Find Instantaneous Center of Rotation for Impulse on a Rod

[AlchemistK]

Homework Statement

Suppose there is a rod on a friction less horizontal plane and an impulse "J" is given to it at a distance "x" from its center.

Using this data I have to find out the instantaneous center of rotation.

The Attempt at a Solution

By impulse-momentum theorem,
J = M*Vcm
(Vcm= Velocity of center of mass)

and by angular impulse momentum theorem,
J*x = I*ω

What do I do next?

 

[kushan]

Is the length l and mass m ?

 

[AlchemistK]

Well, the source of the question isn't very reliable, but no that data is not given.

But if it can't be done without knowing that, go on ahead and use l and m.

 

[kushan]

See I think ( I am not sure :( )

J(x+r)= ( Ml^2 /12 + mr^2 )( v/(x+r))

 

[kushan]

Here I have assumed instantaneous axis of rotation is r units below centre of mass

 

[AlchemistK]

J(x+r)= ( Ml^2 /12 + mr^2 )( v/(x+r))

v here is Vcm right?

And I just rechecked the question, the length is given as L but the mass isn't given.

The answer is given as (L^2)12x

 

[kushan]

L is given ? and j is not coming in the answer?

 

[AlchemistK]

Yes, L is given and nope, no J.

In fact, when we solve this equation : J(x+r)= ( Ml^2 /12 + Mr^2 )( v/(x+r))

we get a quadratic: (on substituting Vcm as J/M by linear momentum impulse theorem)

x^2 + 2xr - (L^2)/12 = 0

Which unfortunately doesn't give the correct answer, but I think we're close.

 

[kushan]

ohk let me try once again

 

[kushan]

add conservation of moentum also to find v

 

[AlchemistK]

On further speculation, had it been v/r instead of v/(x+r) we'd have got the correct answer, without a quadratic forming.

 

[kushan]

yea i guess it should be v/r , i am not still sure

 

[AlchemistK]

Why should it be v/r ?

 

[AlchemistK]

ω = v/(perpendicular distance) = v/(x+r) as you earlier said.

 

[kushan]

even if we put v/r answer is not coming

 

[AlchemistK]

J(x+r)= ( Ml^2 /12 + Mr^2 )( v/r)

Now J = M*v

J(x+r)= ( Ml^2 /12 + Mr^2 )*J/(Mr)

r(x+r)= l^2 /12 + r^2

xr = l^2 /12

Whoops! I realized I made an error typing the correct answer in my earlier post! I'm very sorry!

 

[kushan]

i have some figures

 

[kushan]

And the last one

 

[AlchemistK]

Oh yes x.x distance from the com! Thanks a lot!

 

[kushan]

your welcome :D
keep coming back to PF

 

[tiny-tim]

this is all very complicated

just use v_c.o.m = J/m, Iω = Jx, v_c.o.m = -rω

 

[kushan]

hey tiny tim
Should I be about com or instantaneous axis?

 

[tiny-tim]

you can choose either (but nowhere else)

c.o.m. is much simpler

 

[kushan]

lol , yea it seems easy now , but can you please wish to go through what I have done ,
Is it correct ?

 

[AlchemistK]

What difference would it have made if the rod was standing vertical?

 

[tiny-tim]

i'm not looking through pictures

AlchemistK's post #16 looks correct

(but using the c.o.m. would have avoided those extra r²s)​