An object consists of two particles, of masses m1 and m2, connected by a light rigid rod of length L. Neglecting the mass of the rod, find the moment of inertia of this system for rotations of this object about an axis perpendicular to the rod and a distance x from m1, where x < L.
Hence, show that the moment of inertia is minimum when the axis is located at the centre of mass.
I have tried to solve the first part of the problem and found that:
I = (m1+m2)*x^2 + m2*L*(L-2x) , where I is the moment of inertia.
But I don't know how to show "the moment of inertia is minimum when the axis is located at the centre of mass."
Thanks for answering.
Physics question about moment of inertia and centre of mass.?
2014-11-04 6:59 pm
回答 (2)
2014-11-04 7:50 pm
Since the rod is massless, we can ignore it.
I_point mass = m*r^2
I_m1 = m1*x^2
I_m2 = m2*(L-x)^2 = m2*(L^2 - 2*L*x + x^2) = m2*L^2 - 2*m2*L*x + m2*x^2
I_total = I_m1 + I_m2
I_total = m1*x^2 + m2*L^2 - 2*m2*L*x + m2*x^2
I_total = (m1+m2)*x^2 - (2*m2*L)*x +m2*L^2
Take the derivative with respect x to find where the mximum occurs. Maximum occurs at I' = 0
I_total' = 0 = 2*x*(m1+m2) - 2*m2*L
x = m2*L/(m1+m2)
Location of the CG... Set m1 at 0
CG_measured from m1 = (m1*r1 + m2*r2)/(m1+m2)
CG = (m1*0 + m2*L) / (m1+m2)
CG = (m2*L)/(m1+m2)
The CG and x at I' = 0 are the same.
I_point mass = m*r^2
I_m1 = m1*x^2
I_m2 = m2*(L-x)^2 = m2*(L^2 - 2*L*x + x^2) = m2*L^2 - 2*m2*L*x + m2*x^2
I_total = I_m1 + I_m2
I_total = m1*x^2 + m2*L^2 - 2*m2*L*x + m2*x^2
I_total = (m1+m2)*x^2 - (2*m2*L)*x +m2*L^2
Take the derivative with respect x to find where the mximum occurs. Maximum occurs at I' = 0
I_total' = 0 = 2*x*(m1+m2) - 2*m2*L
x = m2*L/(m1+m2)
Location of the CG... Set m1 at 0
CG_measured from m1 = (m1*r1 + m2*r2)/(m1+m2)
CG = (m1*0 + m2*L) / (m1+m2)
CG = (m2*L)/(m1+m2)
The CG and x at I' = 0 are the same.
2014-11-04 7:39 pm
> "I = (m1+m2)*x^2 + m2*L*(L-2x) "
That's correct, but I think it's more simply written in this (equivalent) way:
I = m1*x² + m2(L-x)²
> "But I don't know how to show..."
Calculus is good for showing maxima and minima:
* You can vary the position of the axis by varying the value of "x"
* Differentiate "I" with respect to "x"
* Figure out the value of "x" when the derivative is zero.
So:
dI/dx = d/dx (m1*x² + m2(L-x)²)
= 2*m1*x − 2*m2*(L-x)
Now set that to zero:
2*m1*x − 2*m2*(L-x) = 0
Solve for "x":
x = L(m2/(m1+m2))
Technically, that means that "I" reaches either a "minumum" or a "maximum" or an "inflection point" where x = L(m2/(m1+m2)). We can determine which of those three is the the case by taking the 2nd derivative of "I":
d²I/dx² = 2(m1+m2)
Clearly this value is positive. That means the graph of "I" vs. "x" is "concave-up" everywhere, which means x = L(m2/(m1+m2)) is a MINIMUM (good!)
Finally, we show that this value of "x" corresponds to the center of mass. Let "Xc" be the distance from m1 that corresponds to the center of mass. We know that, by definition of "center of mass", the following must be true:
m1*(m1's distance from the cm) = m2*(m2's distance from the cm)
And note that "m1's distance from the cm" is Xc, while "m2's distance from the cm" is (L−Xc), so:
m1*Xc = m2*(L-Xc)
Solve for Xc:
Xc = L(m2/(m1+m2))
And this is just the value for "x" that we found before (the value that minimizes "I"). So we can say that "I" is minimum when the axis is at the CM.
That's correct, but I think it's more simply written in this (equivalent) way:
I = m1*x² + m2(L-x)²
> "But I don't know how to show..."
Calculus is good for showing maxima and minima:
* You can vary the position of the axis by varying the value of "x"
* Differentiate "I" with respect to "x"
* Figure out the value of "x" when the derivative is zero.
So:
dI/dx = d/dx (m1*x² + m2(L-x)²)
= 2*m1*x − 2*m2*(L-x)
Now set that to zero:
2*m1*x − 2*m2*(L-x) = 0
Solve for "x":
x = L(m2/(m1+m2))
Technically, that means that "I" reaches either a "minumum" or a "maximum" or an "inflection point" where x = L(m2/(m1+m2)). We can determine which of those three is the the case by taking the 2nd derivative of "I":
d²I/dx² = 2(m1+m2)
Clearly this value is positive. That means the graph of "I" vs. "x" is "concave-up" everywhere, which means x = L(m2/(m1+m2)) is a MINIMUM (good!)
Finally, we show that this value of "x" corresponds to the center of mass. Let "Xc" be the distance from m1 that corresponds to the center of mass. We know that, by definition of "center of mass", the following must be true:
m1*(m1's distance from the cm) = m2*(m2's distance from the cm)
And note that "m1's distance from the cm" is Xc, while "m2's distance from the cm" is (L−Xc), so:
m1*Xc = m2*(L-Xc)
Solve for Xc:
Xc = L(m2/(m1+m2))
And this is just the value for "x" that we found before (the value that minimizes "I"). So we can say that "I" is minimum when the axis is at the CM.
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