Center of mass of an object in $mathbb{R^3}$
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A few days ago I asked this question. I have solved what I was asking there. The mass of the given object is $$M=displaystyleint_{0}^{2pi}displaystyleint_{0}^{pi/3}displaystyleint_{0}^{1}kdisplaystylefrac{r}{3}cos(theta)displaystylefrac{r^2}{6}sin(theta)drdtheta dphi$$$$+displaystyleint_{0}^{2pi}displaystyleint_{pi/3}^{pi/2}displaystyleint_{0}^{2costheta}kdisplaystylefrac{r}{3}cos(theta)displaystylefrac{r^2}{6}sin(theta)drdtheta dphi=displaystylefrac{5pi k}{432}$$
Is it correct that, if the center of mass is $(bar{x},bar{y},bar{z})$, $z=frac{r}{3}cos(theta)$ and the Jacobian of the transformation is $frac{r^2sin^2(theta)}{6}$
$$bar{z}=displaystylefrac{1}{M}displaystyleint_{0}^{2pi}displaystyleint_{0}^{pi/3}displaystyleint_{0}^{1}displaystylefrac{r}{3}cos(theta)displaystylefrac{r^2}{6}sin(theta)drdtheta dphi+displaystyleint_{0}^{2pi}displaystyleint_{pi/3}^{pi/2}displaystyleint_{0}^{2costheta}displaystylefrac{r}{3}cos(theta)displaystylefrac{r^2}{6}sin(theta)drdtheta dphi$$$$=frac{1}{k}$$
This don't make sense to me, since the density of mass of the object is directly proportional to the plane $xy$, shouldn't the z-axis of the center of mass be directly proportional to $k$?
multivariable-calculus definite-integrals volume
asked Dec 21 '18 at 12:55
John KeeperJohn Keeper
541315
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add a comment |
$begingroup$
A few days ago I asked this question. I have solved what I was asking there. The mass of the given object is $$M=displaystyleint_{0}^{2pi}displaystyleint_{0}^{pi/3}displaystyleint_{0}^{1}kdisplaystylefrac{r}{3}cos(theta)displaystylefrac{r^2}{6}sin(theta)drdtheta dphi$$$$+displaystyleint_{0}^{2pi}displaystyleint_{pi/3}^{pi/2}displaystyleint_{0}^{2costheta}kdisplaystylefrac{r}{3}cos(theta)displaystylefrac{r^2}{6}sin(theta)drdtheta dphi=displaystylefrac{5pi k}{432}$$
Is it correct that, if the center of mass is $(bar{x},bar{y},bar{z})$, $z=frac{r}{3}cos(theta)$ and the Jacobian of the transformation is $frac{r^2sin^2(theta)}{6}$
$$bar{z}=displaystylefrac{1}{M}displaystyleint_{0}^{2pi}displaystyleint_{0}^{pi/3}displaystyleint_{0}^{1}displaystylefrac{r}{3}cos(theta)displaystylefrac{r^2}{6}sin(theta)drdtheta dphi+displaystyleint_{0}^{2pi}displaystyleint_{pi/3}^{pi/2}displaystyleint_{0}^{2costheta}displaystylefrac{r}{3}cos(theta)displaystylefrac{r^2}{6}sin(theta)drdtheta dphi$$$$=frac{1}{k}$$
This don't make sense to me, since the density of mass of the object is directly proportional to the plane $xy$, shouldn't the z-axis of the center of mass be directly proportional to $k$?
multivariable-calculus definite-integrals volume
asked Dec 21 '18 at 12:55
John KeeperJohn Keeper
541315
$endgroup$
$begingroup$
You have to take the density, and $k$ with it, into the integral
$endgroup$
– Rafa Budría
Dec 21 '18 at 13:48
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@Rafa Budría Could you elaborate? Isnt the z-axis of the center of mass defined as $frac{1}{M}int_VzdV$?
$endgroup$
– John Keeper
Dec 21 '18 at 14:17
$begingroup$
The density $kz$ has to be into the integral, multiplying the coordinate $z$. The center of mass is the weighted mean of the masses.
$endgroup$
– Rafa Budría
Dec 21 '18 at 14:22
add a comment |
$begingroup$
A few days ago I asked this question. I have solved what I was asking there. The mass of the given object is $$M=displaystyleint_{0}^{2pi}displaystyleint_{0}^{pi/3}displaystyleint_{0}^{1}kdisplaystylefrac{r}{3}cos(theta)displaystylefrac{r^2}{6}sin(theta)drdtheta dphi$$$$+displaystyleint_{0}^{2pi}displaystyleint_{pi/3}^{pi/2}displaystyleint_{0}^{2costheta}kdisplaystylefrac{r}{3}cos(theta)displaystylefrac{r^2}{6}sin(theta)drdtheta dphi=displaystylefrac{5pi k}{432}$$
Is it correct that, if the center of mass is $(bar{x},bar{y},bar{z})$, $z=frac{r}{3}cos(theta)$ and the Jacobian of the transformation is $frac{r^2sin^2(theta)}{6}$
$$bar{z}=displaystylefrac{1}{M}displaystyleint_{0}^{2pi}displaystyleint_{0}^{pi/3}displaystyleint_{0}^{1}displaystylefrac{r}{3}cos(theta)displaystylefrac{r^2}{6}sin(theta)drdtheta dphi+displaystyleint_{0}^{2pi}displaystyleint_{pi/3}^{pi/2}displaystyleint_{0}^{2costheta}displaystylefrac{r}{3}cos(theta)displaystylefrac{r^2}{6}sin(theta)drdtheta dphi$$$$=frac{1}{k}$$
This don't make sense to me, since the density of mass of the object is directly proportional to the plane $xy$, shouldn't the z-axis of the center of mass be directly proportional to $k$?
