Moment of Inertia of a hollow cone about its base

[Potatochip911]

Homework Statement

Show that the moment of inertia of a hollow cone of mass M, radius R, and height h about its base is $\frac{1}{4}M(R^2+2h^2)$

Homework Equations

$I=\int r^2dm$
where r is the perpendicular distance from the axis

Surface Area of a cone $= \pi R (R^2+h^2)^{1/2}$

The Attempt at a Solution

[/B]
Mass density $\sigma = \frac{M}{SA} = \frac{M}{\pi R(R^2+h^2)^{1/2}}$

Now $I = \int r^2dm = \int r^2\sigma dA = \sigma \int r^2dA$

Now the distance from its base is given by $r = r(x) = h(1-x/R)$ and from the image I made it appears the area element is simply $dA = 2\pi xds$, and $ds = \sqrt{dx^2+dy^2} = \sqrt{1+y'(x)^2}dx$. Trivially $y'(x)^2 = (h/R)^2$ giving $dA = 2\pi x \sqrt{1+(h/R)^2}dx$ and $$I = \sigma \int r^2dA =\sigma \int_0^R h^2\left(1-\frac{x}{R}\right)^22\pi x \sqrt{1+\left(\frac{h}{R}\right)^2}dx \\ I = 2\pi\sigma h^2\sqrt{1+\left(\frac{h}{R}\right)^2}\int_0^R \left(1-\frac{x}{R}\right)^2 xdx$$

which clearly isn't going to give the answer since there's no way $\sqrt{1+(h/R)^2}$ will be cancelled

 

[BvU]

You sure ? What about the integral to calculate $\sigma$ ?

 

[Ray Vickson]

Homework Statement

Show that the moment of inertia of a hollow cone of mass M, radius R, and height h about its base is $\frac{1}{4}M(R^2+2h^2)$

Homework Equations

$I=\int r^2dm$
where r is the perpendicular distance from the axis

Surface Area of a cone $= \pi R (R^2+h^2)^{1/2}$

The Attempt at a Solution

[/B]
Mass density $\sigma = \frac{M}{SA} = \frac{M}{\pi R(R^2+h^2)^{1/2}}$

Formulas for surface area are poorly presented on the web, and some of them seem wrong. The more credible sources give
$$\text{SA} = \pi R \left( R + \sqrt{R^2+h^2} \right), $$
but one should make absolutely sure by calculating the relevant integral.

 

[Potatochip911]

Formulas for surface area are poorly presented on the web, and some of them seem wrong. The more credible sources give
$$\text{SA} = \pi R \left( R + \sqrt{R^2+h^2} \right), $$
but one should make absolutely sure by calculating the relevant integral.

You sure ? What about the integral to calculate $\sigma$ ?

Whoops, I forgot to mention it's an open ended cone. I.e. There is no base portion $pi R^2$

 

[BvU]

Basically you have to calculate $$\int r^2 dm\over \int dm $$

 

[Potatochip911]

Basically you have to calculate $$\int r^2 dm\over \int dm $$

Could you elaborate on why I need to calculate this quantity when moment of inertia is given by $I = \int r^2 dm$?

 

[Ray Vickson]

Homework Statement

Show that the moment of inertia of a hollow cone of mass M, radius R, and height h about its base is $\frac{1}{4}M(R^2+2h^2)$

Homework Equations

$I=\int r^2dm$
where r is the perpendicular distance from the axis

Surface Area of a cone $= \pi R (R^2+h^2)^{1/2}$

The Attempt at a Solution

[/B]
Mass density $\sigma = \frac{M}{SA} = \frac{M}{\pi R(R^2+h^2)^{1/2}}$

Now $I = \int r^2dm = \int r^2\sigma dA = \sigma \int r^2dA$
View attachment 109197

Now the distance from its base is given by $r = r(x) = h(1-x/R)$ and from the image I made it appears the area element is simply $dA = 2\pi xds$, and $ds = \sqrt{dx^2+dy^2} = \sqrt{1+y'(x)^2}dx$. Trivially $y'(x)^2 = (h/R)^2$ giving $dA = 2\pi x \sqrt{1+(h/R)^2}dx$ and $$I = \sigma \int r^2dA =\sigma \int_0^R h^2\left(1-\frac{x}{R}\right)^22\pi x \sqrt{1+\left(\frac{h}{R}\right)^2}dx \\ I = 2\pi\sigma h^2\sqrt{1+\left(\frac{h}{R}\right)^2}\int_0^R \left(1-\frac{x}{R}\right)^2 xdx$$

which clearly isn't going to give the answer since there's no way $\sqrt{1+(h/R)^2}$ will be cancelled

Yes, it will cancel because it appears as well in the denominator of your equation for $\sigma$.

 

[Potatochip911]

Yes, it will cancel because it appears as well in the denominator of your equation for $\sigma$.

Hmm, evaluating that integral gives $I= Mh^2/6$ which unfortunately isn't the correct answer