
#1
Apr312, 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. 



#2
Apr412, 02:59 PM

Sci Advisor
HW Helper
Thanks
P: 26,167

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 


Register to reply 
Related Discussions  
Coefficient of Friction when applied force is not horizontal to surface?  Introductory Physics Homework  3  
Friction (horizontal) Applied Greater Force than Equilibrium  Introductory Physics Homework  1  
Angular acceleration of a rod pivoted at an end, at angle below the horizontal  Introductory Physics Homework  2  
applied force on two blocks on a horizontal frictional plane  Introductory Physics Homework  6  
Horizontal Force on Pivot Point  Introductory Physics Homework  1 