Radius of gyration or gyradius of a body about the axis of rotation is defined as the radial distance to a point which would have a moment of inertia the same as the body's actual distribution of mass, if the total mass of the body were concentrated there.
Mathematically the radius of gyration is the root mean square distance of the object's parts from either its center of mass or a given axis, depending on the relevant application. It is actually the perpendicular distance from point mass to the axis of rotation. One can represent a trajectory of a moving point as a body. Then radius of gyration can be used to characterize the typical distance travelled by this point.
Suppose a body consists of
n
{\displaystyle n}
particles each of mass
m
{\displaystyle m}
. Let
r
1
,
r
2
,
r
3
,
…
,
r
n
{\displaystyle r_{1},r_{2},r_{3},\dots ,r_{n}}
be their perpendicular distances from the axis of rotation. Then, the moment of inertia
I
{\displaystyle I}
of the body about the axis of rotation is
I
=
m
1
r
1
2
+
m
2
r
2
2
+
⋯
+
m
n
r
n
2
{\displaystyle I=m_{1}r_{1}^{2}+m_{2}r_{2}^{2}+\cdots +m_{n}r_{n}^{2}}
If all the masses are the same (
m
{\displaystyle m}
), then the moment of inertia is
I
=
m
(
r
1
2
+
r
2
2
+
⋯
+
r
n
2
)
{\displaystyle I=m(r_{1}^{2}+r_{2}^{2}+\cdots +r_{n}^{2})}
.
Since
m
=
M
/
n
{\displaystyle m=M/n}
(
M
{\displaystyle M}
being the total mass of the body),
I
=
M
(
r
1
2
+
r
2
2
+
⋯
+
r
n
2
)
/
n
{\displaystyle I=M(r_{1}^{2}+r_{2}^{2}+\cdots +r_{n}^{2})/n}
From the above equations, we have
M
R
g
2
=
M
(
r
1
2
+
r
2
2
+
⋯
+
r
n
2
)
/
n
{\displaystyle MR_{g}^{2}=M(r_{1}^{2}+r_{2}^{2}+\cdots +r_{n}^{2})/n}
Radius of gyration is the root mean square distance of particles from axis formula
R
g
2
=
(
r
1
2
+
r
2
2
+
⋯
+
r
n
2
)
/
n
{\displaystyle R_{g}^{2}=(r_{1}^{2}+r_{2}^{2}+\cdots +r_{n}^{2})/n}
Therefore, the radius of gyration of a body about a given axis may also be defined as the root mean square distance of the various particles of the body from the axis of rotation. It is also known as a measure of the way in which the mass of a rotating rigid body is distributed about its axis of rotation.
Hello,
I'm given a problem of a mass rolling down an incline with mass 'm', radius 'r', and radius of gyration Rg, and I need to write the Lagrangian for the motion. I'm confused on why both r and Rg are given. Don't I just need one of the two for the moment of inertia?
Thanks!
Homework Statement
Homework Equations
Centre of gravity: X=m1x1-m2x2/m1-m2
MOI rectangle: 1/3ml^2
MOI triangle: 1/18md^2
Radius of gyration: Ixx=mk^2
The Attempt at a Solution
Mass of body 1: b*l*p = 0.8*1*10=8kg
Mass of body 2: 1/2b*h*p = 1/2(0.4)*0.6*10=1.2kg1.1
X=m1x1-m2x2/m1-m2...
Find the moments of inertia about the x-axis, y-axis and the origin. Also, find the radius of gyration about the x-axis and y-axis.
y = 0, y = b, x = 0, x = a
Rho = ky
1. Is ky the density function?
2. Do I integrate over dxdy or dydx?
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Homework Statement
A 30 kg wheel has a center of mass 0.1 m left from the center of the wheel and radius of gyration KG = 0.15 m. Find the angular acceleration if the wheel is originally at rest. The radius of the wheel is 0.25m.
Homework Equations
I=mk^2
T=f*d
M=I*a
Fn acting bottom in Y...
Hi There
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Homework Statement
the theory behind Parallel axis theorem
Homework Equations
parallel axis theorem:Io=Ig+md²
Radius of Gyration=Ig=m*K²
The Attempt at a Solution
ok folks If I understand this theory correctly the radius of gyration is the radius or distance from the axis of rotation to the...
Homework Statement
Stacy is throwing a discus. During the throw, she applies an average torque of 90Nm to the discus for 0.3 seconds. The discus has a mass of 1.0kg, and has a radius of gyration of 0.1m about its spin axis. If the initial angular velocity of the discus was zero, what was the...
Homework Statement
The Attempt at a Solution
Alright, so the solution manual shows that the intertia of the disk is:
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Homework Statement
Homework Equations
radius of gyration:
r = root (I/m)
I = moment of inertia
m = mass
parallel axis theorem given above
The Attempt at a Solution
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Hey Guys,
I having trouble with understanding radius of gyration, could someone please explain what it is? I have just never understood it's full meaning. So for example, the radius of gyration of a spinning wheel of a car is ...some value... What does that mean?
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Homework Statement
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Homework Statement
Determine the polar moment of inertia and the polar radius of gyration of the shaded area shown with respect to point P.
http://imgur.com/8Kc1S
Homework Equations
Jp = Ix + Iy
Ix = &int y^2dA
Iy = &int X^2dA
The Attempt at a Solution
A = 2(a/2)(a) +...
Homework Statement
Homework Equations
i guess its r=sqrt(I/A)
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The Attempt at a Solution
I guess I'm just having trouble getting I. A is 33pi
Homework Statement
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Homework Statement
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Hey -
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