Question
Asked Dec 19, 2019
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A light rod of length l= 1.00 m
rotates about an axis perpendicular
to its length and passing
through its center as in
Figure P8.51. Two particles of
masses m1 = 4.00 kg and m2 =3.00 kg are connected to the
ends of the rod. (a) Neglecting
the mass of the rod, what is the
system’s kinetic energy when
its angular speed is 2.50 rad/s?
(b) Repeat the problem, assuming
the mass of the rod is taken
to be 2.00 kg.

тg
х
т
Figure P8.51
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тg х т Figure P8.51

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Expert Answer

Step 1

Given:

m1 = 4.00 kg

m2 =3.00 kg

Length of the rod, L = 1 m

Angular velocity of rod = 2.5 rad/s

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-x-

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Step 2

Calculating the moment of inertia of the system given that rod is of negligible mass:

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Iystam = m,r +m,r From the diagram, r== 2 Hence, Iatem = (m, +m, system 1m =1.75 kgm? Substituting the values, Iatam = (4kg +3kg) system Iyuem =1.75 kgm² וב

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Step 3

Calculating the kinetic en...

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Kinetic energy of the system: K.E= system (1.75kgm )(2.5rad /s)? = 2.2J Substituting the values, K.E KE = 2.2J

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Science

Physics

Angular Motion

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