- #1

- 13

- 0

y=sqrt(r^2-x^2)

where r is the radius

is revolved around the x-axis

I dont even know where to begin =[

- Thread starter sunniexdayzz
- Start date

- #1

- 13

- 0

y=sqrt(r^2-x^2)

where r is the radius

is revolved around the x-axis

I dont even know where to begin =[

- #2

berkeman

Mentor

- 58,768

- 8,893

Welcome to the PF. Start with the definition of the Moment of Inertia. What is it in the general case?

y=sqrt(r^2-x^2)

where r is the radius

is revolved around the x-axis

I dont even know where to begin =[

I'm also having a little trouble visualizing the shape... is there any way you can sketch it?

- #3

- 13

- 0

with m=mass and r=radius

but I don't know what it is for a curved rod. The rod in the problem is a semi circle about the origin in quadrants I and II with radius r

- #4

- 47

- 0

[tex]

I\ =\ \int dm\ r^2\

[/tex]

Some tips: dm can be expressed in terms of linear density [tex]dm\ = \rho d\theta[/tex], with theta in play due to it being easier in this case to use polar co-ordinates.

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