for M, u=omega*L/2, the angular velocity*the radius(to the centre of mass)

so mv=(1/2)M(omega)L + mu

now i have the energy equation from before, so
u=3gL[(2m+M)/(3m+M)](1-cos(alpha))

but from there what do i do,

chaoseverlasting, you said that the linear and angular momentum are conserved, how is the angular momentum conserved, at the beginning, the angular momentum is 0 since omega is 0.

i desperately need help with this one,