Question
Asked Dec 19, 2019
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A light rod of length 2L is free to rotate in a vertical
plane about a frictionless pivot through its center. A particle
of mass m1 is attached at one end of the rod, and a mass m2
is at the opposite end, where m1 >m2. The system is released
from rest in the vertical position shown in Figure P8.84a, and
at some later time, the system is rotating in the position shown
in Figure P8.84b. Take the reference point of the gravitational
potential energy to be at the pivot. (a) Find an expression for
the system’s total mechanical energy in the vertical position.
(b) Find an expression for the total mechanical energy in the
rotated position shown in Figure P8.84b. (c) Using the fact
that the mechanical energy of the system is conserved, how
would you determine the angular speed v of the system in the
rotated position? (d) Find the magnitude of the torque on
the system in the vertical position and in the rotated position.

Is the torque constant? Explain what these results imply
regarding the angular momentum of the system. (e) Find an
expression for the magnitude of the angular acceleration
of the system in the rotated position. Does your result make
sense when the rod is horizontal? When it is vertical? Explain.

т
m2
тg
a
Figure P8.84
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т m2 тg a Figure P8.84

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Expert Answer

Step 1

Given,

The masses are m1, m2 and the total length of the rod is 2L.

Step 2

(a)

The total mechanical energy of the system about the vertical position is

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E = m;gL – m,gL gis the accelerati on due to gravity

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Step 3

(b)

The total mechanical energy of the rotating system is the sum of...

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E' = KE'+PE'

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