# Homework Help: Impulse on a rod.

1. Feb 23, 2012

### AlchemistK

1. The problem statement, all variables and given/known data
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.

3. 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?

2. Feb 23, 2012

### kushan

Is the length l and mass m ?

3. Feb 23, 2012

### AlchemistK

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

But if it cant be done without knowing that, go on ahead and use l and m.

4. Feb 23, 2012

### kushan

See I think ( I am not sure :( )

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

5. Feb 23, 2012

### kushan

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

6. Feb 23, 2012

### AlchemistK

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

7. Feb 23, 2012

### kushan

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

8. Feb 23, 2012

### 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.

9. Feb 23, 2012

### kushan

ohk let me try once again

10. Feb 23, 2012

### kushan

add conservation of moentum also to find v

11. Feb 23, 2012

### AlchemistK

12. Feb 23, 2012

### kushan

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

13. Feb 23, 2012

### AlchemistK

Why should it be v/r ?

14. Feb 23, 2012

### AlchemistK

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

15. Feb 23, 2012

### kushan

even if we put v/r answer is not coming

16. Feb 23, 2012

### 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!

17. Feb 23, 2012

### kushan

i have some figures

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18. Feb 23, 2012

### kushan

And the last one

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19. Feb 23, 2012

### AlchemistK

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

20. Feb 23, 2012

### kushan

keep coming back to PF

21. Feb 23, 2012

### tiny-tim

this is all very complicated

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

22. Feb 23, 2012

### kushan

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

23. Feb 23, 2012

### tiny-tim

you can choose either (but nowhere else)

c.o.m. is much simpler

24. Feb 23, 2012

### kushan

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

25. Feb 23, 2012

### AlchemistK

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