multivariable-calculus definite-integrals volume
asked Dec 21 '18 at 12:55
John KeeperJohn Keeper
541315
$endgroup$
A few days ago I asked this question. I have solved what I was asking there. The mass of the given object is $$M=displaystyleint_{0}^{2pi}displaystyleint_{0}^{pi/3}displaystyleint_{0}^{1}kdisplaystylefrac{r}{3}cos(theta)displaystylefrac{r^2}{6}sin(theta)drdtheta dphi$$$$+displaystyleint_{0}^{2pi}displaystyleint_{pi/3}^{pi/2}displaystyleint_{0}^{2costheta}kdisplaystylefrac{r}{3}cos(theta)displaystylefrac{r^2}{6}sin(theta)drdtheta dphi=displaystylefrac{5pi k}{432}$$
Is it correct that, if the center of mass is $(bar{x},bar{y},bar{z})$, $z=frac{r}{3}cos(theta)$ and the Jacobian of the transformation is $frac{r^2sin^2(theta)}{6}$
$$bar{z}=displaystylefrac{1}{M}displaystyleint_{0}^{2pi}displaystyleint_{0}^{pi/3}displaystyleint_{0}^{1}displaystylefrac{r}{3}cos(theta)displaystylefrac{r^2}{6}sin(theta)drdtheta dphi+displaystyleint_{0}^{2pi}displaystyleint_{pi/3}^{pi/2}displaystyleint_{0}^{2costheta}displaystylefrac{r}{3}cos(theta)displaystylefrac{r^2}{6}sin(theta)drdtheta dphi$$$$=frac{1}{k}$$
This don't make sense to me, since the density of mass of the object is directly proportional to the plane $xy$, shouldn't the z-axis of the center of mass be directly proportional to $k$?
multivariable-calculus definite-integrals volume
multivariable-calculus definite-integrals volume
asked Dec 21 '18 at 12:55
John KeeperJohn Keeper
541315
asked Dec 21 '18 at 12:55
John KeeperJohn Keeper
541315
asked Dec 21 '18 at 12:55
John KeeperJohn Keeper
541315
asked Dec 21 '18 at 12:55
John KeeperJohn Keeper
541315
asked Dec 21 '18 at 12:55
John KeeperJohn Keeper
541315
541315
$begingroup$
You have to take the density, and $k$ with it, into the integral
$endgroup$
– Rafa Budría
Dec 21 '18 at 13:48
$begingroup$
@Rafa Budría Could you elaborate? Isnt the z-axis of the center of mass defined as $frac{1}{M}int_VzdV$?
$endgroup$
– John Keeper
Dec 21 '18 at 14:17
$begingroup$
The density $kz$ has to be into the integral, multiplying the coordinate $z$. The center of mass is the weighted mean of the masses.
$endgroup$
– Rafa Budría
Dec 21 '18 at 14:22
add a comment |
$begingroup$
You have to take the density, and $k$ with it, into the integral
$endgroup$
– Rafa Budría
Dec 21 '18 at 13:48
$begingroup$
@Rafa Budría Could you elaborate? Isnt the z-axis of the center of mass defined as $frac{1}{M}int_VzdV$?
$endgroup$
– John Keeper
Dec 21 '18 at 14:17
$begingroup$
The density $kz$ has to be into the integral, multiplying the coordinate $z$. The center of mass is the weighted mean of the masses.
$endgroup$
– Rafa Budría
Dec 21 '18 at 14:22
$begingroup$
You have to take the density, and $k$ with it, into the integral
$endgroup$
– Rafa Budría
Dec 21 '18 at 13:48
$begingroup$
You have to take the density, and $k$ with it, into the integral
$endgroup$
– Rafa Budría
Dec 21 '18 at 13:48
$begingroup$
@Rafa Budría Could you elaborate? Isnt the z-axis of the center of mass defined as $frac{1}{M}int_VzdV$?
$endgroup$
– John Keeper
Dec 21 '18 at 14:17
$begingroup$
@Rafa Budría Could you elaborate? Isnt the z-axis of the center of mass defined as $frac{1}{M}int_VzdV$?
$endgroup$
– John Keeper
Dec 21 '18 at 14:17
$begingroup$
The density $kz$ has to be into the integral, multiplying the coordinate $z$. The center of mass is the weighted mean of the masses.
$endgroup$
– Rafa Budría
Dec 21 '18 at 14:22
$begingroup$
The density $kz$ has to be into the integral, multiplying the coordinate $z$. The center of mass is the weighted mean of the masses.
$endgroup$
– Rafa Budría
Dec 21 '18 at 14:22
add a comment |
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$begingroup$
You have to take the density, and $k$ with it, into the integral
$endgroup$
– Rafa Budría
Dec 21 '18 at 13:48
$begingroup$
@Rafa Budría Could you elaborate? Isnt the z-axis of the center of mass defined as $frac{1}{M}int_VzdV$?
$endgroup$
– John Keeper
Dec 21 '18 at 14:17
$begingroup$
The density $kz$ has to be into the integral, multiplying the coordinate $z$. The center of mass is the weighted mean of the masses.
$endgroup$
– Rafa Budría
Dec 21 '18 at 14:22