Moment of Inertia of a hollow cone about its base

  • #1

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##
MOI.png


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
 

Answers and Replies

  • #2
BvU
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You sure ? What about the integral to calculate ##\sigma## ?
 
  • #3
Ray Vickson
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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.
 
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  • #4
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##
 
  • #5
BvU
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Basically you have to calculate $$\int r^2 dm\over \int dm $$
 
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  • #6
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##?
 
  • #7
Ray Vickson
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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##.
 
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  • #8
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
 

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