1) Find the angular momentum vector about the origin of a particle of mass = 1.2

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 )

Explanation / 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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