Man moving on the disc!

Classical Mechanics Level 5

Let a uniform circular disc of mass m and radius R is free to rotate in Horizontal Plane about a Point "O" at the edge ( Hinge at the Point "O" ) . A Man Having Same Mass m is standing at the point "A" which is just opposite diametric end point with respect to point "O" .

Now at time t=0 , man Starts moving with Constant velocity " ${ u }_{ rel }$ " with respect to disc.

Then Find The angular displacement of the disc ( ${ \theta }_{ disc }$ ) in the time interval when Man reaches again at point "A" at the disc.

If Your answer is ${ \theta }_{ disc }$ and it is expressed as :

$\\ { \theta }_{ disc }\quad =\quad \pi \quad -\quad \pi \quad \times \sqrt { \cfrac { a }{ b } }$ .

Then Compute " a + b " ?

Details

$\bullet$ 'a' and 'b' are co-prime positive integers.

$\bullet$ Man moves on circumference of disc only.

Original

This is part of my set Deepanshu's Mechanics Blasts

The answer is 14.

1 solution

Deepanshu Gupta
Nov 14, 2014

Image

Here we can't apply Conservation of linear momentum and energy conservation . Since system is hinge about "O"

So let us Consider "Man + disc" as a system , So Net Angular momentum of the system is conserved about the Hinge Point "O".

$\sum { { L }_{ (intial),o } } \quad =\quad \sum { { L }_{ (final),o } } \quad \quad (\quad anti\quad clockwise\quad +ve\quad )\\ \\ 0\quad +\quad 0\quad =\quad { L }_{ (disc),o }\quad +\quad { L }_{ (man),o }\\ \\ \quad \quad \quad \quad 0\quad \quad =\quad (\cfrac { m{ R }^{ 2 } }{ 2 } \quad +\quad m{ R }^{ 2 }\quad )(-{ \omega }_{ d })\quad +\quad mx(\quad { u }_{ rel }\cos { \cfrac { { \theta }_{ r } }{ 2 } } \quad -{ \quad \omega }_{ d }x\quad )\quad \quad .\quad .\quad .\quad (1)$ .

$x\quad =\quad 2R\cos { \cfrac { { \theta }_{ r } }{ 2 } } \quad \quad .\quad .\quad .\quad .\quad (2)$ .

from 1 and 2 we get :

${ \omega }_{ d }\quad =\quad \frac { 4{ u }_{ rel }{ (\cos { \cfrac { { \theta }_{ r } }{ 2 } } ) }^{ 2 } }{ 3R\quad +\quad 8R{ (\cos { \cfrac { { \theta }_{ r } }{ 2 } } ) }^{ 2 } } \quad \quad \quad .\quad .\quad .\quad .\quad .\quad (3)$ .

And

${ \omega }_{ d }\quad =\quad \cfrac { d{ \theta }_{ d } }{ dt } \quad \quad .\quad .\quad .\quad (4)$ .

Also man moves in circle with centre 'C' with angular velocity ${ \omega }_{ rel }$ Such that :

${ u }_{ rel }\quad =\quad { \omega }_{ rel }R\quad$ .

$\\ \cfrac { { u }_{ rel } }{ R } \quad =\quad { \omega }_{ rel }\quad =\quad \cfrac { d{ \theta }_{ r } }{ dt } \quad \quad \quad .\quad .\quad .\quad .\quad (5)$ .

using all equations we get relation :

$\int _{ 0 }^{ { \theta }_{ d } }{ \quad d{ \theta }_{ d } } \quad =\quad \int _{ { \theta }_{ r }=0 }^{ { \theta }_{ r }=2\pi }{ \quad } \frac { 4{ (\cos { \cfrac { { \theta }_{ r } }{ 2 } } ) }^{ 2 } }{ 3\quad +\quad 8{ (\cos { \cfrac { { \theta }_{ r } }{ 2 } } ) }^{ 2 } } \quad d{ \theta }_{ r }$ .

Now integrating this simple integral by change limits from 0 to pi/2 and by putting ${ \theta }_{ r }\quad =\quad 2t$ and then change it into tan(t) and putt again $tan(t)=Z$ and using standard result that : $\int { \cfrac { dx }{ { a }^{ 2 }+{ x }^{ 2 } } } =\quad \cfrac { 1 }{ a } \arctan { \cfrac { x }{ a } }$ . we get Final answer :

$\boxed { { \theta }_{ disc }\quad =\quad \pi \quad -\quad \pi \quad \times \sqrt { \cfrac { 3 }{ 11 } } }$ .

Q.E.D

Simple question did it the same way.

Ronak Agarwal - 6 years, 6 months ago

Did it the same way. Simple but creative!

Aniket Sanghi - 4 years, 2 months ago

Just to make it more clear it must be mentioned constant speed not velocity.But often go differently.

Spandan Senapati - 4 years ago

Well I am a bit dumb...the qs was tough by my standards but ur solution was simply mindblowing sir...hats off...upvoted u sir for such a nice solution and presentation!!!!

rajdeep brahma - 3 years ago

Man moving on the disc!

Original

This is part of my set Deepanshu's Mechanics Blasts

The answer is 14.

1 solution

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