A rod of mass M and length L is initiallyat reat in a vertical position on a frictionless table, as shown below. !CM 0 y CM 22 If the stick starts falling, find the speed of the center of mass of as a function of the angle that the stick makes with the virtical. L We can find the speed of the center of mass by using the conservation of energy method. The only force acting o the stick is the gravitational force Mg in the vertical downward direction. Since there is no horizontal force acting on the rod, the center of mass falls vertically downward as shown. Let y be the speed of the center of mass. Initially, the body is at reat, it has zero kinetic energy, and its potential energy is MgL/2. E₁ = Ki + V₁ = MgL after the center of mass has fallen through a height y, the rod has rotated through an angle 0. The final potential energy is then, V₁= and the final kinetic energy, noting the vertical speed is ỷ and the rotational velocity is w = 0, is Kf = My² + 1,0² (Hint: Io = f pr² dr) where lo is the moment of inertial of the rod about an axis through the center of mass: The final total energy is From the conservatnon of energy Ę; = Ef, we get lo= E₁ = K₁ + V₁ = My² + ½±10² + Mg( ± − y) To obtain ý, we must eliminate ė. From the figure, we find: whick gives: and therfore: Make the appropriate substitustions to find: Ў y= y = = }My +ID +Mg( ; –3)
A rod of mass M and length L is initiallyat reat in a vertical position on a frictionless table, as shown below. !CM 0 y CM 22 If the stick starts falling, find the speed of the center of mass of as a function of the angle that the stick makes with the virtical. L We can find the speed of the center of mass by using the conservation of energy method. The only force acting o the stick is the gravitational force Mg in the vertical downward direction. Since there is no horizontal force acting on the rod, the center of mass falls vertically downward as shown. Let y be the speed of the center of mass. Initially, the body is at reat, it has zero kinetic energy, and its potential energy is MgL/2. E₁ = Ki + V₁ = MgL after the center of mass has fallen through a height y, the rod has rotated through an angle 0. The final potential energy is then, V₁= and the final kinetic energy, noting the vertical speed is ỷ and the rotational velocity is w = 0, is Kf = My² + 1,0² (Hint: Io = f pr² dr) where lo is the moment of inertial of the rod about an axis through the center of mass: The final total energy is From the conservatnon of energy Ę; = Ef, we get lo= E₁ = K₁ + V₁ = My² + ½±10² + Mg( ± − y) To obtain ý, we must eliminate ė. From the figure, we find: whick gives: and therfore: Make the appropriate substitustions to find: Ў y= y = = }My +ID +Mg( ; –3)
Physics for Scientists and Engineers: Foundations and Connections
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ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter10: Systems Of Particles And Conservation Of Momentum
Section: Chapter Questions
Problem 75PQ
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Transcribed Image Text:A rod of mass M and length L is initiallyat reat in a vertical position on a frictionless table, as shown below.
!CM
0
y
CM
22
If the stick starts falling, find the speed of the center of mass of as a function of the angle that the stick makes with the virtical.
L
We can find the speed of the center of mass by using the conservation of energy method. The only force acting o the stick is the gravitational force Mg in
the vertical downward direction. Since there is no horizontal force acting on the rod, the center of mass falls vertically downward as shown. Let y be the
speed of the center of mass.
Initially, the body is at reat, it has zero kinetic energy, and its potential energy is MgL/2.
E₁ = Ki + V₁ = MgL
after the center of mass has fallen through a height y, the rod has rotated through an angle 0. The final potential energy is then,
V₁=
and the final kinetic energy, noting the vertical speed is ỷ and the rotational velocity is w = 0, is
Kf
=
My² + 1,0²
(Hint: Io = f pr² dr)
where lo is the moment of inertial of the rod about an axis through the center of mass:
The final total energy is
From the conservatnon of energy Ę; = Ef, we get
lo=
E₁ = K₁ + V₁ = My² + ½±10² + Mg( ± − y)
To obtain ý, we must eliminate ė. From the figure, we find:
whick gives:
and therfore:
Make the appropriate substitustions to find:
Ў
y=
y =
= }My +ID +Mg( ; –3)
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