Rotational Mechanics question

A disc of mass m and radius R is attached to a rectangular plate of the same mass breadth and length 2R . Find moment of inertia of this system about diameter of of disc. Note by Zahid Shekh Mohammed
6 years, 6 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
• Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

• bulleted
• list

1. numbered
2. list

1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

Hey you have mentioned that it is a $\underline{rectangular}$ $\underline{plate}$ of same mass, breadth and length 2R which actually contradicts that it is a rectangle. I have solved the problem taking it to be a square.

Here is how I got it: $\I_{of disc about diameter}$=$\frac {1}{4}$M$R^{2}$ and $\I_{of square about diameter of plate}$= $\frac {4}{3}$M$R^{2}$ . So $\I_{required}$=$\frac {19}{12}$M$R^{2}$

- 6 years, 6 months ago

Is it attached about its length or width ?

- 6 years, 6 months ago

- 6 years, 6 months ago

Hey dear, please post complete question because incomplete ques. causes undesired anxiety. You have not mentioned about which diameter ? Please post the complete ques. once more.

- 6 years, 5 months ago

Can I email the photo of question please give your email

- 6 years, 5 months ago

but answer is 31MRsqaure\12

- 6 years, 6 months ago

No actually breadth is R and length is 2R

- 6 years, 6 months ago

Ip=I (com) + Md^2 , p= moment of inertia of axis through p , parallel to centre of mass and at a distance d from centre of mass. Icom = M/6(R)^2 for rectangular sheet of breadth R Therefore , Taking p is the centre of disk and axis parallel to centre of mass of rectangular sheet axis, We get Ip=M/6R^2 + M(3R/2)^2 {AS DISTANCE BETWEEN COM OF RECTANGLE AND CIRCLE IS 3R/2}
Therefore, ip = 29Mr^2/12 Adding this to Mr^/4 , we get 32mr^2/12

- 5 years, 5 months ago

please specify how it is attached as it would be of great help in solving the problem...

- 6 years, 6 months ago

the disc and rectangular are attached about their ends

- 6 years, 6 months ago