Piece-wise Uniform Rod Inertia

Classical Mechanics Level 3

Consider a uniform rod of length $b$ . One half of this rod has a linear mass density of $\lambda_1 = 9$ while the other half has a linear mass density of $\lambda_2 = 1$ . Consider this composite rod to be lying on the XY plane. The moment of inertia of the rod about an axis perpendicular to the XY plane and passing through its center of mass can be expressed as:

$I_{cm} = \frac{pb^3}{q}$

Note that $p$ and $q$ are positive co-prime integers. Enter your answer as $\boxed{p+q}$ .

The answer is 73.

1 solution

Steven Chase
Oct 11, 2019

I will spare the gory details of the number crunching and focus on the concepts.

Masses of the two parts:

$M_1 = \frac{9 b}{2} \\ M_2 = \frac{b}{2}$

Coordinate of center of mass:

$x_{CM} = \frac{M_1 \frac{b}{4} + M_2 \frac{3 b}{4} }{M_1 + M_2} = \frac{3 b}{10}$

Moment of inertia (using parallel axis theorem):

$I = \frac{M_1 (b/2)^2}{12} + \frac{M_2 (b/2)^2}{12} + M_1 (b/4 - x_{CM})^2 + M_2 (3b/4 - x_{CM})^2 = \frac{13 \, b^3}{60}$

Thanks for the solution. I intended to post a single problem composed of many smaller computations. Instead, I am posting it in parts. More will be online soon.

Karan Chatrath - 1 year, 8 months ago

I have uploaded follow-ups. I hope you find them interesting. Request you to take a look when you can.

Karan Chatrath - 1 year, 8 months ago

Indeed, I will try them tomorrow. They look pretty interesting. Thanks

Steven Chase - 1 year, 8 months ago

I have tried the energy problem, and I get an expression with a form almost identical to yours, but it seems to require another constant in the numerator above your "C" constant. I incorporated translational kinetic energy of the COM, rotational kinetic energy about the COM, and gravitational potential energy, assuming the center of the circle to be the potential reference.

Steven Chase - 1 year, 8 months ago

I have rechecked my work and I cannot seem to find an error. You can report the problem and show your working following which I will re-assess my steps.

Just as a test:

Taking $b = 0.5$ and $R = 2$ , I get:

$T = \frac{475}{96}\dot{\theta}^2$

I used these numbers in the follow up to the energy problem.

Karan Chatrath - 1 year, 8 months ago

@Karan Chatrath – Never mind, I spotted an arithmetic error I made. I have since solved and posted a solution. Thanks

Steven Chase - 1 year, 8 months ago

Piece-wise Uniform Rod Inertia

The answer is 73.

1 solution

0 pending reports