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Pivoted rod - impulse applied so no horizontal force on pivot

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Aaron7
#1
Apr3-12, 07:29 PM
P: 14
1. The problem statement, all variables and given/known data
There is a rod of length L pivoted on one end. It is originally at rest horizontally then released. When vertical, an impulse is applied to bring the rod to rest.

Throughout the question I have worked out I = 1/3 mL^2

w= √(3g/L) at the vertical position and an impulse of m√(gL/3) is the minimum required to achieve bringing the rod to rest.

I am attempting to find the impulse and distance from the pivot so that no horizontal force is exerted at the pivot.


2. Relevant equations

G = I dw/dt
v=rw
a=r dw/dt
Angular impulse J = r x (Impulse) = change in angular momentum
Angular momentum L = Iw

3. The attempt at a solution

I have tried: impulse required = mv = mrw
=> m√(gL/3) = mr√(3g/L)
=> r = L/3

I am confused on how to find the impulse and its distance so that the horizontal force on the pivot is 0. Any help is much appreciated.

Many thanks.
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tiny-tim
#2
Apr4-12, 02:59 PM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,148
Hi Aaron7!

Find the change in (linear) momentum of the centre of mass that gives you the magnitude of the impulse

then find the change in angular momentum of the whole rod that give you the torque of that impulse


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