Question

3. A certain thin one-dimensional rod has a length of L and a non-uniform density, which is given by the following function (assume x=0 corresponds to the left end of the rod): ?(x) = x^2 / 7 Q1) Find the rod's center of mass as a function of L. Q2) Find the moment of inertia for the rod if it is rotated about its midpoint (L/2).

          3. A certain thin one-dimensional rod has a length of L and a non-uniform density, which is given by the following function (assume x=0 corresponds to the left end of the rod):
?(x) = x^2 / 7

Q1) Find the rod's center of mass as a function of L.

Q2) Find the moment of inertia for the rod if it is rotated about its midpoint (L/2).

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3. A certain thin one-dimensional rod has a length of L and a non-uniform density, which is given by the following function (assume x=0 corresponds to the left end of the rod):
?(x) = x^2 / 7

Q1) Find the rod's center of mass as a function of L.

Q2) Find the moment of inertia for the rod if it is rotated about its midpoint (L/2).

Added by Ignacio R.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Supreeta N

06/16/2023

Step 1

To do this, we can integrate the density function over the length of the rod: $$ m = \int_0^L p(x) dx = \int_0^L x^2 dx $$ Now, we can find the center of mass by dividing the first moment of mass by the total mass: $$ x_{cm} = \frac{1}{m} \int_0^L x p(x) dx = Show more…

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A certain thin one-dimensional rod has a length of L and a non-uniform density, which is given by the following function (assume x=0 corresponds to the left end of the rod): p(x) = x^2 / 7 Q1) Find the rod's center of mass as a function of L. Q2) Find the moment of inertia for the rod if it is rotated about its midpoint (L/2).

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Transcript

00:01 Hello students, we are given a road.

00:04 So road of length l is there and its density function is given as density of x is equal to x square by 7.

00:14 So it's a function of position density is changing with respect to position now to calculate the first thing, which is the center of mass.

00:22 We can calculate the center of mass, which is equal to 1 by m, which is the total mass times integral of x times row x dx.

00:35 Okay, so that is the center of mass.

00:37 So m is just m is just equal to so m will be equal to integral of row of x dx.

00:48 So that is equal to integral of row of x is equal to x square by 7 dx.

00:55 So that will be equal to from 0 to l right 0 to l.

01:01 That is the length.

01:01 So it will be equal to l cube by 3 times 7 that is 21.

01:08 So that will be the value.

01:11 So we got the m.

01:12 So center of mass is equal to 1 1 by l cube by so that is 21 by l cube times integral of x times row of x is equal to x square by x square by 7 times dx with the length of 0 to l.

01:31 So it will be equal to 21 by l cube times x cube that is x raised to 4 by 4.

01:39 So that is l raised to 4 by 28.

01:44 So that will give you a value which will be equal to this will be 7 by sorry.

01:51 Okay, you can keep it 21 by 28 times l.

01:55 So that is the center of mass.

02:00 The center of mass will be at the point where it is 21 by 28 that will be the center of mass point.

02:08 Now to calculate the moment of inertia.

02:11 So this is the center of mass...

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