- #1

zenterix

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- Homework Statement
- (Apostol, Vol II, 2nd Ed., Chapter. 12.10, Problem 10)

A homogenous paper rectange of base ##2\pi a## and altitude ##h## is rolled to form a circular cylindrical surface ##S## of radius ##a##. Calculate the moment of inertia of ##S## about an axis through a diameter of the circular base.

- Relevant Equations
- ##I=\int\int_S r^2 dm##

Here is the homogenous paper rectangle

And if we roll it we get a cylinder with base radius ##a##.

It is not clear to me what "an axis through a diameter of the circular base means".

Let's imagine such as axis is ##\alpha## in the following figure

Then we have

$$I_{\alpha}=\int_0^h\int_0^{2\pi} z^2 \cdot \frac{M}{4\pi a^2}ad\theta dz = \frac{Mh^3}{6a}\tag{1}$$

where I have used a mass density of ##\frac{M}{4\pi a^2}## for the cylindrical surface.

Now, technically, if we were to use the following axis instead

then it also "passes through" a diameter of the circular base. In this case we have

$$I_{\alpha}=\int_0^h\int_0^{2\pi} a^2\cdot \frac{M}{4\pi a^2} a d\theta dz=\frac{Mah}{2}$$

Neither one of these calculations matches the answer at the end of the book, which is ##\pi a^3h+\frac{2}{3}\pi ah^3##.

What am I doing wrong here?

And if we roll it we get a cylinder with base radius ##a##.

It is not clear to me what "an axis through a diameter of the circular base means".

Let's imagine such as axis is ##\alpha## in the following figure

Then we have

$$I_{\alpha}=\int_0^h\int_0^{2\pi} z^2 \cdot \frac{M}{4\pi a^2}ad\theta dz = \frac{Mh^3}{6a}\tag{1}$$

where I have used a mass density of ##\frac{M}{4\pi a^2}## for the cylindrical surface.

Now, technically, if we were to use the following axis instead

**
**then it also "passes through" a diameter of the circular base. In this case we have

$$I_{\alpha}=\int_0^h\int_0^{2\pi} a^2\cdot \frac{M}{4\pi a^2} a d\theta dz=\frac{Mah}{2}$$

Neither one of these calculations matches the answer at the end of the book, which is ##\pi a^3h+\frac{2}{3}\pi ah^3##.

What am I doing wrong here?