1use energy conservation.
mg(x+L)=1/2 m v^2 + 1/2 k x^2 + 1/2 m(x+L)^2 w^2
from free body diagram
mg+m v^2/(x+L)= k x.
we have to solve them to get x
One end of a spring of unstretched length L and force constant k
is fixed to a pivot, and a body of mass m is attached to its other end
The spring is released from an unstretched, horizontal position, as
shown in figure
What is the length of the spring when it reaches a vertical position?
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8 Answers
1
1357yes i wud agree priyam..
basically power.. u are probably repeating energy terms !
62you need to define w clearly and v clearly..
what is the direction of v..
and about which point is this w.
1yeah you are right, if mass of spring were given then rotational KE would come or else itwouldn't.
so it should be mg(x+L)=1/2 m v^2 + 1/2 k x^2
62Good work guys.. i will try to post the final solution.. I should have last nite.. but din get time to..
use energy conservation.
mg(x+L)=1/2 mv2 + 1/2 k x2
from free body diagram
mv2/(x+L)= kx - mg
And now again i feel like .. eeks who will!!!