1) Find the angular momentum vector about the origin of a particle of mass = 1.2
ID: 1607424 • Letter: 1
Question
1) Find the angular momentum vector about the origin of a particle of mass = 1.20 kg that is moving with a velocity (2.13 i + 1.22 j) m/s at the instant when its location is given by the position vector, r = (1.00 i – 0.724 j) m.
Draw a sketch of all the vectors. Label them properly.
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2) A thin rod of mass M and length L is struck at one end by a ball of clay of mass m, moving with speed v as shown in the figure. The ball sticks to the rod.
What is the angular momentum of the ball of clay about the pivot point A before collision?
After the collision, what is the angular momentum of the clay-rod system about A, the midpoint of the rod? Draw a sketch of the system after the collision on the right. ( I of a thin rod about the C.M. is (1/12) ML2 )
909Explanation / Answer
1)
Given
Velocity v=2.13i+1.22j
Position vector r=1.00i-0.724j
L=rxp
i j k
1.00 -0.724 0
2.13 1.22 0
2.13x-0.724=-1.54 , 1x1.22=1.22
=>ad-bc
= 1.22-(-1.54)
rxv=2.76
L=(m)(rxv)
= (1.2)x(2.76)
= 3.312 kgm2/s
2)
There are no external forces on the system
So the total angular momentum is conserved
Therefore angular momentum about point A after collision
Massm=m
Length=l
I=(1/12)x(m)l2 + m(l/2)2
I=(1/3)xml2
Li=Lf
mv(l/2)=(1/3)ml2 f
v=(2/3)(lf)
3v/2l=f
L=I
L=(1/3)ml2 x
L=((1/3)ml2 )x(3v/2l)
L=(1/2)xmlv
That is mvl/2
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