Rotating Rod

Classical Mechanics Level 5

A uniform rod of mass $m = 2015.2016 \text{ gm}$ is placed on a rough table with $\dfrac13$ of its portion on the table. A point mass of equal mass is attched to its end that is in air. Then the system is left alone.

Observation : We see that the rod starts sliding when it makes an angle $\theta$ with the horizontal (measured clockwise).

What is the coefficient of friction $\mu$ of the table.

Details and Assumptions :

The length of the rod is $L = 2016.2017\text{ cm}$ .
Given, $\tan\theta = \dfrac {1}{12}$ .
The point mass is attached to the right end of the rod.
Take $g = 9.8\text{ m/s}^2$ .

Inspiration: Aditya Kumar.

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The answer is 0.5.

1 solution

Spandan Senapati
Sep 23, 2016

So lution goes like this.e answer is $0.5$ .what we can think of is that friction is static until the body arrives at the critical condition(the condition of slipping).so just at the moment of slipping friction is limiting.apply energy conservation to get the required angular velocity of the system .we can treat the system after we find the location of its centre of mass.resolve weight in suitable direction and treat the centripetal and tangential acceleration of the centre of mass.find a relation among friction and normal reaction .it comes out that $f=4.5mgsin@$ and $N=0.75mgcos@$ .and $tan@=1/12$

Rotating Rod

Inspiration: Aditya Kumar.

Try more here

The answer is 0.5.

1 solution

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