Question

A rod of length l=2.2m and mass M= 9.7kg joins two particles
with masses m_{1} =12.9kg and m_{2} = 5.0kg, at its
ends. The combination rotates in the xy-plane about a pivot through
the center of the rod with the linear speed of the masses of v=
12.9 m/s. (Moment of inertia of a uniform rod rotating about its
center of mass

* I*=

1 |

12 |

*M**l*^{2}

)

a) Calculate the total moment of inertia of the system

I = kg m ^{2}

b) What is the magnitude of the angular momentum about the
origin?

L= kg m^{2} /s

c)If the length of the rod reduces to l=1.1m. What will be the
final angular speed *ω*_{f}?

*ω*_{f} = rad/s

Answer #1

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with masses m1 =4.8kg and m2 = 2.8kg, at its ends. The combination
rotates in the xy-plane about a pivot through the center of the rod
with the linear speed of the masses of v= 3.5 m/s. (Moment of
inertia of a uniform rod rotating about its center of mass I= 1 12
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m/s. (Moment of inertia of a uniform rod rotating about its center
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I=
1
12
M l2
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m2 = 3.00 kg,
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about a pivot through the center of the rod (see figure below).
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8%)
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